Section 13.1
13.1.1: Part (a):
f (1, −2) · 1 + f (2, −2) · 1 + f (1, −1) · 1 + f (2, −1) · 1 + f (1, 0) · 1 + f (2, 0) · 1 = 198.
Part (b):
f (2, −1) · 1 + f (3, −1) · 1 + f (2, 0) · 1 + f (3, 0) · 1 + f (2, 1) · 1 + f (3, 1) · 1 = 480.
The average of the two answers is 335, fairly close to the exact value 312 of the integral.
13.1.2: Part (a):
f (1, −1) · 1 + f (2, −1) · 1 + f (1, 0) · 1 + f (2, 0) · 1 + f (1, 1) · 1 + f (2, 1) · 1 = 144.
Part (b):
f (2, −2) · 1 + f (3, −2) · 1 + f (2, −1) · 1 + f (3, −1) · 1 + f (2, 0) · 1 + f (3, 0) · 1 = 570.
The average of the two answers is 357, fairly close to the exactly value 312 of the integral. The computations
shown here can be automated in computer algebra systems. For example, in Mathematica, after defining
f (x, y) = 4x3 + 6xy 2 , you could proceed as follows.
x[i ] := i + 1; y[j ] := j - 2; deltax = x[1] - x[0]; deltay = y[1] - y[0];
(∗ Part (a):
∗) xstar[i ] := x[i-1]; ystar[j ] := y[j]
Sum[ Sum[ f[xstar[i],ystar[j]]∗deltax∗deltay, {j, 1, 3}, {i, 1, 2} ]
144
(∗ Part (b):
∗) xstar[i ] := x[i]; ystar[j ] := y[j-1]
Sum[ Sum[ f[xstar[i],ystar[j]]∗deltax∗deltay, {j, 1, 3}, {i, 1, 2} ]
570
The idea is that to work another such problem, all you need to do is redefine f , xstar and ystar, and the
limits on i and j.
13.1.3: We omit ∆x and ∆y from the computation because each is equal to 1.
f
1
1
2, 2
+f
3
1
2, 2
+f
1
3
2, 2
1654
+f
3
3
2, 2
= 8.
This is also the exact value of the iterated integral.
13.1.4: We omit ∆x and ∆y from the computation because each is equal to 1.
f
1
1
2, 2
+f
3
1
2, 2
+f
1
3
2, 2
+f
3
3
2, 2
= 4.
This is also the exact value of the iterated integral. In a Mathematica solution similar to the one in Problem
2, we would use
xstar[i ] := (x[i] + x[i-1])/2; ystar[j ] := (y[j] + y[j-1])/2
13.1.5: The Riemann sum is
f (2, −1) · 2 + f (4, −1) · 2 + f (2, 0) · 2 + f (4, 0) · 2 = 88.
The true value of the integral is
416
≈ 138.666666666667.
3
13.1.6: We omit ∆x = 1 and ∆y = 1 from the computation.
f (1, 1) + f (2, 1) + f (1, 2) + f (2, 2) + f (1, 3) + f (2, 3) = 43.
The true value of the integral is 26. The midpoint approximation gives the fairly close Riemann sum 25.
13.1.7: We factor out of each term in the sum the product ∆x · ∆y = 14 π 2 . The Riemann sum then takes
the form
1 2 1
π · f 4 π,
4
1
4π
+f
3
1
4 π, 4 π
+f
1
3
4 π, 4 π
+f
3
3
4 π, 4 π
= 12 π 2 ≈ 4.935.
The true value of the integral is 4.
13.1.8: We factor out of each term in the sum the product ∆x · ∆y =
1
6 π.
The Riemann sum then takes
the form
1
6π
· f 14 ,
1
6π
+f
3
1
4, 6 π
+f
1
1
4, 2 π
+f
3
1
4, 2 π
+f
1
5
4, 6 π
+f
3
5
4, 6 π
= 12 π ≈ 1.571.
Mathematica reports that the true value of the integral is
1 CosIntegral[4π] + 1 (EulerGamma + Log[4π])
-4
4
and when we asked for a numerical value with the command N[%], it returned the approximation
0.77858913775068568
1655
13.1.9: Because f (x, y) = x2 y 2 is increasing in both the positive x-direction and the positive y-direction
on [1, 3] × [2, 5], L M U .
13.1.10: Because f (x, y) =
100 − x2 − y 2 is decreasing in both the positive x-direction and the positive
y-direction on [1, 4] × [2, 5], U M L.
13.1.11: We integrate first with respect to x, then with respect to y:
4
2
2 4
2
2
3 2
x + 4xy dy =
(3x + 4y) dx dy =
(24 + 16y) dy = 24y + 8y 2 = 80.
2
0
0
0
0
0
0
13.1.12: We integrate first with respect to x, then with respect to y:
2
3
3 2
3
3
4 2
1 3
8
2
x y dy =
y dy =
y
x y dx dy =
= 12.
3
3
0
0
0
0 3
0
0
13.1.13: We integrate first with respect to y, then with respect to x:
2
3
2 3
2
2
7
(2x − 7y) dy dx =
(4x − 28) dx = 2x2 − 28x
= −48 − 30 = −78.
2xy − y 2 dx =
2
−1 1
−1
−1
1
−1
13.1.14: We integrate first with respect to y, then with respect to x:
4
1
1 4
1
1
1 2 4
2 3
2
3
x y
x y dy dx =
dx =
60x dx = 20x
= 180.
−2 2
−2 4
−2
2
−2
13.1.15: We integrate first with respect to x, then with respect to y:
3
3 3
3
1
7
(xy + 7x + y) dx dy =
xy + x2 + x2 y dy
2
2
0
0
0
0
3
3
1
3
513
(5y + 21) dy =
(15y 2 + 126y) =
= 128.25.
=
2
4
4
0
0
13.1.16: We integrate first with respect to x, then with respect to y:
4
2 4
2
1 3 2
2 2
x y − 17x dy
(x y − 17) dx dy =
3
0
2
0
2
2
2
2
2
164
(28y 2 − 51) dy =
(28y 3 − 153y) = −
≈ −18.222222222222.
=
3
9
9
0
0
13.1.17: We integrate first with respect to y, then with respect to x:
2
2 2
2
2 3 3 2 2
2
2
xy − x y
(2xy − 3x y) dy dx =
dx
2
−1 −1
−1 3
−1
2
=
−1
2
9
3
3 3
9
2
2
(4x − 3x ) dx = 3x − x
= 0 − = − = −4.5.
2
2
2
2
−1
1656
13.1.18: We integrate first with respect to y, then with respect to x:
3
1
−1
−3
3
3
(x y − xy ) dy dx =
3
1 3 2 1 4
x y − xy
2
4
1
3
=
1
−1
dx
−3
3
(20x − 4x3 ) dx = 10x2 − x4 = 9 − 9 = 0.
1
13.1.19: We integrate first with respect to x, then with respect to y:
π/2
π/2
π/2
π/2
− cos x cos y
sin x cos y dx dy =
0
0
0
π/2
=
dy
0
cos y dy =
π/2
sin y
0
0
= 1 − 0 = 1.
13.1.20: This is merely Problem 19 with x and y interchanged, so the answer should be the same.
π/2
π/2
π/2
0
0
0
π/2
− cos x cos y
cos x sin y dy dx =
π/2
=
dx
0
cos x dx =
π/2
sin x
0
0
= 1 − 0 = 1.
13.1.21: We integrate first with respect to y, then with respect to x:
1
1
1
y
xe dy dx =
0
0
1
xe
y
dx
0
0
1
=
0
1
(e − 1)x2
2
(ex − x) dx =
1
=
0
1
(e − 1) ≈ 0.8591409142295226.
2
13.1.22: We integrate first with respect to x, then with respect to y:
1
2
2 y
1
x e dx dy =
0
−2
0
=
0
1
1 3 y
x e
3
2
dy
−2
16 y
e dy =
3
1
16 y
e
3
=
0
16
(e − 1) ≈ 9.1641697517815746.
3
13.1.23: We integrate first with respect to y, then with respect to x:
0
1
π
ex sin y dy dx =
0
1
0
=
1
π
− ex cos y
dx
0
2ex dx =
1
2ex
0
0
1657
= 2e − 2 ≈ 3.436563656918.
13.1.24: We integrate first with respect to x, then with respect to y:
1
1 1
1
1
y+1
x+y
x+y
e
e
dx dy =
dy =
− ey dy
e
0
0
0
0
0
1
= ey+1 − ey
= (e2 − e) − (e − 1) = (e − 1)2 ≈ 2.9524924420125598.
0
13.1.25: We integrate first with respect to x, then with respect to y:
π
π π
π
π
1 2
1 2
x y − cos x
(xy + sin x) dx dy =
dy =
2 + π y dy
2
2
0
0
0
0
0
π
1
1 4
= 2y + π 2 y 2
=
π + 8π ≈ 30.635458065680.
4
4
0
13.1.26: We integrate first with respect to x, then with respect to y:
π/2
π/2 π/2
π/2 (y − 1) cos x dx dy =
dy =
(y − 1) sin x
0
0
0
=
0
0
1 2
y −y
2
π/2
=
0
π/2
(y − 1) dy
1 2
(π − 4π) ≈ −0.3370957766587268.
8
13.1.27: We integrate first with respect to x, then with respect to y:
e
π/2 e
π/2 sin y
dx dy =
(ln x) sin y dy
x
0
1
0
1
π/2
=
π/2
− cos y
sin y dy =
0
0
= 0 − (−1) = 1.
13.1.28: We integrate first with respect to y, then with respect to x:
e
e
e
e e
e
ln y
1
1
dy dx =
dx = ln x = 1 − 0 = 1.
dx =
x 1
1
1 xy
1
1 x
1
13.1.29: We integrate first with respect to x, then with respect to y:
1
1 1
1
1
1
x
1
1
+
+ ln(x + 1) dy =
+ ln 2 dy
dx dy =
x+1 y+1
y+1
y+1
0
0
0
0
0
1
= ln(y + 1) + y ln 2 = 2 ln 2 − 0 = 2 ln 2 ≈ 1.3862943611198906.
0
13.1.30: We integrate first with respect to y, then with respect to x:
3
2 2
2
2 3
x y
y
4
+
dy dx =
+ x ln y dx =
+ x ln 3 dx
y
x
2x
x
1
1
1
1
1
2
1 2
3
=
x ln 3 + 4 ln x = 4 ln 2 + ln 3 ≈ 4.4205071552419458.
2
2
1
1658
13.1.31: The first evaluation yields
2 1
(2xy − 3y 2 ) dx dy =
−2
−1
2
−2
2
=
1
x2 y − 3xy 2
(−6y 2 ) dy =
dy
−1
− 2y 3
−2
The second yields
1
−1
2
−2
2
(2xy − 3y ) dy dx =
1
xy − y
−1
2
3
dx
(−16) dx =
−π/2
0
π/2
1
− 16x
= −16 − 16 = −32.
dy
0
π/2
=
−1
π
− cos x cos y
−π/2
= −16 − 16 = −32.
−2
−1
13.1.32: The first evaluation yields
π/2 π
sin x cos y dx dy =
−2
2
1
=
2
π/2
2 cos y dy = 2 sin y
−π/2
−π/2
= 2 − (−2) = 4.
The second yields
π
π/2
π
sin x cos y dy dx =
0
−π/2
π/2
dx
sin x sin y
0
−π/2
π
=
π
− 2 cos x
2 sin x dx =
0
0
= 2 − (−2) = 4.
13.1.33: The first evaluation yields
1
2 1
2
2
2
2
2 3/2
1/2
3/2
3/2
(x + y)
(y + 1) − y
(x + y)
dx dy =
dy =
dy
3
3
3
1
0
1
1
0
2
√
4 5/2
4 √
4
5/2
(y + 1) −
y
9 3 − 8 2 + 1 ≈ 1.406599671769.
=
=
15
15
15
1
The second yields
1 2
1/2
(x + y)
dy dx =
0
2
2
3/2
3/2
(x + 2) − (x + 1)
dx =
dx
3
3
0
0
1
1
√
4
4
4 √
5/2
5/2
(x + 2) −
(x + 1)
9 3 −8 2 +1 .
=
=
15
15
15
0
1
1
ln 3
2
(x + y)3/2
3
2
1
13.1.34: The first evaluation yields
0
ln 3
0
ln 2
ex+y dx dy =
0
ln 2
ex+y
dy =
0
0
1659
ln 3
(2ey − ey ) dy =
ln 3
ey
0
= 3 − 1 = 2.
The second yields
ln 2
ln 3
e
0
x+y
ln 2
dy dx =
ln 3
e
0
x+y
0
ln 2
dx =
0
0
(3e − e ) dx =
x
x
ln 2
2e
x
0
= 4 − 2 = 2.
13.1.35: We may assume that n 1 and, if you wish, even that n is a positive integer. Then
1
1
1
n n
xn+1 y n
n+1
x y dx dy =
0
0
0
Therefore
1
lim
n→∞
0
1
1
1
dy =
0
0
yn
dy =
n+1
xn y n dx dy = lim
n→∞
0
y n+1
(n + 1)2
1
=
0
1
.
(n + 1)2
1
= 0.
(n + 1)2
13.1.36: Note that whatever the choice of (xi , yi ), f (xi , yi ) = k, and hence f (xi , yi ) ∆Ai is equal to
the product of k and the area a(Ri ) of Ri for each i. Hence
n
n
n
f (xi , yi ) ∆Ai =
k · a(Ri ) = k
a(Ri ) = k · a(R) = k(b − a)(d − c).
i=1
i=1
i=1
13.1.37: Let a(R) denote the area of R. If 0 x π and 0 y π, then 0 f (x, y) sin 12 π = 1.
Hence every Riemann sum lies between 0 · a(R) and 1 · a(R). Therefore
0
π
0
π
0
√
sin xy dx dy a(R) = π 2 ≈ 9.869604401.
The exact value of the integral is
0
π
π
√
sin xy dx dy = 4
0
0
π
sin t
dt ≈ 7.4077482079298646814442134806319654533832.
t
13.1.38: The corresponding relation between Riemann sums is
n
n
cf (xi , yi ) · ∆Ai = c
i=1
f (xi , yi ) · ∆Ai .
i=1
13.1.39: The corresponding relation among Riemann sums is
n
[f (xi , yi ) + g(xi , yi )] · ∆Ai =
i=1
n
n
f (xi , yi ) · ∆Ai +
g(xi , yi ) · ∆Ai .
i=1
i=1
13.1.40: The corresponding relation between Riemann sums is this: If f (x, y) g(x, y) at each point of
R, then
n
i=1
f (xi ,
yi )
· ∆Ai n
i=1
1660
g(xi , yi ) · ∆Ai .
Section 13.2
13.2.1:
1
0
1
x
0
2x
2
0
1
0
1
1
0
y
1
=
0
13.2.4:
2
1
2
0
y/2
2
0
13.2.5:
1
x2
1
xy dy dx =
0
13.2.6:
0
0
1
√
1 2
xy
2
y
1
(x + y) dx dy =
0
0
y
13.2.7:
√
x
(2x − y) dy dx =
0
1
0
13.2.8:
2
√
√
− 2y
0
2y
2
(3x + 2y) dx dy =
0
x4
0
x
(y − x) dy dx =
1
0
1
0
13.2.10:
2
y+2
2
−y
√x
dx
x
2
√2y
√
− 2y
dy
2
√ 8√
64
3/2
5/2
4 2 y
.
dy =
2 y
=
5
5
0
x
dx
x4
1
1
1
1 9
1
1
1
x
=− .
− x2 + x5 − x8 dx = − x3 + x6 −
2
2
6
6
18
18
0
2
(x + 2y ) dx dy =
−1
dy
y
3 2
x + 2xy
2
1 2
y − xy
2
=
√y
1
3 2
1
1
1 2 4 5/2 1 3
3/2
.
=
− x + 2x − x dx = − x + x − x
2
2
4
5
2
20
0
0
1
1
1 6
1 5
1
x dx =
x
.
=
2
12
12
0
1
1
1
1 2 2 5/2 1 3
3
3
y + y 3/2 − y 2 dy =
y + y
.
− y
=
2
2
4
5
2
20
0
=
dy
y/2
0
13.2.9:
2
2
28
.
(2x + 2x2 ) dx = x2 + x3 =
3
3
0
1
1
2xy − y 2
2
=
0
5
.
6
dy
0
0
x
=
y
dx =
1
1
1
x 2
0
1
1 2 1 3
x + x
2
3
2
1
1
5
1
5 3
4
+ y − y 2 dy =
y + y2 −
y
= .
2
8
2
2
24
3
0
=
2
0
0
1 2
x + xy
2
1
1
1
1
3
1
+ y − y 2 dy =
(y + y 2 − y 3 ) = .
2
2
2
2
0
=
(x + x2 ) dx =
dx =
1 2
x + xy
2
(x + y) dx dy =
0
2x
1 2
x + xy
2
(x + y) dx dy =
0
1 2
y
2
y+
1
0
y=0
(1 + y) dy dx =
0
dx =
y + xy
0
2
13.2.3:
x
(1 + x) dy dx =
13.2.2:
−1
1 2
x + 2xy 2
2
1661
y+2
dy
−y
2
4
=
(2 + 2y + 4y + 4y ) dy = 2y + y + y 3 + y 4
3
−1
13.2.11:
1
x3
e
0
y/x
1
dy dx =
2
x 3
xe
0
0
1
exp(x2 ) − x2
=
2
13.2.12:
π
sin x
π
y dy dx =
0
13.2.13:
0
3
0
0
y
1 2
y
2
(y 2 + 16)1/2 dx dy =
sin x
0
1
dx =
0
0
y/x
=
0
3
0
x(y 2 + 16)1/2
y
dy
0
3
y(y 2 + 16)1/2 dy =
0
13.2.14:
e2 1/y
e
xy
dx dy =
0
1
e2
1
= 36.
−1
e−2
≈ 0.3591409142295226.
2
0
=
2
π
1
1
π
sin2 x dx =
(2x − sin 2x) = .
2
8
4
0
π
dx =
0
2
x exp(x2 ) − x dx
1
3
1 xy
e
y
1/y
e2
dy =
1
0
1 2
(y + 16)3/2
3
3
=
0
61
125 64
−
=
.
3
3
3
e2
e−1
dy = (e − 1) ln y
= 2(e − 1).
y
1
13.2.15: The following sketch of the graphs of y = x2 and y ≡ 4 is extremely helpful in finding the limits
of integration.
y
4
2
-2
-1
1
2
x
Answer:
2
4
2
xy dy dx =
−2
x2
13.2.16:
−2
√
6
√
− 6
2−x2
2
1 2
xy
2
x dy dx =
−4
4
x2
2
1 6
32 32
1 5
2
x
−
= 0.
dx =
=
8x − x
dx = 4x −
2
12
3
3
−2
−2
√
6
√
− 6
2
2
2−x2
x y
dx =
−4
1
= 2x − x5
5
3
√ 6
√
=
− 6
1662
√
6
√
− 6
(6x2 − x4 ) dx
48 √
6 ≈ 23.5151015307185097.
5
13.1.17: The following diagram, showing the graphs of y = x2 and y = 8 − x2 , is useful in finding the
limits of integration. (Solve y = x2 and y = 8 − x2 simultaneously to find where the two curves cross.)
y
8
6
4
2
-2
-1
1
2
x
Answer:
2
8−x2
2
x dy dx =
13.2.18:
1
−2
x2
1−y 2
1
y dx dy =
y 2 −1
−1
1−y2
xy
−1
2
1
(8x − 2x3 ) dx = 4x2 − x4
= 8 − 8 = 0.
2
−2
−2
dx =
xy
x2
−2
8−x2
y 2 −1
2
1
1
dy =
(2y − 2y ) dy = y − y 4
2
−1
3
2
1
=
−1
1 1
− = 0.
2 2
13.1.19: The following graph of y = sin x is helpful in determining the limits of integration.
y
1
0.5
1
Answer:
π
sin x
π
x dy dx =
0
0
2
sin x
0
π
dx =
xy
x sin x dx =
0
0
x
3
π
sin x − x cos x
= π.
0
13.2.20: To determine the order of integration and to determine the limits of integration, it is helpful to
draw the domain of the double integral. The value of the integral is
π/2
cos x
π/2
sin x dy dx =
−π/2
0
cos x
y sin x
−π/2
0
π/2
1
sin2 x
dx =
sin x cos x dx =
2
−π/2
1663
π/2
=
−π/2
1 1
− = 0.
2 2
13.2.21: A Mathematica command to draw the domain of the double integral is
ParametricPlot[ {{t,1}, {E,t}, {t,t}}, {t,1,E}, AxesOrigin → {0,0},
PlotRange → {{-0.3,3.3}, {-0.3,3.3}} ];
The graph produced by this command is shown next.
y
3
2
1
1
Answer:
e
1
x
1
1
dy dx =
y
e
x
1
1
3
x
e
dx =
ln y
1
2
e
ln x dx = −x + x ln x
1
= 0 − (−1) = 1.
13.2.22: A Mathematica command to draw the domain of the double integral is
ParametricPlot[ { Cos[t], Sin[t] }, { t, 0, Pi/2 }, AspectRatio → Automatic ];
The value of the double integral is
1 √1−x2
xy dy dx =
0
0
0
1
1 2
xy
2
√1−x2
dx =
0
0
1
1
(x − x3 ) dx =
2
1 2 1 4
x − x
4
8
13.2.23: A Mathematica command to draw the domain of the double integral is
Plot[ {x, -x/2, 1}, {x, -2, 1}, AspectRatio → Automatic,
PlotRange → {{-2,1}, {0,1}} ];
the resulting figure is next.
y
1
-2
-1
1
1664
x
1
=
0
1
.
8
The value of the double integral is
1
0
y
−2y
(1 − x) dx dy =
1
0
1
x − x2
2
y
1
dy =
0
−2y
1
1
3 2
2
3
(3y + y ) = 2 − 0 = 2.
3y + y
dy =
2
2
0
13.2.24: The value of the double integral is
3
0
9−x
2x
(9 − y) dy dx =
3
0
1
9y − y 2
2
9−x
3
dx =
0
2x
3
1 3
3 2
2
(x − 12x + 27) dx =
(x − 18x + 81x) = 54.
2
2
0
13.2.25: The domain of the double integral can be plotted by executing the following Mathematica command:
Plot[ {x∧2, 4}, {x, -2, 2} ];
and the resulting figure is next.
y
4
3
2
1
-2
-1
1
2
x
When the order of integration is reversed, we obtain
4
√
y
√
− y
0
2
4
x y dx dy =
0
1 3
x y
3
√y
√
− y
dy =
0
4
4
4 7/2
2 5/2
512
y
y
≈ 24.380952380952.
dy =
=
3
21
21
0
13.2.26: To draw the domain of the double integral, execute the Mathematica command
Plot[ {x, x∧4}, {x, 0, 1}, AspectRatio → Automatic ];
Reversal of the order of integration yields
0
1
y 1/4
(x − 1) dx dy =
y
1
0
=
0
1
1 2
x −x
2
y1/4
dy
y
1
1 2 1 3 1 3/2 4 5/4
1
2
(2y − y 2 + y 1/2 − 2y 1/4 ) dy =
y − y + y
− y
=− .
2
2
6
3
5
15
0
1665
13.2.27: The Mathematica command
Plot[ { x∧2, 2∗x + 3 }, { x, -1, 3 } ];
produces a figure showing the domain of the double integral; it appears next.
y
8
6
4
2
-1
1
2
3
x
When the order of integration is reversed, two integrals are required—one for the part of the region for which
0 y 1, the other for the part when 1 y 9. The reason is that the lower limit of integration for x
changes where y = 1. The first integral is
I1 =
1
√
√
− y
0
y
1
x dx dy =
0
1 2
x
2
√y
√
− y
1
dy =
0 dy = 0.
0
The second integral is
I2 =
9
√
y
9
x dx dy =
1
(y−3)/2
1
=
1
9
1 2
x
2
√y
dy
(y−3)/2
9
1
1
32
2
2
3
(10y − 9 − y ) dy =
(15y − 27y − y ) =
.
8
24
3
1
13.2.28: When the order of integration is reversed, two integrals are required. The one on the left half of
the domain of the double integral is
I1 =
0
√
√
− 4+x
−4
4+x
0
y dy dx =
−4
1 2
y
2
√4+x
√
− 4+x
0
=
0 dx = 0
−4
and the one of the right half of the domain is
I2 =
0
4
√
4−x
√
− 4−x
y dy dx =
0
4
1 2
y
2
√4−x
√
− 4−x
dx =
4
0 dx = 0.
0
13.2.29: To see the domain of the double integral, execute the Mathematica command
1666
Plot[ { 2∗x, 4∗x - x∧2 }, { x, 0, 2 } ];
the result is shown next.
y
4
2
1
2
x
When the order of integration is reversed, the given integral becomes
4 y/2
4 y/2
1 dx dy =
dy
x
√
√
0
2− 4−y
0
=
0
2− 4−y
4
1
y + (4 − y)1/2 − 2
2
dy =
1 2
2
y − 2y − (4 − y)3/2
4
3
4
=
0
4
.
3
13.2.30: The domain of the given integral is bounded above by the line y = x, below by the x-axis, and on
the right by the line x = 1. When the order of integration is reversed, we obtain
x
1 x
1
1
exp(−x2 ) dy dx =
x exp(−x2 ) dx
y exp(−x2 ) dx =
0
0
0
=
−
0
0
1
exp(−x2 )
2
Because the antiderivative
t
F (t) =
1
=
0
e−1
≈ 0.3160602794142788.
2e
exp(−x2 ) dx
0
is known to be nonelementary, the original integral cannot be evaluated by hand using the fundamental
theorem of calculus.
13.2.31: The domain of the given integral is bounded above by the line y = π, on the left by the y-axis,
and on the right by the line y = x. When the order of integration is reversed, we obtain
π
y
π y
π
π
x sin y
sin y
dx dy =
dy =
sin y dy = − cos y
= 2.
y
y
0
0
0
0
0
0
If the improper integral is disturbing, merely define the integrand to have the value 1 at y = 0. Then it will
be continuous and the integral will no longer be improper. Because the antiderivative
t
sin y
dy
F (t) =
y
0
1667
is known to be nonelementary, the only way to evaluate the given integral by hand is first to reverse the
order of integration.
13.2.32: The domain of the given integral is bounded above by the line y = x, on the right by the line
√
x = π , and below by the x-axis. When the order of integration is reversed, we obtain
√
π
x
2
√
π
sin x dy dx =
0
0
2
x
√
π
2
dx =
y sin x
0
x sin x dx =
0
0
1
− cos x2
2
√ π
=
0
1 1
+ = 1.
2 2
13.2.33: To generate a figure showing the domain of the given integral, use the Mathematica command
ParametricPlot[ {{t,t}, {1,t}, {t,0}}, {t,0,1}, AspectRatio → Automatic,
PlotRange → {{-0.1, 1.1}, {-0.1, 1.1}} ];
The result is shown next.
y
1
0.5
0.5
1
x
When the order of integration is reversed, we obtain
1
0
x
0
1
dy dx =
1 + x4
1
0
y
1 + x4
x
1
dx =
0
0
x
dx =
1 + x4
1
arctan x2
2
1
=
0
π
.
8
The integration can be carried out in the order given in the textbook, but finding the partial fraction
decomposition of the integrand is long and complex if machine aid is not available.
13.2.34: The domain of the given integral is the plane region bounded above by the graph of y = tan x,
below by the x-axis, and on the right by the line x = π/4. When the order of integration is reversed, the
result is
π/4 tan x
0
0
π/4
tan x
y sec x
sec x dy dx =
0
π/4
dx =
sec x tan x dx =
0
0
13.2.35: We used Mathematica in this problem. First we entered
1668
π/4
sec x
0
= −1 +
√
2.
Solve[ x∧3 + 1 == 3∗x∧2, x ]
and the computer returned the exact answers. Then we asked for the numerical values to 40 places, and we
found that the curves intersect at the three points with approximate coordinates
(a, 3a2 ) ≈ (−0.5320888862379561, 0.8493557485738457),
(b, 3b2 ) ≈ (0.6527036446661393, 1.270661432813855),
and
(c, 3c2 ) ≈ (2.8793852415718168, 24.8725781081447688).
We then entered the command
Plot[ {x∧3 + 1, 3∗x∧2}, {x, a, b}, PlotRange → {-1, 25} ];
and thereby discovered that the cubic graph is above the quadratic on (a, b) but below it on (b, c). Thus
we needed to compute two integrals:
i1 = Integrate[ Integrate[ x, {y, 3∗x∧2, x∧3 + 1} ], {x, a, b} ]
and
i2 = Integrate[ Integrate[ x, {y, x∧3 + 1, 3∗x∧2} ], {x, b, c} ]
Results:
I1 ≈ 0.0276702414879754
and
I2 ≈ 7.9240769481663325.
All the digits are correct or correctly rounded because we used at least 32 decimal digits in every computation.
The answer, therefore, is
I1 + I2 ≈ 7.9517471896543079.
13.2.36: The Mathematica command
Solve[ x∧4 == x + 4, x ]
yielded the exact solution—two real, two complex non-real. We asked for the real solutions to 40 places and
found that the two curves intersect in the two points
(a, a4 ) ≈ (−1.2837816658635382, 2.7162183341364618)
and
(b, b4 ) ≈ (1.5337511687552043, 5.5337511687552043).
Clearly the quartic lies under the linear graph on (a, b). Hence the only integral we need compute is
1669
Integrate[ Integrate[ x, {y, x∧4, x + 4} ], {x, a, b} ]
and the computer reported that its value is approximately 1.8930263071804474. All digits are correct or
correctly rounded because we carried at least 32 decimal digits in every computation.
13.2.37: We began with the Mathematica command
Solve[ x∧2 - 1 == 1/(1 + x∧2), x ]
and were rewarded with the exact solutions—two real, two complex non-real. The two real solutions are
a = −21/4 and b = 21/4 , so we used these exact values in the following computations. The double integral
has the value
Integrate[ Integrate[ x, {y, x∧2 - 1, 1/(x∧2 + 1)}, {x, a, b} ]
√ 1 √ √
√
1
+
= 0.
1 − 2 − ln 1 + 2
−1 + 2 + ln 1 + 2
2
2
13.2.38: We used Mathematica, beginning with the command
Solve[ x∧4 - 16 == 2∗x - x∧2, x ]
and the computer returned the exact answer—two real solutions, two complex non-real solutions. We
approximated the real solutions to 40 decimal digits; thus we found that the two curves cross at
(a, 2a − a2 ) ≈ (−1.7521717788841865, −6.5744495004865478)
and
(b, 2b − b2 ) = (2, 0).
We carried at least 32 decimal digits in our computations, so the value of the integral given here is correct
or correctly rounded:
Integrate[ Integrate[ x, {y, x∧4 - 16, 2∗x - x∧2}, {x, a, b} ] ]
8.871348510800994831161862.
13.2.39: We used Mathematica to automate Newton’s method for solving the equation x2 = cos x:
f[x ] := x∧2 - Cos[x]
g[x ] := N[x - f[x]/f[x], 60]
The function g carries out the iteration of Newton’s method, carrying 60 digits in its computations. A graph
indicated that the positive solution of f (x) = 0 is close to 4/5, hence we entered the successive commands
1670
g[4/5]
g[%]
(Recall that % refers to the “last output.”)
g[%]
After six iterations the results agreed to over 50 decimal degits, and thus we find that the graphs cross at
the two points
(b, b2 ) ≈ (0.8241323123025224, 0.6791940681811024)
and (a, a2 )
where a = −b. Hence the approximate value of the double integral is
Integrate[ Integrate[ x, {y, x∧2, Cos[x]}], {x, a, b} ]
0.0 × 10−58
A moment’s thought about Riemann sums, symmetry, and the sign of the integrand reveals that the exact
value of the integral is zero.
13.2.40: We let f (x) = x2 − 2x − sin x. Clearly f (0) = 0. We used the function g of the solution of
Problem 39 to implement Newton’s method for finding the positive solution. Beginning with the initial
approximation x0 = 2.2, six iterations yielded 50-place accuracy; the graphs cross at (0, 0) and at
(b, b2 − 2b) ≈ (2.3169342886237398, 0.7343159205529156).
The Mathematica command
Integrate[ Integrate[ x, {y, x∧2 - 2∗x, Sin[x]} ], {x, a, b} ]
3.3945384440042540571563864737
yielded a good approximation to the value of the double integral. All digits shown are correct or correctly
rounded because we carried at least 32 decimal digits in each computation.
13.2.41: The integral is zero. Each term f (xi , yi ) ∆Ai = xi ∆Ai in every Riemann sum is cancelled by
a similar term in which xi has the opposite sign.
13.2.42: By symmetry around both coordinate axes, the value of the integral is
1
4
0
0
1−x
x2 dy dx = 4
0
1
x2 y
1−x
dx = 4
0
0
1671
1
(x2 − x3 ) dx = 4
1 3 1 4
x − x
3
4
1
=
0
1
.
3
13.2.43: Every term of the form f (xi , yi ) ∆Ai in every Riemann sum is cancelled by a similar term in
which xi has the opposite sign (but yi is the same). Therefore the value of the integral is zero.
13.2.44: Clearly the double integral of y 2 over the square is the same as the double integral of x2 , so the
answer is double the answer to Problem 42.
13.2.45:
Suppose that the rectangle R consists of those points (x, y) for which both a x b and
c y d. Suppose that k is a positive constant and that f is a function continuous on R. Then
I=
f (x, y) dA
R
exists. Suppose that > 0 is given. Then there exists a number δ1 > 0 such that, for every partition
P = {R1 , R2 , . . . Rn } of R such that |P | < δ1 and for every selection (xi , yi ) in Ri (i = 1, 2, . . . , n),
n
f (xi , yi ) ∆Ai − I <
2k
i=1
(where ∆Ai is the area a(Ri ) of Ri ). Consequently,
n
f (xi , yi ) ∆Ai − kI < .
k
2
i=1
(1)
Moreover, kf is continuous on R, and hence
J=
kf (x, y) dA
R
exists. So there exists a number δ2 > 0 such that, for every partition P = {R1 , R2 , . . . Rn } of R such that
|P | < δ2 and every selection (xi , yi ) in Ri ( i = 1, 2, . . . n),
n
kf (xi , yi ) ∆Ai − J < ;
2
i=1
that is,
n
f (xi , yi ) ∆Ai − J < .
k
2
i=1
Let δ be the minimum of δ1 and δ2 . Then, for every partition P of R with |P | < δ and every selection
(xi , yi ) in Ri (i = 1, 2, . . . n), we have both
n
f (xi , yi ) ∆Ai <
kI − k
2
i=1
n
f (xi , yi ) ∆Ai − J < .
k
2
i=1
1672
(by (1)) and
Add the last two inequalities. Then, by the triangle inequality (Theorem 1 of Appendix A, page A-2),
|kI − J | <
+ = .
2 2
Because is an arbitrary positive number, this proves that J = kI; that is, we have shown that
kf (x, y) dA = k
R
f (x, y) dA.
R
The proof is similar in the case k < 0, and if k = 0 there is nothing to prove.
For a shorter proof, one that exploits both the continuity of f and the fact that R is a rectangle with
sides parallel to the coordinate axes, choose a continuous function F such that Fx = f . Then choose a
continuous function P such that Py = F . Then
d
b
kf (x, y) dA =
d
kf (x, y) dx dy =
c
R
a
b
dy
kF (x, y)
c
a
d
d
[kF (b, y) − kF (a, y)] dy = kP (b, y) − kP (a, y)
=
c
c
= kP (b, d) − kP (a, d) − kP (b, c) + kP (a, c) = k [P (b, d) − P (a, d) − P (b, c) + P (a, c)]
d
= k P (b, y) − P (a, y) = k
d
b
f (x, y) dx dy = k
c
a
d b
dy = k
F (x, y)
c
[F (b, y) − F (a, y)] dy
c
c
=k
d
a
f (x, y) dA.
R
13.2.46: Suppose that R is a plane rectangle with sides parallel to the coordinate axes, so that R consists
of those points (x, y) for which both a x b and c y d for some numbers a, b, c, and d. Suppose
that f and g are functions both of which are continuous on R. Then there exist continuous functions F
and G such that Fx = f and Gx = g on R. Moreover, there exist continuous functions P and Q such that
Py = F and Qy = G on R. Then
f (x, y) dx dy =
c
R
d b
f (x, y) dA =
=
a
d
d
b
dy
F (x, y)
c
a
d
[F (b, y) − F (a, y)] dy = P (b, y) − P (a, y)
c
c
= P (b, d) − P (a, d) − P (b, c) + P (a, c).
Similarly,
g(x, y) dA = Q(b, d) − Q(a, d) − Q(b, c) + Q(a, c).
R
1673
Finally,
[f (x, y) + g(x, y)] dA =
d
[f (x, y) + g(x, y)] dx dy =
c
R
d b
a
b
dy
F (x, y) + G(x, y)
c
a
d
[F (b, y) + G(b, y) − F (a, y) − G(a, y)] dy
=
c
d
= P (b, y) + Q(b, y) − P (a, y) − Q(a, y)
c
= P (b, d) + Q(b, d) − P (a, d) − Q(a, d) − P (b, c) − Q(b, c) + P (a, c) + Q(a, c).
We compare these three results; it follows immediately that
[f (x, y) + g(x, y)] dA =
f (x, y) dA +
g(x, y) dA.
R
R
R
13.2.47: Suppose that R is a plane rectangle with sides parallel to the coordinate axes, so that R consists
of those points (x, y) for which both a x b and c y d for some numbers a, b, c, and d. Suppose
that f is continuous on R and that m f (x, y) M for all (x, y) in R.
Let g(x, y) ≡ m and h(x, y) ≡ M for (x, y) in R. Then g(x, y) f (x, y) h(x, y) for all points
(x, y) in R. Let P = {R1 , R2 , . . . , Rn } be a partition of R and let (xi , yi ) be a selected point in Ri for
1 i n. As usual, let ∆Ai = a(Ri ) for 1 i n. Then for each integer i, 1 i n, we have
g(xi , yi ) f (xi , yi ) h(xi , yi );
thus
g(xi , yi ) ∆Ai f (xi , yi ) ∆Ai h(xi , yi ) ∆Ai
for 1 i n. Add these inequalities to find that
n
m · ∆Ai n
i=1
Therefore
m·
f (xi , yi ) ∆Ai i=1
n
a(Ri ) n
i=1
that is,
m · a(R) M · ∆Ai .
i=1
f (xi , yi ) ∆Ai M ·
i=1
n
n
n
a(Ri );
i=1
f (xi , yi ) ∆Ai M · a(R)
(1)
i=1
for every partition P of R and every selection (xi , yi ) for P. That is, the inequalities in (1) hold for every
Riemann sum for f on R. Because the double integral of f on R is the limit of such sums, we may conclude
that
m · a(R) f (x, y) dA M · a(R).
R
1674
13.2.48: Suppose that R1 and R2 are rectangles with sides parallel to the coordinate axes and that the
right-hand edge of R1 coincides with the left-hand edge of R2 . Suppose that R1 consists of those points
(x, y) in the plane for which a x b and r y s and that R2 consists of those points (x, y) for which
b x c and r y s. Let R be the union of R1 and R2 , so that R consists of those points (x, y) for
which a x c and r y s.
Suppose that f is continuous on R, and thus on R1 and R2 . Choose a continuous function F such
that Fx = f on R and a function P such that Py = F on R. Note that these equations hold on R1 and
R2 as well. Then
s
b
f (x, y) dA =
s
f (x, y) dx dy =
r
R1
a
b
r
s
[F (b, y) − F (a, y)] dy
dy =
F (x, y)
r
a
s
= P (b, y) − P (a, y)
= P (b, s) − P (a, s) − P (b, r) + P (a, r).
r
Similarly,
f (x, y) dA = P (c, s) − P (b, s) − P (c, r) + P (b, r)
R2
and
f (x, y) dA = P (c, s) − P (a, s) − P (c, r) + P (a, r).
R
Therefore
f (x, y) dA +
R1
f (x, y) dA
R2
= P (b, s) − P (a, s) − P (b, r) + P (a, r) + P (c, s) − P (b, s) − P (c, r) + P (b, r)
= P (c, s) − P (a, s) − P (c, r) + P (a, r) =
f (x, y) dA.
R
13.2.49: Suppose that R is a rectangle with sides parallel to the coordinate axes, that f (x, y) g(x, y)
for all (x, y) in R, and that both
I=
f (x, y) dA
and
R
J=
g(x, y) dA
R
exist. Suppose by way of contradiction that J < I. Let = I − J. Note that /3 > 0. Choose δ > 0 so
small that if P = {R1 , R2 , . . . , Rn } is a partition of R with |P | < δ and (xi , yi ) is a selection for P with
(xi , yi ) in Ri for 1 i n, then
n
f (xi , yi ) ∆Ai − I <
3
i=1
and
1675
n
g(xi , yi ) ∆Ai − J < .
3
i=1
With such a partition and such a selection, note that
n
g(xi , yi ) ∆Ai < J +
i=1
and thus
n
n
< I−
<
f (xi , yi ) ∆Ai ,
3
3
i=1
[f (xi , yi ) − g(xi , yi )] ∆Ai > 0.
i=1
Because ∆Ai > 0 for 1 i n, it follows that
f (xj , yj ) > g(xj , yj )
for some j, 1 j n, contrary to hypothesis. Therefore I J.
13.2.50: Let m = f (x0 , y0 ) and M = f (x1 , y1 ) where (x0 , y0 ) and (x1 , y1 ) are points of R. Let r(t) be
a continuous parametric curve lying entirely in R such that r(0) = (x0 , y0 ) and r(1) = (x1 , y1 ). By Eq. (8)
of Section 13.2,
f (x, y) dA M · a(R).
m · a(R) R
Note that g(t) = f (r(t)) · a(R) is continuous on [0, 1] and that g(0) = m · a(R) and g(1) = M · a(R).
Therefore
g t =
f (x, y) dA
(1)
R
for some t in [0, 1]. Let (x, y ) = g t . Then (x, y ) is a point of R, and by Eq. (1),
f (x, y ) · a(R) =
f (x, y) dA.
R
13.2.51:
Recall that R is the region in the first quadrant bounded by the circle x2 + y 2 = 1 and the
coordinate axes. Hence
(x + y) dA =
1
1
=
0
=
1−x2
(x + y) dy dx
x=0
R
√
y=0
1
xy + y 2
2
√1−x2
1
dx =
0
0
1
1
1
− (1 − x2 )3/2 + x − x3
3
2
6
1
x 1 − x2 + (1 − x2 ) dx
2
1
=
0
2
1 1 1
− + = .
2 6 3
3
13.2.52: Here is one way to use Mathematica to solve this problem. First define f (x, y) = xy and set
n = 5. Then the Riemann sum for the induced partition using the midpoint of each small rectangle (actually,
a square) is
1676
( 1/n∧2)∗Sum[ f[ (i − 1/2)/n, (y − 1/2)/n ], { j, 1, n − 1 },
{ i, 1, IntegerPart[ Sqrt[ n∧2 − j∧2 ] ] } ]
207
2500
(You may need to substitute Floor for IntegerPart in the Mathematica command shown here.) So the
midpoint approximation to the integral is 0.0828. Its actual value is
Integrate[ f[ x, y ], { y, 0, 1 }, { x, 0, Sqrt[ 1 − y∧2 ] } ]
1
8
Therefore the exact value of the integral is 0.125. You can use ideas illustrated by the limits of summation
in the first command to verify the entries in the second column of the table in Fig. 13.2.3.
13.2.53: The domain of the integral and the partition using n = 5 subintervals of equal length in each
direction is shown next.
y
1
0.8
0.6
0.4
0.2
0.2 0.4 0.6 0.8
1
x
The midpoints of the subrectangles of the inner partition are indicated with “bullets” in the figure. Let
f (x, y) = xy exp y 2 . Then the corresponding midpoint sum for the given integral is
1 S = 2 f (0.1, 0.1) + f (0.3, 0.1) + f (0.5, 0.1) + f (0.7, 0.1) + f (0.3, 0.1) + f (0.3, 0.3)
n
+ f (0.3, 0.5) + f (0.3, 0.7) + f (0.5, 0.1) + f (0.5, 0.3) + f (0.5, 0.5) + f (0.5, 0.7)
+ f (0.7, 0.1) + f (0.7, 0.3) + f (0.7, 0.5)
1
=
25
4 0.01 12 0.09 4 0.25
63 0.49
e
e
+ e
+ e
+
25
25
5
100
≈ 0.109696.
The exact value of the integral is
1 √1−y2
f (x, y) dA =
xy exp y 2 dx dy
R
y=0
x=0
1677
1
=
0
=
1 2
x y exp y 2
2
2 − y2
exp y 2
4
√1−y2
1
1
(y − y 3 ) exp y 2 dy
2
dy =
0
0
1
=
0
e−2
≈ 0.1795704571147613088400718678.
4
If you prefer the other order of integration—which avoids the integration by parts—it is
1 √1−x2
f (x, y) dA =
xy exp y 2 dy dx
x=0
R
1
y=0
=
0
=
1
x exp y 2
2
√1−x2
1
dx =
0
0
1
1
− x2 − exp 1 − x2
4
4
1
1
x exp 1 − x2 − x dx
2
e−2
.
4
=
0
Section 13.3
13.3.1: The area is
1
y
A=
1
1 dx dy =
x=y 2
y=0
1
=
y
x
y=0
(y − y 2 ) dy =
y=0
1 2 1 3
y − y
2
3
dy
x=y 2
1
=
y=0
1
1 1
− = .
2 3
6
To find the limits of integration, it is very helpful to sketch the domain of the double integral. The figure is
next; it was produced by Mathematica via the command Plot[ {x, Sqrt[x]}, {x, 0, 1} ].
y
1
0.5
0.5
13.3.2: The area is
1
y=x
A=
1 dy dx =
0
y=x4
0
1
(x − x4 ) dx =
1678
1
x
1 2 1 5
x − x
2
5
1
=
0
3
.
10
13.3.3: The graphs cross where x2 = 2x + 3; that is, where x = −1 and where x = 3. A sketch of the
domain of the integral is next; it was produced by Mathematica via the command Plot[ {x∧2, 2∗x + 3},
{x, -1, 3} ].
y
9
6
3
-1
1
2
3 x
The area of the region is
3
3
1
32
.
(3 + 2x − x2 ) dx = 3x + x2 − x3
=
3
3
−1
−1
2x+3
1 dy dx =
A=
−1
y=x2
3
13.3.4: The graphs cross where 2x + 3 = 6x − x2 ; that is, where x = 1 and where x = 3. The area they
enclose is
3
6x−x2
A=
3
1 dy dx =
1
1
y=2x+3
1
(4x − 3 − x ) dx = 2x − 3x − x3
3
2
2
3
=
1
4
.
3
13.3.5: The graphs y = x2 and y = 2 − x cross where x2 = 2 − x; that is, where x = −2 and where x = 1.
And the x-axis is also part of the boundary of the region, so the figure below indicates the specified region
whose area is sought. It was produced by the Mathematica command
Plot[ {x∧2, 2 - x, 0}, {x, -0.4, 2.2} ]
The area of the roughly triangular-shaped region bounded by all three graphs is
1
A=
y=0
2−y
√
x= y
1 dx dy =
0
1
1
2 3/2 1 2
5
√
(2 − y − y) dy = 2y − y
− y
= .
3
2
6
0
1679
y
1
1
x
2
13.3.6: The region is bounded on the northwest by the graph of y = (x + 1)2 , on the northeast by the
graph of y = (x − 1)2 , and below by the x-axis. To avoid radicals we will integrate first with respect to y,
then with respect to x, even though this entails computing two integrals. The area of the part of the region
to the right of the y-axis is
1
(x−1)2
A1 =
1
1 dy dx =
x=0
0
y=0
1
1
1
(x − 1)2 dx = x − x2 + x3 = .
3
3
0
The area of the part of the region to the left of the y-axis is
0
(x+1)2
A2 =
x=−1
y=0
0
1
1 dy dx =
(x + 1) dx = x + x + x3
3
−1
2
2
Therefore the total area bounded by all three of the given curves is A1 + A2 =
0
=
−1
1
.
3
2
.
3
13.3.7: The graphs cross where x2 + 1 = 2x2 − 3; that is, where x = −2 and where x = 2. The area
between them is
2
1 dy dx =
y=2x2 −3
x=−2
2
1
32
.
(4 − x2 ) dx = 4x − x3
=
3
3
−2
−2
x2 +1
A=
2
13.3.8: The graphs cross where x2 + 1 = 9 − x2 ; that is, where x = −2 and where x = 2. The area between
them is
2
A=
x=−2
9−x2
2
2
1 dy dx =
(8 − 2x ) dx = 8x − x3
3
2
−2
y=x +1
2
2
=
−2
64
.
3
13.3.9: The first-quadrant region bounded by these three curves is shown next; the figure was generated
using the Mathematica command Plot[ {x, 2x, 2/x}, {x, -0.2, 2.2} ].
1680
y
2
1
1
x
2
√ √ 2 , 2 . The area they bound (in
The bounding curves cross at the origin and at the points (1, 2), and
the first quadrant) is
1
A=
√
2x
2
1 dy dx +
x=0
1 2
x
=
2
y=x
1
2/x
1 dy dx =
x=1
1
+ 2 ln x − x2
2
0
x dx +
0
y=x
√2
=
1
√
1
1
2
2
−x
x
dx
1
1
+ ln 2 − = ln 2 ≈ 0.693147180559945309417232.
2
2
13.3.10: The curves cross where
x2 =
2
;
1 + x2
x4 + x2 − 2 = 0;
(x2 + 2)(x2 − 1) = 0;
thus where x = −1 and where x = 1. The area of the region they bound is
1
2/(1+x2 )
1
1 dy dx =
x=−1
y=x2
−1
2
− x2
1 + x2
13.3.11: The volume is
1 1
V =
(1 + x + y) dy dx =
x=0
y=0
x=0
13.3.12: The volume is
2 3
V =
(2x + 3y) dx dy =
y=0
x=0
x=0
1
2
1
= π − ≈ 2.474925986923.
dx = 2 arctan x − x3
3
3
−1
1
y + xy + y 2
2
2
2
y=0
1
dx =
0
y=0
x + 3xy
3
2
y=0
13.3.13: The volume is
2 1
V =
(y + ex ) dx dy =
y=0
1
dy =
0
x=0
1
xy + ex
2
dy =
x=0
1681
0
2
1
1
3
1 2
3
x+ x
= 2.
x+
dx =
2
2
2
0
9
(9 + 9y) dy = 9y + y 2
2
2
= 36.
0
2
1
(e + y − 1) dy = ey + y 2 − y = 2e.
2
0
13.3.14: The volume is
π
π
V =
π
(3 + cos x + cos y) dy dx =
x=0
y=0
x=0
π
=
π
π
(3π + π cos x) dx = 3πx + π sin x
0
dx
3y + y cos x + sin y
0
y=0
= 3π 2 ≈ 29.608813203268075856503473.
13.3.15: The domain of the integral can be drawn by using the Mathematica command Plot[ 1 - x, {x,
0, 1} ] and the result is shown next.
y
1
0.5
0.5
1
x
The volume is
1
1−x
V =
1
(x + y) dy dx =
x=0
y=0
x=0
1
xy + y 2
2
1−x
1
dx =
0
y=0
1
(1 − x2 ) dx =
2
1
1
x − x3
2
6
13.3.16: The volume is
4
(4−x)/2
V =
4
(3x + 2y) dy dx =
x=0
4
=
0
y=0
x=0
3xy + y 2
(4−x)/2
dx
y=0
4
5 3
64
5
x
.
=
4 + 4x − x2 dx = 4x + 2x2 −
4
12
3
0
13.3.17: The domain of the integral can be drawn using the Mathematica command
ParametricPlot[ {{1,t}, {t,0}, {t,t∧2}}, {t,0,1}, AspectRatio → 1 ];
and the result is shown next.
1682
1
=
0
1
.
3
y
1
0.5
0.5
x
1
The volume is
1
x2
V =
1
(1 + x + y) dy dx =
x=0
x=0
y=0
=
1
y + xy + y 2
2
x 2
1 3 1 4
1 5
x + x +
x
3
4
10
1
dx =
0
y=0
1
=
0
1 4
2
3
x +x + x
dx
2
41
≈ 0.683333333333.
60
13.3.18: The domain of the integral can be drawn using the Mathematica command
ParametricPlot[ {{0,t}, {t,1}, {Sqrt[t],t}}, {t,0,1}, AspectRatio → 1 ];
and the volume of the solid is
1 1
V =
(2x + y) dy dx =
x=0
y=x2
1
x=0
1
2xy + y 2
2
1
1
dx =
0
y=x2
1
1 5
1
x + x2 − x4 −
x
=
2
2
10
1
=
0
1
1
+ 2x − 2x3 − x4
2
2
dx
9
.
10
13.3.19: The domain of the integral is shown next.
y
1
-1
The volume of the solid is
1 1
2
x dy dx =
V =
x=−1
y=x2
1
x=−1
1
2
1
x y
y=x2
1
x
1 3 1 5
x − x
dx =
(x − x ) dx =
3
5
−1
1683
2
4
1
=
−1
4
.
15
13.3.20: The domain of the integral can be drawn using the Mathematica command
ParametricPlot[ {{t∗t,t}, {4,t}}, {t, -2, 2} ];
and the volume of the solid is
2 V =
y=−2
4
y 2 dx dy =
x=y 2
13.3.21: The volume of the solid is
2 1
2
2
V =
(x + y ) dx dy =
x=0
2
y=0
1 3
x + xy 2
3
13.3.22: The volume is
1 2−x2
V =
(1 + x2 + y 2 ) dy dx =
1
=
−2
=
4
2
128
≈ 8.533333333333.
15
2
dy =
0
x=0
(4y 2 − y 4 ) dy
=
−2
1
x=−2
y=x
2
−2
x=y 2
1
dy =
4 3 1 5
y − y
3
5
=
x=−2
xy 2
y=−2
y=0
2
1
y + x y + y3
3
2
1
+ y2
3
1
(y + y 3 )
dy =
3
2
=
0
2−x2
dx
y=x
1
14
1
1
1
1
1 7
(14 − 3x − 9x2 − 4x3 + 3x4 − x6 ) dx =
x − x2 − x3 − x4 + x5 −
x
3
3
2
3
5
21
−2
837
≈ 11.9571428571428571.
70
13.3.23: The domain of the integral can be drawn by executing the Mathematica command
ParametricPlot[ {{3, 2∗t/3}, {t, 0}, {t, 2∗t/3}}, {t, 0, 3} ];
and the result is shown next.
y
2
1
1
The volume of the solid is
3 2x/3
V =
(9 − x − y) dy dx =
x=0
y=0
x=0
3
0
3
9y − xy −
1 2
y
2
3
=
2
x
2x/3
dx
y=0
3
8 3
8
x
= 19.
6x − x2 dx = 3x2 −
9
27
0
1684
10
.
3
13.3.24: The volume of the solid is
1
√
x
V =
x=0
(10 + y − x ) dy dx =
1
x=0
1
10y − x y + y 2
2
2
√x
dx
y=x2
1
20 3/2 1 2 10 3 2 7/2
1
1 4
1 5
1/2
2
5/2
x + x −
x − x +
x
10x + x − 10x − x + x
dx =
2
2
3
4
3
7
10
0
1
=
0
=
y=x2
2
1427
≈ 3.3976190476190476.
420
13.3.25: The volume is
1
2−2x
2
2
1
(4x + y ) dy dx =
V =
x=0
1
=
0
y=0
x=0
8
(1 − 3x + 6x2 − 4x3 ) dx =
3
1
4x y + y 3
3
2
2−2x
dx
y=0
16 3 8 4
8
x − 4x2 +
x − x
3
3
3
1
=
0
4
.
3
13.3.26: The volume is
1
x2
1
(2x + 3y) dy dx =
V =
y=x3
x=0
x=0
=
3
2xy + y 2
2
x 2
1
dx =
0
y=x3
1 4
1 5
3 7
x −
x −
x
2
10
14
1
=
0
1 4 3 6
3
2x − x − x
dx
2
2
13
≈ 0.1857142857142857.
70
13.3.27: The volume is
2
(6−3x)/2
V =
x=0
(6 − 3x − 2y) dy dx =
y=0
2
6y − 3xy − y
x=0
2
=
0
2
(6−3x)/2
dx
y=0
2
3
9
9
9 − 9x + x2 dx = 9x − x2 + x3 = 6.
4
2
4
0
13.3.28: The volume is
2
(4−y)/2
V =
y=0
x=y/2
2
(8 − 4x − 2y) dx dy =
8x − 2x2 − 2xy
y=0
2
=
0
(4−y)/2
dy
x=y/2
2 3
y − 4y 2 + 8y
(2y − 8y + 8) dy =
3
2
2
=
0
16
.
3
13.3.29: The triangular domain of the integral can be drawn by executing the Mathematica command
ParametricPlot[ {{1, 2 + 2∗t}, {1 + 4∗t, 2}, {1 + 4∗t, 4 - 2∗t}},
{t, 0, 1}, PlotRange → {{-0.5, 5.5}, {-0.5, 4.5}},
1685
AspectRatio → Automatic, AxesOrigin → {0, 0} ];
and the result is shown next.
y
4
3
2
1
1
The volume of the solid is
V =
5
(9−x)/2
4
3
5
xy dy dx =
x=1
2
5
=
1
y=2
x=1
65
9
1
x − x2 + x3
8
4
8
x
5
1 2
xy
2
(9−x)/2
dx
y=2
65 2 3 3
1 4
x − x +
x
dx =
16
4
32
5
= 24.
1
13.3.30: To generate the triangular base of the solid, execute the Mathematica command
ParametricPlot[ {{-3, -4 + 8∗t}, {-3 + 8∗t, 4 - 4∗t}, {-3 + 8∗t, -4 + 4∗t}},
{t, 0, 1} ];
and the result is shown next.
y
4
2
-2
x
4
2
-2
-4
The top of the triangle has equation y =
1
2 (5
− x) and the bottom has equation y = 12 (x − 5). Hence the
volume of the solid is
5 (5−x)/2
2
2
(25 − x − y ) dy dx =
V =
x=−3
5
=
−3
y=(x−5)/2
1375 75
25 2 13 3
−
x−
x +
x
12
4
4
12
5
x=−3
1
25y − x y − y 3
3
2
(5−x)/2
dx
y=(x−5)/2
1375
75 2 25 3 13 4
x−
x −
x +
x
dx =
12
8
12
48
1686
5
=
−3
1792
.
3
13.3.31: The volume is
√1−y2
1
V =
y=−1
x=−
√
(x + 1) dx dy.
1−y 2
This integral can be evaluated exactly with a single command in Mathematica, but we will evaluate it one
step at a time. As usual, the Mathematica output is rewritten slightly for more clarity.
Integrate[ x + 1, x ]
x+
1 2
x
2
x → Sqrt[1 - y∗y]) - (% /.
(% /.
2
x → -Sqrt[1 - y∗y])
1
1
1 − y 2 + (1 − y 2 ) + (y 2 − 1)
2
2
Simplify[ % ]
2
1 − y2
Integrate[ %, y ]
y
1 − y 2 + arcsin y
y → 1) - (% /.
(% /.
y → -1)
π
N[ %, 60 ]
3.14159265358979323846264338327950288419716939937510582097494
13.3.32: The volume is
3
V =
√
9−x2
√
y=− 9−x2
x=−3
(9 − x2 − y 2 ) dy dx.
We used Mathematica much as in the solution of Problem 31 to obtain the numerical value of V :
3
V =
x=−3
=
1
9y − x y − y 3
3
4
(9 − x2 )1/2
3
2
√9−x2
√
y=− 9−x2
45
1
x − x3
8
4
13.3.33: The volume is
1
V =
x=−1
3
dx =
−3
4
(9 − x2 )3/2 dx
9
x 3
81
81
arcsin
π ≈ 127.2345024703866262.
=
+
2
3
2
−3
√
1−x2
√
y=− 1−x2
2 4 − x2 − y 2 dy dx.
1687
We used Derive to evaluate this integral in a step-by-step fashion much as in the solution of Problem 31.
The results:
1
√1−x2
+ y 4 − x2 − y 2
√
y
dx
(4 − x2 ) arctan 4 − x2 − y 2
y=− 1−x2
√ √
1
√ 3 1 − x2
2
2
=
+ 2 3 1 − x dx
2(4 − x ) arctan
3
−1
√
√ √
2
16
3 1 − x2
3 (2x + 1)
2
x(12 − x ) arctan
+
arctan √
=
3
3
3
3 1 − x2
√
√
1
√
3 (2x − 1)
16
2 3 2
arctan √
x 1−x
+
− 4 3 arcsin x +
3
3
3 1 − x2
−1
√
π
32 − 12 3 ≈ 11.7447292674805137.
=
3
V =
x=−1
In Section 13.4 we will find that this integral is quite easy to evaluate if we first convert to polar coordinates.
If so, the integral takes the form
2π
θ=0
1
2
2r 4 − r dr dθ =
r=0
2π
2
− (4 − r2 )3/2
3
θ=0
1
2π
dθ =
r=0
0
√
16
−2 3
3
dθ = 2π
√
16
−2 3
3
.
13.3.34: The volume is
1
V =
x=−1
√
1−x2
√
y=− 1−x2
2 − x2 − y 2 − (x2 + y 2 ) dy dx
We used Derive to evaluate V , one step at a time much as in the solution of Problem 31. The results:
√1−x2
1 2 − x2
y
1 3
y 2 − x2 − y 2
2
arctan −x y− y
+
V =
√
2
2
3
2 − x2 − y 2
x=−1
y=− 1−x2
1
=
−1
=
1 − 4x2 2
2
1 − x dx
(2 − x ) arctan 1 − x +
3
2
√
√
x(6 − x2 )
2 2
x 2 +1
7
2
arctan 1 − x + arcsin x +
arctan √
3
6
3
1 − x2
√
√
1
2 2
x 2 −1
x
x
2 3/2
2
+
arctan √
1−x
+ (1 − x ) −
3
3
6
1 − x2
−1
=
π √
8 2 − 7 ≈ 2.2586524883563962.
6
The volume would be much easier to evaluate using polar coordinates (as in Section 14.4, coming up next).
We would thereby obtain
V = 2π
1
(r
0
1
1
1
π √
8 2 −7 .
2 − r2 − r3 ) dr = 2π − (2 − r2 )3/2 − r4 =
3
4
6
0
1688
13.3.35: Given: the plane with Cartesian equation
x y z
+ +
= 1
a
b
c
cutting off a tetrahedron in the first octant. We set z = 0 and solve for
y=
b
(a − x);
a
the triangular region in the first quadrant bounded by this line and the coordinate axes is the domain of the
volume integral. Hence the volume of the tetrahedron is
b(a−x)/a
a 2abcy − 2bcxy − acy 2
x y
dy dx =
c 1− −
dx
a
b
2ab
x=0 y=0
x=0
y=0
2
a
a
3a bcx − 3abcx2 + bcx3
bc(a − x)2
abc
.
dx
=
=
=
2
2
2a
6a
6
0
0
a
b(a−x)/a
V =
13.3.36: The volume is
a
V =
a
2
= 2h
−a
√a2 −x2
a
(x + h) dy dx =
√
y=− a2 −x2
x=−a
√
a2 −x2
xy + hy
x=0
2 1/2
(a − x )
a
dx +
−a
2
2 1/2
2x(a − x )
√
y=− a2 −x2
a
dx =
−a
2(x + h)(a2 − x2 )1/2 dx
2
dx = h · πa + − (a2 − x2 )3/2
3
2
a
= πa2 h.
−a
We evaluated the first integral in the second line by observing that it is the area of a semicircle of radius a.
13.3.37: The volume is
1 √1−y2 1 − y 2 dx dy =
V =
y=0
√1−y2
1
1
1
2
dy =
(1 − y 2 ) dy = y − y 3 = .
x 1 − y2
3
3
y=0
0
0
x=0
x=0
1
As the text indicates, the other order of integration provides more difficulties. You obtain
√
1
1−x2
V =
x=0
1 − y 2 dy dx =
1
x=0
y=0
1 1
y 1 − y 2 + arcsin y
2
2
√1−x2
dx
y=0
1 1
x 1 − x2 + arcsin
1 − x2
dx
2
2
0
1
1 2 2 1
2
x −
1 − x2 + x arcsin
1 − x2
= .
=
6
3
2
3
0
1
=
13.3.38: The volume is
π
sin x
V =
π
2 sin x dy dx =
x=0
=
0
π
y=− sin x
sin x
dx
2y sin x
x=0
y=− sin x
π
4 sin2 x dx = 2x − sin 2x = 2π ≈ 6.2831853071795865.
0
1689
13.3.39: We integrate to find the volume of an eighth of the sphere, then multiply by 8. Thus the volume
of a sphere of radius a is
a
√
a2 −x2
V =8
x=0
a2 − x2 − y 2 dy dx.
y=0
Let y = (a2 − x2 )1/2 sin θ. Then dy = (a2 − x2 )1/2 cos θ dθ. This substitution yields
a
π/2
V =8
x=0
π/2
=8
x=0
=8
a/2
0
a
1/2
(a2 − x2 )1/2 cos θ dθ dx
(a2 − x2 ) cos2 θ dθ dx
θ=0
π/2
=8
x=0
(a2 − x2 ) − (a2 − x2 ) sin2 θ
θ=0
a
(a2 − x2 ) ·
θ=0
1 + cos 2θ
dθ dx = 8
2
π/2
1
1
(a2 − x2 ) θ + sin 2θ
dx
2
4
x=0
0
a
a
π 2
1
2
4
(a − x2 ) dx = 2π a2 x − x3 = 2π · a3 = πa3 .
4
3
3
3
0
13.3.40: Given: the ellipsoid with equation
x2
y2
z2
+ 2 + 2 = 1;
2
a
b
c
(1)
we assume that a, b, and c are all positive. Set z = 0 in Eq. (1) to find that the ellipsoid intersects the
xy-plane in the ellipse with equation
y2
x2
+ 2 = 1;
2
a
b
that is,
y=
b 2
(a − x2 )1/2
a
(we take the positive root because we plan to integrate over the quarter of the ellipse that lies in the first
quadrant). Finally, we solve Eq. (1) for
z=
c 2 2
(a b − b2 x2 − a2 y 2 )1/2 .
ab
We integrate to find the volume of the eighth of the ellipsoid that lies in the first octant, then multiply by
8. Hence the volume of the ellipsoid is
a
(b/a)(a2 −x2 )1/2
V =8
x=0
y=0
c 2 2
(a b − b2 x2 − a2 y 2 )1/2 dy dx.
ab
Let
y=
b 2
(a − x2 )1/2 sin θ;
a
then
1690
dy =
b 2
(a − x2 )1/2 cos θ dθ.
a
This substitution yields
a V =8
1/2 b 2
c 2 2
(a b − b2 x2 )(1 − sin2 θ)
· (a − x2 )1/2 cos θ dθ
a
x=0 θ=0 ab
a π/2
a π/2
8bc
8bc
1 + cos 2θ
dθ dx
= 2
(a2 − x2 ) cos2 θ dθ dx = 2
(a2 − x2 ) ·
a x=0 θ=0
a x=0 θ=0
2
π/2
θ sin 2θ
8bc a
8bc a π 2
= 2
+
(a − x2 ) dx
(a2 − x2 )
dx = 2
a x=0
2
4
a
4
0
θ=0
a
1
2πbc 2
4
2πbc
= 2 a2 x − x3 = 2 · a3 = πabc.
a
3
a
3
3
0
13.3.41:
π/2
We integrate over the quarter-circle of radius 5 and center (0, 0) in the first quadrant, then
multiply by 4. Hence the volume is
5 √25−x2
2
2
V =4
(25 − x − y ) dy dx = 4
x=0
5
=
0
=
5
x=0
y=0
1
25y − x2 y − y 3
3
√25−x2
dx
y=0
x 5
8
125
1 3
8
2 3/2
2
(25 − x )
x − x + 625 arcsin
dx =
25 − x
3
3
8
4
5
0
625
π ≈ 981.747704246810387019576057.
2
The techniques of Section 14.4 will transform this problem into one that is remarkably simple.
13.3.42:
When we solve the equations of the paraboloids simultaneously,
we find that
x2 + y 2 = 4. Thus the intersection of the paraboloids is a curve that lies in this cylinder. So an appropriate domain for a double integral will be the circular disk of radius 2 centered at the origin. Therefore
the volume of the intersection of the two paraboloids is
2 √4−x2
2
2
(12
−
3x
−
3y
)
dy
dx
=
V =
√
y=− 4−x2
x=−2
2
12y − 3x y − y
x=−2
3
√4−x2
√
y=− 4−x2
dx
x 2
4(4 − x2 )3/2 dx = (10x − x3 ) 4 − x2 + 24 arcsin
= 24π ≈ 75.398223686155.
2
−2
−2
=
2
2
13.3.43: Suppose that the cylinder is the one with equation x2 + z 2 = R2 and that the square hole is
centered on the z-axis and its sides are parallel to the coordinate planes. Thus the hole meets the xy-plane
in the square with vertices at ± 21 R, ± 21 R . We will integrate over the quarter of that square that lies in
the first quadrant, then multiply by 4. Hence the volume of material removed by the drill is
R/2
R/2 R/2 R/2 2
2
2
2
2 R − x dy dx = 4
dx
V =4
2y R − x
x=0
y=0
x=0
R/2
2
2 1/2
y=0
R/2
1
x
1 3
2
2 1/2
R(R − x )
dx = 4 Rx(R − x ) − R arctan
=4
2
2
(R2 − x2 )1/2
0
0
√
1√
1√
3 3 + 2π
· R3 ≈ (1.913222954981)R3.
=
3 + 2 arctan
3
· R3 =
2
3
6
1691
13.3.44: When the equations of the elliptical paraboloid and the parabolic cylinder are solved simultaneously, one consequence is that x2 + 4y 2 = 4. Hence this elliptical cylinder contains the curve of intersection
of the two surfaces, and the ellipse x2 + 4y 2 = 4 in the xy-plane is an appropriate domain for a double
integral. Hence the volume bounded by the two surfaces is
2
(1/2)(4−x2 )1/2
2
V =
x=−2
2
=
−2
y=−(1/2)(4−x2 )1/2
2
2
(4 − x − 4y ) dy dx =
x=−2
4
4y − x y − y 3
3
2
(1/2)(4−x2 )1/2
dx
y=−(1/2)(4−x2 )1/2
x 2
2
1
= 4π ≈ 12.566370614359.
(4 − x2 )3/2 dx =
(4 − x2 )1/2 (10x − x3 ) + 4 arcsin
3
6
2
−2
13.3.45: The region bounded by the parabolas y = x2 and y = 8 − x2 in the xy-plane is a suitable domain
for a double integral that gives the volume of the solid. Hence the volume of the solid is
2
8−x2
V =
y=x2
x=−2
8 3 2 5
x − x
3
5
=
2
2
(2x − x ) dy dx =
2
2
8−x2
x y
=
−2
2
−2
y=x2
x=−2
2
dx =
(8x2 − 2x4 ) dx
256
≈ 17.066666666667.
15
13.3.46: We used Mathematica in the usual way; the volume of the solid is
π/2
cos x
V =
x=−π/2
π/2
=
−π/2
=
y=− cos x
(4 − x2 − y 2 ) dy dx =
π/2
4y − x2 y −
x=−π/2
1 3
y
3
cos x
dx
y=− cos x
π/2
2
1
3
2
(8 − 2x ) cos x − cos x dx =
207 sin x − 72x cos x − 36x sin x − sin 3x
3
18
−π/2
2
208 − 9π 2
≈ 13.2415067100217525.
9
13.3.47: We used Mathematica in the usual way; the volume of the solid is
π/2
cos x
V =
π/2
cos y dy dx =
x=−π/2
y=− cos x
cos x
dx
sin y
x=−π/2
y=− cos x
π/2
=
−π/2
2 sin(cos x) dx ≈ 3.57297496390010467337.
Mathematica reports that the exact value of the integral is
3 3
1
,
4∗HypergeometricPFQ {1},
, − .
2 2
4
13.3.48: A Mathematica solution:
I1 = Integrate[ Sin[x]∗Cos[y], { y, 0, Cos[x] } ]
1692
(sin x) sin(cos x)
V = 4∗Integrate[ I1, { x, 0, Pi/2 } ]
4[1 − cos(1)]
N[V]
1.83879
13.3.49: A Mathematica solution:
eq1 = z == 2∗x + 3;
eq2 = z == x∧2 + y∧2;
Eliminate[ { eq1, eq2 }, z ]
y 2 = −x2 + 2x + 3
This is the circle (x − 1)2 + y 2 = 4 with center (1, 0) and radius 2. Therefore the volume of the solid is
Integrate[ 3 + 2∗x − x∧2 - y∧2, { x, −1, 3 },
{ y, −Sqrt[ 3 + 2∗x − x{∧2 ], Sqrt[ 3 + 2∗x − x∧2 ] } ]
8π
13.3.50: A Mathematica solution:
eq1 == z == 4∗x + 4∗y;
eq2 = z == x∧2 + y∧2 − 1;
Eliminate[ { eq1, eq2 }, z ]
−y 2 + 4y + 1 == x2 − 4x
This is the circle (x − 2)2 + (y − 2)2 = 9 with center (2, 2) and radius 3. Hence the volume of the solid
bounded by the two surfaces is
Integrate[ 1 + 4∗x + 4∗x - x∧2 - y∧2, { x, −1, 5 },
{ y, 2 − Sqrt[ 9 − (x − 2)∧2 ], 2 + Sqrt[ 9 − (x - 2)∧2 ] } ]
81π
2
The answer is correct in spite of the typographical error in the last Mathematica command, which should be
1693
Integrate[ 1 + 4∗x + 4∗y - x∧2 - y∧2, { x, −1, 5 },
{ y, 2 − Sqrt[ 9 − (x − 2)∧2 ], 2 + Sqrt[ 9 − (x - 2)∧2 ] } ]
How do you explain that?
13.3.51: A Mathematica solution:
eq1 = z == −16∗x − 18∗y;
eq2 = z == 11 − 4∗x∧2 − 9∗y∧2;
Eliminate[ { eq1, eq2 }, z ]
−9y 2 + 18y + 11 = 4x2 − 16x
This is the ellipse 4(x − 2)2 + 9(y − 1)2 = 36 with center (2, 1) and semiaxes a = 3 and b = 2. Hence the
volume of the solid is
Integrate[ 11 − 4∗x∧2 − 9∗y∧2 + 16∗x + 18∗y, { x, −1, 5 },
{ y, 1 − 1/3∗Sqrt[ 36 − 4∗(x − 2)∧2 ], 1 + 1/3∗Sqrt[ 36 − 4∗(x − 2)∧2 ] } ]
108π
13.3.52: A Mathematica solution:
I1 = Simplify[ 8∗Integrate[ Sqrt[ 4 − x∧2 − y∧2 ], y ] ]
4y 4 − x2 − y 2 − 4(x2 − 4) tan−1
y
4 − x2 − y 2
(I1 /. y → 1) − (I1 /. y → 0)
4
1
3 − x2 − 4(x2 − 4) tan−1 √
3 − x2
(Assuming that no one really wants to see the antiderivative of the preceding expression, let’s jump immediately to the double integral that gives the volume of the hole.)
V = 8∗Integrate[ Integrate[
Sqrt[ 4 − x∧2 - y∧2 ], { y, 0, 1 } ], { x, 0, 1 } ]
4 √
1
1
5
2 2 + 11 tan−1 √
+ 19 tan−1 √
− 8 tan−1 √
3
3
2
2
N[V]
1694
14.5755
100∗%/(4∗Pi∗8/3)∗percent
43.4954 percent
13.3.53: First let’s put the center of the sphere at the point (−2, 0, 0). A Mathematica solution:
V = FullSimplify[ 2∗Integrate[ Integrate[
Sqrt[ 16 − (x + 2)∧2 − y∧2 ], { y, −1, 1 } ], { x, −1, 1 } ] ]
√ √ √ √
2 √
6 6 − 2 14 + 29 cot−1 6 + 41 cot−1 14 − 47 csc−1 15
3
+ 47 sin−1
3/5 + 20 tan−1
3/2 − 108 tan−1 9 3/2
√ √ − 20 tan−1 11/ 14 + 108 tan−1 19/ 14
N[V]
26.7782
100∗%/(4∗Pi∗64/3)∗percent
9.98878 percent
Section 13.4
13.4.1: The circle with center (0, 0) and radius a > 0 has polar description r = a, 0 θ 2π. Therefore
its area is
2π
a
A=
2π
r dr dθ =
θ=0
r=0
θ=0
1 2
r
2
a
dθ =
r=0
0
2π
2π
1 2
1 2
a dθ =
a θ
= πa2 .
2
2
0
13.4.2: The circle with polar equation r = 3 sin θ has area
3 sin θ
π π
π 3 sin θ
1 2
9
r
sin2 θ dθ
r dr dθ =
dθ =
A=
2
θ=0 r=0
θ=0
0 2
r=0
π
π
9
9
9
(1 − cos 2θ) dθ =
=
2θ − sin 2θ = π.
4
8
4
0
0
13.4.3: The area bounded by the cardioid with polar description r = 1 + cos θ, 0 θ 2π, is
1+cos θ
2π 1+cos θ
2π 2π 1 2
1
1
A=
r
+ cos θ + cos2 θ dθ
r dr dθ =
dθ =
2
2
θ=0 r=0
θ=0 2
0
r=0
2π
1
3
=
= π.
6θ + 8 sin θ + sin 2θ
8
2
0
1695
13.4.4: The area bounded by one loop of the four-leaved rose with polar equation r = 2 cos 2θ is
2 cos 2θ
π/4 2 cos 2θ
π/4
π/4
1 2
r
A=
r dr dθ =
dθ =
2 cos2 2θ dθ
2
θ=−π/4 r=0
θ=−π/4
−π/4
r=0
π/4
1
1
= π ≈ 1.5707963267948966.
4θ + sin 4θ
4
2
−π/4
=
13.4.5: To see the two circles, execute the Mathematica command
ParametricPlot[ {{Cos[t], Sin[t]}, {2∗Sin[t]∗Cos[t], 2∗Sin[t]∗Sin[t]}},
{t, 0, 2∗Pi}, AspectRatio → Automatic ];
the result is shown next.
y
2
1
1 x
-1
-1
To find where the two circles meet, solve their equations simultaneously:
2 sin θ = 1;
sin θ =
1
;
2
θ=
1
5
π, θ = π.
6
6
To find the area between them, two integrals are required. Each is doubled because we are actually finding
the area of the right half of the intersection of the circles.
π/6
2 sin θ
A1 = 2
π/6
r dr dθ = 2
θ=0
π/2
0
r=0
1
A2 = 2
π/2
r dr dθ =
θ=π/6
r=0
√
π/6
2π − 3 3
;
2 sin2 θ dθ = 2θ − sin 2θ
=
6
0
1 dθ =
θ=π/6
π
.
3
Therefore the total area enclosed by both circles is
√
4π − 3 3
A = A1 + A2 =
≈ 1.2283696986087568.
6
1696
13.4.6: We are to find the area within the limaçon r = 2 + cos θ and outside the circle r = 2. Their graphs
are shown next.
y
2
1
-2
-1
1
2
3
x
-1
-2
To find where they intersect, solve their equations simultaneously for θ = ± 12 π. The area we seek can be
found with a single integral:
π/2
2+cos θ
π/2
r dr dθ =
A=
θ=−π/2
r=2
θ=−π/2
1 2
r
2
2+cos θ
π/2
dθ =
−π/2
r=2
1
(2 + cos θ)2 − 2
2
dθ
π/2
1
π + 16
≈ 4.7853981633974483.
=
=
2θ + 16 sin θ + sin 2θ
8
4
−π/2
13.4.7: To find where the limaçon r = 1 − 2 sin θ passes through the origin, solve 1 − 2 sin θ = 0 for θ = 16 π,
θ = 56 π. The small loop of the limaçon is generated by the values of θ between these two angles, and is thus
given by
5π/6
1−2 sin θ
A=
5π/6
r dr dθ =
θ=π/6
r=0
θ=π/6
=
1 2
r
2
1−2 sin θ
5π/6
dθ =
π/6
r=0
3
1
θ + 2 cos θ − sin 2θ
2
2
5π/6
=
π/6
1
(1 − 2 sin θ)2 dθ
2
√
2π − 3 3
≈ 0.5435164422364773.
2
13.4.8: In polar coordinates the integral becomes
3
2π 3
1
81
π ≈ 127.2345024703866262.
r3 dr dθ = 2π r4 =
I=
4
2
θ=0 r=0
0
Because the inner integral does not involve θ, it is constant with respect to θ. Therefore to integrate it with
respect to θ over the interval 0 θ 2π, simply multiply the inner integral by 2π.
13.4.9: In polar coordinates the integral takes the form
2
2π 2
1
16
π ≈ 16.7551608191455639.
I=
r2 dr dθ = 2π r3 =
3
3
θ=0 r=0
0
1697
Because the inner integral does not involve θ (either in the integrand or in the limits of integration), it is
constant with respect to θ. Therefore to integrate it with respect to θ over the interval 0 θ 2π, simply
multiply the inner integral by 2π. We will use this time-saving technique frequently and without further
comment.
13.4.10: Note that the domain of the integral is generated as θ varies from − 21 π to 12 π. In polar coordinates
the integral is
π/2
2 cos θ
I=
θ=−π/2
r3 dr dθ =
r=0
π/2
θ=−π/2
1 4
r
4
2 cos θ
4 cos4 θ dθ
−π/2
r=0
=
π/2
dθ =
1
12θ + 8 sin 2θ + sin 4θ
8
π/2
=
−π/2
3
π ≈ 4.71238898038468985769.
2
If you prefer, the last integral in the first line of the display may be evaluated using Formula (113) of the
endpapers:
π/2
4
4 cos θ dθ = 8
−π/2
0
π/2
cos4 θ dθ = 8 ·
1 3 π
3
· · = π.
2 4 2
2
13.4.11: Note that the domain of the integral is generated as θ varies from 0 to π. In polar coordinates
the integral takes the form
π
sin θ
I=
θ=0
π
(10 + 2r cos θ + 3r sin θ) · r dr dθ =
r=0
θ=0
5r2 +
1 3
r (2 cos θ + 3 sin θ)
3
sin θ
dθ
r=0
2
2
3
4
=
5 sin θ + sin θ cos θ + sin θ dθ
3
0
π
1
23
π ≈ 9.03207887907065556058.
=
276θ − 8 cos 2θ + 2 cos 4θ − 144 sin 2θ + 3 sin 4θ =
96
8
0
π
The reduction formulas in Problems 53 and 54 of Section 7.3, and the formulas of Problem 58 there and
Formula (113) of the endpapers, may be used if you prefer to avoid trigonometric identities.
13.4.12: The polar form of the integral is
a
2π a
1
πa4
2
2
2 2
4
I=
.
(a − r ) · r dr dθ = 2π (2a r − r ) =
4
2
θ=0 r=0
0
13.4.13: In polar form the given integral becomes
1
π/2 1
1
r
π
1
2
·
ln(1
+
r
dr
dθ
=
)
= π ln 2 ≈ 0.5443965225759005.
J=
2
1
+
r
2
2
4
θ=0
r=0
0
To evaluate the integral in Cartesian coordinates, you would first need to evaluate
√1−y2
1
1 − y2
1
dx = arctan .
1 + x2 + y 2
1 + y2
1 + y2
x=0
1698
(1)
Then you would need to antidifferentiate the last expression in (1). Neither Mathematica nor Derive could
do so.
13.4.14: Conversion of the integral to polar coordinates yields
1
π/2 1
√ r
π
π √
K=
dr dθ = · − 4 − r2
= · 2 − 3 ≈ 0.4208936072384665.
2
2
4 − r2
θ=0
r=0
0
It is possible to evaluate this integral in Cartesian coordinates. You should obtain
√ 2
1
1 √1−x2
1−x
1
y
dy dx =
dx
K=
arctan 4 − x2 − y 2
4 − x2 − y 2
0
0
0
0
√
1
1 − x2
√
=
dx
arctan
3
0
√
1
√
1 − 2x
1 + 2x
1 − x2
√
= x arctan
− 3 arcsin x − arctan √ √
+ arctan √ √
3
3 1 − x2
3 1 − x2
0
√
π
=
2− 3 .
2
We have concealed one complication: The evaluation in the next-to-last line requires l’Hôpital’s rule.
13.4.15: In polar coordinates the integral becomes
2
π/2 2
1 5
16
π
π 32
r
=
π ≈ 10.0530964914873384.
I=
r4 dr dθ = ·
= ·
2
5
2 5
5
θ=0
r=0
0
This integral can be evaluated in Cartesian coordinates. You should obtain
2
√
I=
0
4−x2
(x2 + y 2 )3/2 dy dx
0
2
=
0
√
4−x2
5 2
1 3 2
3
4
x y+ y
x + y 2 + x ln y + x2 + y 2
dx
8
4
8
0
5 2
1
3 4
3 4 2 3/2
2
2
x 4 − x + (4 − x ) − x ln x + x ln 2 + 4 − x
=
dx
4
2
4
8
0
2
x
2
3 3 3 5
32
3 5 16
2
2
x+
x
arcsin
−
x ln x +
x ln 2 + 4 − x
π.
4−x +
=
=
5
20
5
2
20
40
5
0
2
The final equality requires l’Hôpital’s rule.
13.4.16: The polar form of the integral is
π/2 csc θ
J=
r3 cos2 θ dr dθ =
θ=π/4
r=0
π/2
θ=π/4
=
−
1 4
r cos2 θ
4
1
cot3 θ
12
1699
csc θ
r=0
π/2
=
π/4
π/2
dθ =
π/4
1
cot2 θ csc2 θ dθ
4
1
≈ 0.0833333333333333.
12
This integral is easier to evaluate in Cartesian coordinates.
13.4.17: In polar coordinates the integral becomes
π/2
1
π
r sin r dr dθ = ·
2
r=0
K=
θ=0
2
1
− cos r2
2
1
dθ =
0
π
(1 − cos 1) ≈ 0.3610457246892050.
4
Exact evaluation of the given integral in Cartesian coordinates may be impossible. Mathematica reports
that
2
2
sin(x + y ) dx =
where
FresnelS(x) =
π
2
2
2
2
+ (sin y ) · FresnelC x
(cos y ) · FresnelS x
2
π
π
x
sin
0
πt2
2
dt
and
x
cos
FresnelC(x) =
0
πt2
2
dt.
13.4.18: The polar form of the given integral is
π/4
2 cos θ
π/4
1 dr dθ =
I=
θ=0
2 cos θ
r=sec θ
θ=0
π/4
(2 cos θ − sec θ) dθ
dθ =
r
0
r=sec θ
π/4
√
√ = 2 sin θ − ln(sec θ + tan θ)
= 2 − ln 1 + 2 ≈ 0.5328399753535520.
0
This integral can also be evaluated in Cartesian coordinates. You should obtain
2
I=
2
2
ln y + x + y
1
√
2x−x2
2
dx =
1
0
√
2
2x + 2x − x
− ln x + ln
dx
√ √
2 √
√
√ 2 2x − x2
√
− x ln x + x ln
2x + 2x − x2
= 2 − ln 1 + 2 .
= −
x
1
13.4.19: The volume is
2π
1
V =
θ=0
2π
(2 + r cos θ + r sin θ) · r dr dθ =
r=0
=
0
2π
r2 +
θ=0
1 3
r (cos θ + sin θ)
3
1
dθ
r=0
2π
1
1
1
1
= 2π.
1 + cos θ + sin θ dθ = θ − cos θ + sin θ
3
3
3
3
0
13.4.20: The volume is
2π
2
V =
θ=0
r=0
(2 + r cos θ) · r dr dθ =
2π
r2 +
θ=0
1 3
r cos θ
3
2
dθ =
r=0
2π
4+
0
8
cos θ dθ
3
2π
8
= 4θ + sin θ
= 8π ≈ 25.1327412287182349.
3
0
1700
13.4.21: The volume is
π 2 sin θ
V =
(3 + r cos θ + r sin θ) · r dr dθ =
θ=0
π
=
0
=
r=0
π
θ=0
3 2 1 3
r + r (cos θ + sin θ)
2
3
2 sin θ
dθ
r=0
π
8
8
1
6 sin2 θ + sin3 θ cos θ + sin4 θ dθ =
48θ − 4 cos 2θ + cos 4θ − 26 sin 2θ + sin 4θ
3
3
12
0
1 48π − 3
+
= 4π ≈ 12.56637061435917295385.
4
12
13.4.22: The volume is
2π 1+cos θ
(1 + r cos θ) · r dr dθ =
V =
θ=0
2π
=
0
r=0
2π
θ=0
1 2 1 3
r + r cos θ
2
3
1
1
2
3
(1 + cos θ) + (1 + cos θ) cos θ dθ
2
3
1+cos θ
dθ
r=0
2π
1
4 + 10 cos θ + 8 cos2 θ + 2 cos3 θ + cos 2θ + 2 cos θ cos 2θ + cos2 θ cos 2θ dθ
6
0
2π
1
11
=
π ≈ 8.6393797973719314.
=
132θ + 200 sin θ + 44 sin 2θ + 8 sin 3θ + sin 4θ
96
4
0
=
13.4.23: We will find the volume of the sphere of radius a centered at the origin:
a
2π a
2
2
4
V =
2r a2 − r2 dr dθ = 2π · − (a2 − r2 )3/2
= 2π · a3 = πa3 .
3
3
3
θ=0 r=0
r=0
13.4.24: When we solve the equations of the paraboloids simultaneously, we find that x2 + y 2 = 4. Hence
the curve in which the surfaces intersect lies on the cylinder x2 + y 2 = 4. Thus the disk in the xy-plane
bounded by the circle x2 + y 2 = 4 is appropriate for the domain of the volume integral. So the volume of
the solid bounded by the paraboloids is
2
2π 2
3
(12 − 3r2 ) · r dr dθ = 2π · 6r2 − r4
= 24π ≈ 75.3982236861550377.
V =
4
θ=0 r=0
r=0
13.4.25: The volume of the solid is
2π a
(h + r cos θ) · r dr dθ =
V =
θ=0
r=0
a
2π 1 2
1 2
1
1
r h + r2 cos θ
a h + a3 cos θ dθ
dθ =
3
2
3
θ=0 2
0
r=0
2π
1
=
= πa2 h.
3a2 hθ + 2a3 sin θ
6
0
2π
13.4.26: The wedge is bounded above by the plane z = x = r cos θ, below by the plane z = 0, and on the
side by the cylinder r = 2 for − 21 π θ 12 π. Hence the volume of the wedge is
2
π/2 2
π/2 π/2
1 3
8
2
r cos θ
cos θ dθ
V =
r cos θ dr dθ =
dθ =
3
θ=−π/2 r=0
θ=−π/2
−π/2 3
r=0
=
8
sin θ
3
1701
π/2
=
−π/2
16
≈ 5.3333333333333333.
3
13.4.27: When we solve the equations of the paraboloids simultaneously, we find that one consequence is
that x2 + y 2 = 1. Thus the curve in which the paraboloids meet lies on that cylinder, and hence the circular
disk x2 + y 2 1 in the xy-plane is a suitable domain for the double integral that yields the volume between
the paraboloids. That volume is therefore
2π
1
2
(4 − 4r ) · r dr dθ = 2π · 2r − r
V =
θ=0
2
4
1
r=0
dθ = 2π ≈ 6.2831853071795865.
r=0
13.4.28: When we solve the equations of the paraboloids simultaneously, we find that one consequence is
that x2 + y 2 = 1. Thus the curve in which the paraboloids meet lies on that cylinder, and hence the circular
disk x2 + y 2 1 in the xy-plane is a suitable domain for the double integral that yields the volume between
the paraboloids. That volume is therefore
2π
1
(1 − r2 ) · r dr dθ = 2π ·
V =
θ=0
13.4.29:
2
r=0
1 2 1 4
r − r
2
4
1
dθ =
r=0
1
π ≈ 1.5707963267948966.
2
When the equations of the sphere and the cone are solved simultaneously, one consequence is
that x + y 2 =
1 2
2a .
Therefore the circle in which the sphere and cone meet lies on the cylinder with that
equation. Thus a suitable domain for the double integral that gives the volume in question is the circle in
the xy-plane with equation x2 + y 2 = 12 a2 . Hence the volume of the “ice-cream cone” is
√
a/ 2
2π
V =
θ=0
= 2π ·
r=0
√
a/ 2
1 2
(a − r2 )3/2 + r3
a2 − r2 − r · r dr dθ = 2π · −
3
r=0
√ √ 1 1 4 − 2 2 a3 = π 2 − 2 a3 ≈ (0.6134341230070734)a3.
12
3
13.4.30: The volume is
π
2a sin θ
V =
θ=0
=
r=0
r3 dr dθ =
π
θ=0
1 4
a (12θ − 8 sin 2θ + sin 4θ)
8
1 4
r
4
π
=
0
2a sin θ
dθ =
r=0
π
4a4 sin4 θ dθ
0
3 4
πa ≈ (4.7123889803846899)a4.
2
13.4.31: The lemniscate defined by r2 = 2 sin 2θ is shown below, as generated by the Mathematica command
ParametricPlot[ {{(Sqrt[2∗Sin[2∗t]])∗Cos[t], (Sqrt[2∗Sin[2∗t]])∗Sin[t]},
{-(Sqrt[2∗Sin[2∗t]])∗Cos[t], -(Sqrt[2∗Sin[2∗t]])∗Sin[t]}},
{t, 0, Pi/2}, AspectRatio → Automatic ];
1702
y
1
0.5
-1
-0.5
0.5
x
1
-0.5
-1
We find the volume over the first-quadrant loop and under the paraboloid z = r2 to be
π/2
√
2 sin 2θ
3
π/2
r dr dθ =
V =
θ=0
r=0
θ=0
1 4
r
4
√2 sin 2θ
dθ =
r=0
π/2
sin2 2θ dθ
0
π/2
1
1
= π ≈ 0.7853981633974483.
=
4θ − sin 4θ
8
4
0
13.4.32: The volume is
2
2π 2
1
2 3/2
2
V =
2r 18 − 2r dr dθ = 2π · − (18 − 2r )
3
θ=0 r=0
0
√
√
4
= π −5 10 + 27 2 ≈ 93.71319733506050999635.
3
√
The volume of a cylinder of the same radius as the given cylinder but with height 6 2 (the major axis
√
√
of the ellipsoid) is 24π 2 ≈ 106.6292 and the volume of the ellipsoid itself is 36π 2 ≈ 159.9438, so the
answer we obtained is certainly plausible.
13.4.33: Part(a): A cross section of Fig. 13.4.23 in the xz-plane reveals a right triangle with legs b and
a − h and hypotenuse a, and it follows immediately that b2 = 2ah − h2 .
Part (b):
The volume of the
spherical segment is
2π
V =
r
0
=
b
0
b
1
1
1
a2 − r2 − (a − h) dr dθ = 2π − (a2 − r2 )3/2 − ar2 + hr2
3
2
2
0
π
3r2 h − 3r2 a − 2(a2 − r2 )3/2
3
b
=
0
π 2
3b h − 3b2 a − 2(a2 − b2 )3/2 + 2a3
3
π 2
π 2
=
3b h − 3b2 a − 2(a2 − 2ah + h2 )3/2 + 2a3 =
3b h − 3b2 a − 2(a − h)3 + 2a3
3
3
π 2
π 2
2
2
2
3
3b h − 3a(2ah − h2 ) + 6a2 h − 6ah2 + 2h3
= (3b h − 3b a + 6a h − 6ah + 2h ) =
3
3
π
π
π 2
= (3b h − 6a2 h + 3ah2 + 6a2 h − 6ah2 + 2h3 ) = (3b2 h − 3ah2 + 2h3 ) = h(3b2 − 3ah + 2h2 ).
3
3
3
1703
Recall that 2ah = b2 + h2 , so we may substitute
3 2
2b
+ 32 h2 for 3ah in the last expression. Therefore the
volume of the spherical cap is
3 2 3 2
1
1
1
2
2
V = πh 3b − b − h + 2h = πh(6b2 − 3b2 − 3h2 + 4h2 ) = πh(3b2 + h2 ).
3
2
2
6
6
13.4.34: We first convert the given integral to polar form and integrate over the quarter circle with polar
description 0 θ 12 π, 0 r a. Then we let a → +∞. Thus we obtain
Ia =
π/2
0
a
1
r
π
π
1
·
−
dr
dθ
=
=
.
1
−
(1 + r2 )2
2
2(1 + r2 ) 0
4
1 + a2
a
0
Then
∞
0
∞
0
1
(1 +
x2
+
dx dy = lim Ia =
y 2 )2
a→∞
π
.
4
13.4.35: Using the Suggestion, we have
2π(b − x) dA = 2π(b − r cos θ)r dr dθ,
and hence the volume of the torus is
2π
V =
0
0
a
2πr(b − r cos θ) dr dθ =
2π
0
=
2
bπr − πr3 cos θ
3
2
a
1
(3πa2 bθ − 2πa3 sin θ)
3
dθ
0
2π
= 2π 2 a2 b.
0
Read the First Theorem of Pappus in Section 13.5 and apply it to this circular disk of radius a rotated
around a circle of radius b to obtain the same answer in a tenth of the time.
13.4.36:
The plane and the paraboloid meet in the circle x2 + y 2 = 9, z = −3, so the circular disk
x2 + y 2 9, z = 0 is a suitable domain for the volume integral. The volume is therefore
2π
V =
0
0
3
1
(18 − 2r ) · r dr dθ = 2π · 9r − r4
2
2
2
3
0
= 81π ≈ 254.4690049407732523.
13.4.37: The plane and the paraboloid meet in a curve that lies on the cylinder x2 + y 2 = 4, so the circular
disk x2 + y 2 4 in the xy-plane is a suitable domain for the volume integral. The volume of the solid is thus
2π
V =
0
0
2
2
1
(4 − r2 ) · r dr dθ = 2π · 2r2 − r4 = 8π ≈ 25.1327412287183459.
4
0
1704
13.4.38: The circular disk x2 + y 2 4 in the xy-plane is a suitable domain for the volume integral. The
volume of the solid is therefore
2π
2
V =
0
0
2π
=
0
(3 + r cos θ + r sin θ) · r dr dθ =
2π
0
3 2 1 3
r + r (cos θ + sin θ)
2
3
2
dθ
0
2π
8
8
8
= 12π ≈ 37.6991118430775189.
6 + [cos θ + sin θ] dθ = 6θ − cos θ + sin θ
3
3
3
0
13.4.39: When we solve the equations of the paraboloids simultaneously, we find that x2 + y 2 = 4. Hence
the curve in which the surfaces intersect lies on the cylinder x2 + y 2 = 4. Thus the disk in the xy-plane
bounded by the circle x2 + y 2 = 4 is appropriate for the domain of the volume integral. So the volume of
the solid bounded by the paraboloids is
2π
V =
θ=0
2
3
(12 − 3r2 ) · r dr dθ = 2π · 6r2 − r4
= 24π ≈ 75.3982236861550377.
4
r=0
r=0
2
13.4.40: When we solve the equations of the paraboloid and the ellipsoid simultaneously, we find that their
curve of intersection lies on the cylinder x2 + y 2 = 4. Hence the circular disk x2 + y 2 4 in the xy-plane is
a suitable domain for the volume integral. The volume is
2
1 4 2
2
2 3/2
2
( 80 − 4r − 2r ) · r dr dθ = 2π · − r − (20 − r )
V =
2
3
0
0
0
√
80 √
152
16
π 10 5 − 19 ≈ 56.3087300917396928.
= 2π
5 −
=
3
3
3
2π
2
13.4.41: A Mathematica solution:
V = Simplify[ 2∗Integrate[ Integrate[
r∗Sqrt[ b∧2 - r∧2 ], { r, 0, a } ], { θ, 0, 2∗Pi } ] ]
3 3
b − (b2 − a2 )3/2 π
2
V1 = V /. { a → 2, b → 4 }
√ 4
64 − 24 3 π
3
Percent of material removed:
N[ 100∗V1/(4∗Pi∗64/3) ]
35.0481
1705
13.4.42: A Maple solution:
V := 2∗int(int(r∗sqrt(12−4∗r∗cos(t)−r∧2), r=0..1), t=0..2∗Pi):
V := evalf(V);
V := 21.22150986
MATLAB gives
f = inline('r.∗sqrt(12−4∗r.∗cos(t)−r.∧2','r','t');
V = 2∗dblquad(f,0,1,0,2∗pi)
V = 21.2215
The percent of material removed:
N[100∗V/(4∗Pi∗64/3)]
7.91603
13.4.43: A Mathematica solution for the case of a square hole:
n = 4;
V = 4∗n∗Integrate[ Integrate[ r∗Sqrt[ 4 − r∧2 ],
{ r, 0, Cos[ Pi/n ]*Sec[ θ ] } ], { θ, 0, Pi/n } ];
(We suppress the output; it’s not attractive.)
V = Chop[ N[V] ]
7.6562
Percent:
100∗V/(4∗Pi∗8/3)
22.8473
Now for the pentagonal, hexagonal, heptagonal, and 17-sided holes. First, the pentagonal hole:
n = 5;
V = 4∗n∗Integrate[ Integrate[ r∗Sqrt[ 4 − r∧2 ],
1706
{ r, 0, Cos[ Pi/n ]*Sec[ θ ] } ], { θ, 0, Pi/n } ];
V = Chop[ N[V] ]
9.03688
Percent:
100∗V/(4∗Pi*8/3)
26.9675
Next, the hexagonal hole:
n = 6;
V = 4∗n∗Integrate[ Integrate[ r∗Sqrt[ 4 − r∧2 ],
{ r, 0, Cos[ Pi/n ]*Sec[ θ ] } ], { θ, 0, Pi/n } ];
V = Chop[ N[V] ]
9.83041
Percent:
100∗V/(4∗Pi*8/3)
29.3355
For the heptagonal and 17-sided holes, we use Maple.
n:=7;
V:=4∗n∗int(int(r∗sqrt(4−r∧2), r=0..cos(Pi/n)∗sec(t)), t=0..Pi/n);
V:=evalf(V);
V := 10.32346688
Percent:
evalf(100∗V/(4∗Pi∗8/3);
30.80682719
n:=17;
V:=4∗n∗int(int(r∗sqrt(4−r∧2), r=0..cos(Pi/n)∗sec(t)), t=0..Pi/n);
1707
V:=evalf(V);
V := 11.49809060
Percent:
evalf(100∗V/(4∗Pi∗8/3);
34.31208666
Section 13.5
Note: To integrate positive integral powers of sines and cosines (and their products), there are effective
techniques not illustrated in the text. They use the following identities, which are consequences of the
Euler-DeMoivre formula
n
(cos θ + i sin θ)n = eiθ = einθ = cos nθ + i sin nθ.
1. sin2 θ =
1
(1 − cos 2θ).
2
2. sin3 θ =
1
(3 sin θ − sin 3θ).
4
3. sin4 θ =
1
(3 − 4 cos 2θ + cos 4θ).
8
4. sin5 θ =
1
(10 sin θ − 5 sin 3θ + sin 5θ).
16
5. sin6 θ =
1
(10 − 15 cos 2θ + 6 cos 4θ − cos 6θ).
32
6. sin7 θ =
1
(35 sin θ − 21 sin 3θ + 7 sin 5θ − sin 7θ).
64
7. sin8 θ =
1
(35 − 56 cos 2θ + 28 cos 4θ − 8 cos 6θ + cos 8θ).
128
8. sin9 θ =
1
(126 sin θ − 84 sin 3θ + 36 sin 5θ − 9 sin 7θ + sin 9θ).
256
9. sin10 θ =
1
(126 − 210 cos 2θ + 120 cos 4θ − 45 cos 6θ + 10 cos 8θ − cos 10θ).
512
10. cos2 θ =
1
(1 + cos 2θ).
2
11. cos3 θ =
1
(3 cos θ + cos 3θ).
4
1708
12. cos4 θ =
1
(3 + 4 cos 2θ + cos 4θ).
8
13. cos5 θ =
1
(10 cos θ + 5 cos 3θ + cos 5θ).
16
14. cos6 θ =
1
(10 + 15 cos 2θ + 6 cos 4θ + cos 6θ).
32
15. cos7 θ =
1
(35 cos θ + 21 cos 3θ + 7 cos 5θ + cos 7θ).
64
16. cos8 θ =
1
(35 + 56 cos 2θ + 28 cos 4θ + 8 cos 6θ + cos 8θ).
128
17. cos9 θ =
1
(126 cos θ + 84 cos 3θ + 36 cos 5θ + 9 cos 7θ + cos 9θ).
256
18. cos10 θ =
1
(126 + 210 cos 2θ + 120 cos 4θ + 45 cos 6θ + 10 cos 8θ + cos 10θ).
512
19. sin2 θ cos2 θ =
1
(1 − cos 4θ).
8
20. sin3 θ cos2 θ =
1
(2 sin θ + sin 3θ − sin 5θ).
16
21. sin4 θ cos2 θ =
1
(2 − cos 2θ − 2 cos 4θ + cos 6θ).
32
22. sin5 θ cos2 θ =
1
(5 sin θ + sin 3θ − 3 sin 5θ + sin 7θ).
64
23. sin6 θ cos2 θ =
1
(5 − 4 cos 2θ − 4 cos 4θ + 4 cos 6θ − cos 8θ).
128
24. sin2 θ cos3 θ =
1
(2 cos θ − cos 3θ − cos 5θ).
16
25. sin2 θ cos4 θ =
1
(2 + cos 2θ − 2 cos 4θ − cos 6θ).
32
26. sin2 θ cos5 θ =
1
(5 cos θ − cos 3θ − 3 cos 5θ − cos 7θ).
64
27. sin2 θ cos6 θ =
1
(5 + 4 cos 2θ − 4 cos 4θ − 4 cos 6θ − cos 8θ).
128
28. sin3 θ cos3 θ =
1
(3 sin 2θ − sin 6θ).
32
29. sin4 θ cos3 θ =
1
(3 cos θ − 3 cos 3θ − cos 5θ + cos 7θ).
64
30. sin5 θ cos3 θ =
1
(6 sin 2θ − 2 sin 4θ − 2 sin 6θ + sin 8θ).
128
40. sin6 θ cos3 θ =
1
(6 cos θ − 8 cos 3θ + 3 cos 7θ − cos 9θ).
256
1709
41. sin4 θ cos4 θ =
1
(3 − 4 cos 4θ + cos 8θ).
128
42. sin5 θ cos4 θ =
1
(6 sin θ + 4 sin 3θ − 4 sin 5θ − sin 7θ + sin 9θ).
256
43. sin6 θ cos4 θ =
1
(6 − 2 cos 2θ − 8 cos 4θ + 3 cos 6θ + 2 cos 8θ − cos 10θ).
512
44. sin5 θ cos5 θ =
1
(10 sin 2θ − 5 sin 6θ + sin 10θ).
512
45. sin6 θ cos5 θ =
1
(10 cos θ − 10 cos 3θ − 5 cos 5θ + 5 cos 7θ + cos 9θ − cos 11θ).
1024
46. sin6 θ cos6 θ =
1
(10 − 15 cos 4θ + 6 cos 8θ − cos 12θ).
2048
13.5.1: By the symmetry principle, the centroid is at (x, y) = (2, 3).
13.5.2: By the symmetry principle, the centroid is at (x, y) = (2, 3).
13.5.3: By the symmetry principle, the centroid is at (x, y) = (1, 1).
13.5.4: The mass and moments are
3
m=
0
3
0
3−x
0
0
3
3
0
3−x
3
y dy dx =
0
0
1
(3 − x) dx = 3x − x2
2
x dy dx =
Mx =
3
1 dy dx =
0
My =
3−x
0
(3x − x2 ) dx =
3
=
0
3 2 1 3
x − x
2
3
9
,
2
3
=
0
9
,
2
and
3
1
1
9
(3 − x)2 dx =
(x − 3)3 = .
2
6
2
0
Therefore the centroid is located at (x, y) = (1, 1).
13.5.5: The mass and moments are
4
m=
0
4
0
(4−x)/2
4
x dy dx =
0
Mx =
4
1 dy dx =
0
My =
(4−x)/2
0
4
0
(4−x)/2
4
y dy dx =
0
0
Therefore the centroid is located at (x, y) =
0
4
4
1 2
4−x
dx = 2x − x
= 4,
2
4
0
4
1 3
16
1 2
2
,
=
2x − x
dx = x − x
2
6
3
0
4
1
1
8
(4 − x)2 dx =
(x − 4)3 = .
8
24
3
0
2
3, 3
.
1710
and
13.5.6: Here we have
1
2−y
m=
1
1 dx dy =
0
Mx =
0
y
1
2−y
1
y dx dy =
0
0
y
(2 − 2y) dy = 2y − y
2
1
=1
and
0
1
2
1
(2y − 2y 2 ) dx = y 2 − y 3 = .
3
3
0
By symmetry, x = 1. Therefore the centroid is located at 1, 13 .
13.5.7: The mass and moments are
2 x2
m=
1 dy dx =
0
My =
0
0
2
2
x2
2
x dy dx =
0
Mx =
x2
2
y dy dx =
0
Therefore the centroid is (x, y) =
1 3
x
x dx =
3
x3 dx =
0
0
2
2
0
0
3
6
2, 5
1 4
x dx =
2
2
1 4
x
4
8
,
3
=
0
2
= 4,
and
0
1 5
x
10
2
=
0
16
.
5
.
13.5.8: By symmetry, My = 0; next,
3
m=
−3
Mx =
3
9
3
1
1 dy dx =
(9 − x ) dx = 9x − x3
3
x2
−3
9
3
y dy dx =
−3
2
x2
Therefore the centroid is located at 0,
−3
27
5
3
= 36
and
−3
3
81
1 5
1
972
4
(81 − x ) dx =
x−
x
.
=
2
2
10
5
−3
.
13.5.9: By symmetry, My = 0. Next,
2
−2
Mx =
−2
2
2
=
−2
32
3
and
2
4 3
1
1 5
256
2 2
x − 8x −
x
.
y dy dx =
− (4 − x ) dx =
=−
2
3
10
15
x2 −4
−2
−2
2
2
1
1 dy dx =
(4 − x ) dx = 4x − x3
3
2
x −4
−2
m=
0
0
2
Therefore the centroid is at the point 0, − 85 .
13.5.10: By symmetry, My = 0. Next,
2
m=
2
1 dy dx =
−2
Mx =
x2 +1
2
0
(x + 1) dx =
−2
x2 +1
2
y dy dx =
−2
0
2
−2
1 3
x +x
3
2
=
−2
28
3
and
2
1 5 1 3 1
1 2
206
2
(x + 1) dx =
x + x + x
.
=
2
10
3
2 −2
15
1711
Hence the centroid is located at the point 0,
103
70
≈ (0, 1.4714285714285714).
13.5.11: The mass and moments are
1
1
xy dy dx =
0
My =
1−x
m=
0
1
0
0
1−x
x2 y dy dx =
0
1
1
1
1
(x − 2x2 + x3 ) dx =
,
6x2 − 8x3 + 3x4 =
2
24
24
0
1
0
1 2 2
x y
2
1−x
0
0
=
Mx =
1
1−x
2
1
xy dy dx =
0
0
0
1 3
xy
3
1
dx =
1−x
1
1
1
,
10x3 − 15x4 + 6x5 =
60
60
0
dx =
0
0
1 2
(x − 2x3 + x4 ) dx
2
1
1
(x − 3x2 + 3x3 − x4 ) dx
3
1
1
1
2
3
4
5
.
=
=
10x − 20x + 15x − 4x
60
60
0
Therefore the centroid is located at the point
2
2
5, 5
.
13.5.12: Using Mathematica, the computations to find the mass m and the moments My and Mx of the
lamina can be partially automated in such a way to reduce typing and make clear the individual steps in
the solution, as follows. First we compute the appropriate antiderivatives with respect to y:
{Integrate[ x∧2, y ], Integrate[ x∧3, y ], Integrate[ x∧2∗y, y ]}
x2 y,
3
x y,
1 2 2
x y
2
Then we evaluate these antiderivatives at y = 1 − x and at y = 0:
(% /.
y → 1 - x) - (% /.
(1 − x)x2 ,
(1 − x)x3 ,
y → 0)
1
(1 − x)2 x2
2
Now we integrate with respect to x:
Integrate[ %, x ]
1 3 1 4
x − x ,
3
4
1 4 1 5
x − x ,
4
5
1 3 1 4
1 5
x − x +
x
6
4
10
Then we evaluate these antiderivatives at x = 0 and at x = 1. This will yield, respectively, the values of m,
My , and Mx .
(% /.
x → 1) - (% /.
x → 0)
1712
1
,
12
1
,
20
1
60
The coordinates of the centroid are then
{ %[[2]]/%[[1]], %[[3]]/%[[1]] }
3 1
,
5 5
Answer: Mass m =
1
, centroid (x, y) =
12
3 1
,
.
5 5
13.5.13: The mass and moments of the lamina are
2
2
2 4−x2
1
1
256
(4 − x2 )2 dx =
,
y dy dx =
=
m=
240x − 40x3 + 3x5
2
30
15
−2 0
−2
−2
My =
4−x2
2
xy dy dx =
−2
Mx =
2
2
−2
−2
0
4−x2
y 2 dy dx =
2
−2
0
2
1
1
2 2
2
4
6
x(4 − x ) dx =
= 0,
48x − 12x + x
2
12
−2
and
2
64
16 3 4 5
1
1 7
4096
(4 − x2 )3 dx =
x−
x + x −
x
.
=
3
3
3
5
21
105
−2
Therefore the centroid of the lamina is (x, y) = 0,
16
7
.
13.5.14: The mass and moments of the lamina are
3
3 9−y2
3
1
9 5
1 7
23328
2
2 3
3
m=
(9 − y ) dy = 243y − 27y + y −
y
,
x dx dy =
=
3
5
21
35
−3 0
−3
−3
My =
−3
Mx =
3
3
−3
9−y 2
x3 dx dy =
−3
0
3
9−y
2
x2 y dx dy =
1
(9 − y 2 )4 dy =
4
3
−3
0
6561
243 5 9 7
1 9
y − 243y 3 +
y − y +
y
4
10
7
36
243 2 81 4 3 6
1
1 8
y(9 − y 2 )3 dy =
y −
y + y −
y
3
2
4
2
24
3
= 0.
−3
13.5.15: The mass and moments of the lamina are
1
1 √x
1
1 2
1
1
5
3
6
m=
(x − x ) dx =
,
xy dy dx =
=
2x − x
2
12
12
2
0
x
0
0
My =
√
x
2
1
x y dy dx =
0
Mx =
1
x2
1
√
0
x
2
xy dy dx =
0
x2
0
1
1
1 3
1
3
6
4
7
(x − x ) dx =
,
=
7x − 4x
2
56
56
0
1
1 5/2
1
3
7
7/2
8
(x − x ) dx =
.
=
16x − 7x
3
168
56
0
Hence the centroid of the lamina is at (x, y) =
9
9
14 , 14
.
1713
−3
3
Therefore the centroid of the lamina is (x, y) = (6, 0).
=
139968
,
35
13.5.16: The mass and moments of the lamina are
1 √x
1
1
1 3/2
x + x5/2 − x4 − x6 dx
m=
(x2 + y 2 ) dy dx =
3
3
0
x2
0
1
2 5/2 2 7/2 1 5
1 7
6
=
x + x − x −
x
,
=
15
7
5
21
35
0
1 √x
1
1 5/2
1 7
2
2
7/2
5
My =
x +x −x − x
x(x + y ) dy dx =
dx
3
3
0
x2
0
1
2 7/2 2 9/2 1 6
1 8
55
x + x − x −
x
,
=
=
21
9
6
24
504
0
1 √x
1
1 2 1 3 1 6 1 8
2
2
Mx =
x + x − x − x
(x + y )y dy dx =
dx
4
2
2
4
0
x2
0
1
1 3 1 4
1 7
1 9
55
=
x + x −
x −
x
.
=
12
8
14
36
504
0
Therefore the centroid of the lamina is at (x, y) =
275
275
432 , 432
.
13.5.17: The mass and moments of the lamina are
1
1 2−x2
1
2 3
8
2
m=
y dy dx =
(2 − 2x ) dx = 2x − x
= ,
3
3
2
−1 x
−1
−1
My =
1
2−x2
x2
−1
Mx =
1
2−x2
1
xy dy dx =
(2x − 2x ) dx = x − x4
2
−1
1
3
1 3
y
3
2−x2
2
1
1
= 0,
−1
1
1
(2 − x2 )3 − x6
3
3
2
−1
−1
x
1
8
4
2
2 7
344
x − x3 + x5 −
x
.
=
=
3
3
5
21
105
−1
y 2 dy dx =
x2
−1
1
Hence the centroid of the lamina is at (x, y) = 0,
43
35
dx =
dx
.
13.5.18: The mass and moments of the lamina are
e
e ln x
e
m=
1 dy dx =
ln x dx = − x + x ln x = 1,
1
My =
1
ln x
e
x dy dx =
1
Mx =
0
e
1
ln x
e
y dy dx =
1
0
1
e
1 2
1
e2 + 1
x ln x − x2 =
,
2
4
4
1
e
1
1
e−2
2
2
(ln x) dx = x − x ln x + x(ln x)
.
=
2
2
2
1
x ln x dx =
0
e
1
(The reduction formula in Problem 51 of Section 7.3 is helpful in evaluating the last integral.) The centroid
of the lamina is located at the point
2
e +1 e−2
,
(x, y) =
≈ (2.0972640247326626, 0.3591409142295226).
4
2
1714
13.5.19: The mass and moments of the lamina are
π
sin x
π
1 dy dx =
m=
0
My =
π
0
sin x
Mx =
x sin x dx =
0
π
0
sin x
π
1
sin2 x dx =
2
y dy dx =
0
0
= 2,
0
π
x dy dx =
0
− cos x
sin x dx =
0
π
0
π
sin x − x cos x
= π,
0
1
(2x − sin 2x)
8
π
=
0
1
π.
4
Therefore the centroid of the lamina is located at
(x, y) =
π π
,
≈ (1.5707963267948966, 0.3926990816987242).
2 8
13.5.20: The mass of the lamina is
1
exp(−x2 )
m=
x=−1
1
|xy | dy dx = −
y=0
exp(−x2 )
1
xy dy dx = 2
0
y=0
exp(−x2 )
1
exp(−x2 )
xy dy dx +
x=−1
=2
0
0
xy dy dx
0
y=0
1
x exp(−2x2 ) dx =
2
y=0
1
− exp(−2x2 )
4
1
=
0
e2 − 1
.
4e2
By symmetry, My = 0. Finally,
1
exp(−x2 )
Mx =
x=−1
=
2
3
1
2
1
|x| · y dy dx =
y=0
x=−1
x exp(−3x2 ) dx =
0
2
3
−
1
|x| · exp(−3x2 ) dx
3
1
exp(−3x2 )
6
1
=
0
1
9
e3 − 1
1
.
1− 3 =
e
9e3
Therefore the centroid of the lamina is located at the point
4(e2 + e + 1)
(x, y) = 0,
≈ (0, 0.4884168976896321).
9e(e + 1)
13.5.21: The mass and moments of the lamina are
a
1 2
1 2
2
a + ax dx =
(a x + ax ) = a3 ,
(x + y) dy dx =
m=
2
2
0
0
0
0
a
a a
a
1 2
1 2 2 1 3
2
My =
a x + ax
a x + ax
x(x + y) dy dx =
=
dx =
2
4
3
0
0
0
0
a
a a
a
1 3
1 3 1 2
1
Mx =
a + a x dx =
a x + a2 x2 =
(x + y)y dy dx =
3
2
3
4
0
0
0
0
a
a
a
Therefore its centroid is located at the point (x, y) =
7
7
12 a, 12 a
1715
.
7 4
a ,
12
7 4
a .
12
and
13.5.22: The mass and moments of the lamina are
a
a a−x
1 3
4 3
2
2
2
2
a − a x + 2ax − x
(x + y ) dy dx =
m=
dx
3
3
0
0
0
a
1 2 2 2 3 1 4
1
1 3
= a4 ,
a x − a x + ax − x
=
3
2
3
3
6
0
a a−x
a
1 3
4
a x − a2 x2 + 2ax3 − x4 dx
My =
x(x2 + y 2 ) dy dx =
3
3
0
0
0
a
1 3 2 1 2 3 1 4
4 5
1 5
a x − a x + ax −
x
a , and
=
=
6
3
2
15
15
0
a a−x
a
1 4
3 4
2
2
3
2 2
3
a − a x + 2a x − 2ax + x
(x + y )y dy dx =
Mx =
dx
4
4
0
0
0
a
1 4
1 3 2 2 2 3 1 4
3 5
1 5
a x − a x + a x − ax +
x
a .
=
=
4
2
3
2
20
15
0
Therefore the centroid of the lamina is located at (x, y) =
2
2
5 a, 5 a
.
13.5.23: The mass and moments of the lamina are
2
2
2 4
128
1 4
1 5
x
,
y dy dx =
=
8− x
dx = 8x −
m=
2
10
5
2
−2 x
−2
−2
My =
−2
Mx =
2
2
2
1 6
1 5
2
x
xy dy dx =
= 0,
8x − x
dx = 4x −
2
12
x2
−2
−2
4
4
2
2
2
y dy dx =
−2
x2
−2
64 1 6
− x
3
3
64
1 7
x−
x
dx =
3
21
Thus the centroid of the lamina is located at the point (x, y) = 0,
13.5.24:
20
7
2
=
−2
and
512
.
7
.
The curves cross where x2 = 2x + 3; that is, at x = −1 and at x = 3. Hence the mass and
moments of the lamina are
3 2x+3
x2 dy dx =
m=
−1
3
3
1
1
96
,
(3x2 + 2x3 − x4 ) dx = x3 + x4 − x5
=
2
5
5
−1
−1
x2
2x+3
3
3
3 4 2 5 1 6
x + x − x
4
5
6
2
−1 x
−1
3 2x+3
3
9 2
1
Mx =
x + 6x3 + 2x4 − x6 dx
x2 y dy dx =
2
2
−1 x2
−1
3
3 3 3 4 2 5
1 7
3616
x + x + x −
x
.
=
=
2
2
5
14
35
−1
My =
x3 dy dx =
(3x3 + 2x4 − x5 ) dx =
3
=
−1
Therefore the centroid is at
(x, y) =
17 113
,
9 21
≈ (1.8888888888888889, 5.3809523809523810).
1716
544
,
15
and
13.5.25: The mass and moments are
π
sin x
π
x dy dx =
m=
0
My =
0
π
0
Mx =
0
sin x
x2 dy dx =
0
π
π
0
sin x
0
= π,
0
π
x2 sin x dx = 2 cos x − x2 cos x + 2x sin x = π 2 − 4,
and
0
π
xy dy dx =
0
π
sin x − x cos x
x sin x dx =
0
π
1
1
1
x sin2 x dx =
(2x2 − cos 2x − 2x sin 2x) = π 2 .
2
16
8
0
Consequently the centroid of the lamina is at
(x, y) =
π2 − 4 π
,
π
8
≈ (1.8683531088546306, 0.3926990816987242).
13.5.26: By symmetry, the moment My is zero, and thus x = 0. The mass and other moment are
π
a
r2 sin θ dr dθ =
m=
θ=0
π
0
r=0
a
2
3
Mx =
π
r sin θ dr dθ =
θ=0
0
r=0
1 3
a sin θ dθ =
3
π
−
1 3
a cos θ
3
π
=
0
2 3
a
3
and
π
1 4
1 2 1 − cos 2θ
1
a ·
dθ = a θ − sin θ cos θ
= πa4 .
4
2
8
8
0
3
πa.
16
Therefore the y-coordinate of the centroid is y =
13.5.27: The mass and moments are
π
m=
θ=0
π
a
a
My =
θ=0
π
r3 cos θ dr dθ =
a
π
0
r=0
Mx =
θ=0
1 3
r
r dr dθ = π ·
3
r=0
2
r3 sin θ dr dθ =
0
r=0
π
a
=
0
1 3
πa ,
3
1 4
a cos θ dθ =
4
1 4
a sin θ dθ =
4
1 4
a sin θ
4
−
3a
0,
≈ 0, (0.47746483) · a .
2π
1717
= 0,
and
0
1 4
a cos θ
4
Therefore the centroid of the lamina is located at the point
(x, y) =
π
π
=
0
1 4
a .
2
13.5.28: The mass and moments are
2π
1+cos θ
m=
2
2π
1
(1 + cos θ)3 dθ =
3
r dr dθ =
θ=0
0
r=0
=
2π
1+cos θ
My =
θ=0
2π
1+cos θ
Mx =
θ=0
0
1
1
+ cos θ + cos2 θ + cos3 θ
3
3
dθ
1
(cos θ)(1 + cos θ)4 dθ
4
1
3
1
cos θ + cos2 θ + cos3 θ + cos4 θ + cos5 θ dθ
4
2
4
2π
1
7
= π,
840θ + 1470 sin θ + 480 sin 2θ + 145 sin 3θ + 30 sin 4θ + 3 sin 5θ
960
4
0
=
2π
2π
0
0
r=0
2π
2π
5
1
= π,
30θ + 45 sin θ + 9 sin 2θ + sin 3θ
36
3
0
r3 cos θ dr dθ =
=
r3 sin θ dr dθ =
0
r=0
2π
=
0
=−
2θ
1
(sin θ)(1 + cos θ)4 dθ
4
1
3
1
sin θ + sin θ cos θ + sin θ cos2 θ + sin θ cos3 θ + sin θ cos4 θ dθ
4
2
4
2π
1
= 0.
210 cos θ + 120 cos 2θ + 45 cos 3θ + 10 cos 4θ + cos 5θ
320
0
Therefore the centroid of the lamina is at (x, y) =
21
20 ,
0 .
13.5.29: The following figure, generated by Mathematica, shows the two circles.
y
2
1
-1
1
x
-1
They cross where 2 sin θ = 1; that is, where θ = 16 π and where θ = 56 π. The lamina in question is the region
1718
outside the lower circle and inside the upper circle. Its mass and moments are
5π/6
2 sin θ
m=
2
5π/6
r sin θ dr dθ =
π/6
1
π/6
8
1
sin4 θ − sin θ
3
3
dθ
√
5π/6
1
8π + 3 3
=
,
=
12θ + 4 cos θ − 8 sin 2θ + sin 4θ
12
12
π/6
5π/6
My =
2 sin θ
3
5π/6
r sin θ cos θ dr dθ =
π/6
1
π/6
1
5
4 sin θ cos θ − sin θ cos θ dθ
4
5π/6
1
=
= 0,
6 cos 4θ − cos 6θ − 12 cos 2θ
48
π/6
5π/6
Mx =
π/6
2 sin θ
r3 sin2 θ dr dθ =
1
5π/6
π/6
1
4 sin6 θ − sin2 θ dθ
4
√
5π/6
1
12π + 11 3
=
.
=
54θ − 42 sin 2θ + 9 sin 4θ − sin 6θ
48
16
π/6
Thus its centroid is at the point
√ 36π + 33 3
√
≈ (0, 1.4034060567438982)
0,
32π + 12 3
(x, y) =
and its mass is approximately 2.5274078042854148.
13.5.30: The limaçon and the circle are shown next, in a figure generated by Mathematica. We are to find
the mass and centroid of the region within the limaçon and outside the circle given its density at (r, θ) is r.
y
2
1
-2
-1
1
2
x
-1
-2
The curves intersect where 1 + 2 cos θ = 2; that is, where θ = ± 31 π. The mass and moments of the lamina
1719
are
π/3
1+2 cos θ
2
m=
π/3
r dr dθ =
−π/3
2
−π/3
1
8
(1 + 2 cos θ)3 −
3
3
dθ
π/3
8
7
1
2 cos θ + 4 cos2 θ + cos3 θ −
dθ =
36 sin θ + 9 sin 2θ + 2 sin 3θ − 3θ
3
3
9
−π/3
−π/3
=
π/3
√
45 3 − 2π
≈ 7.96212233704665463687,
=
9
π/3 1+2 cos θ
π/3 1
3
4
My =
(cos θ)(1 + 2 cos θ) − 4 cos θ dθ
r cos θ dr dθ =
4
−π/3 2
−π/3
15
cos θ dθ
2 cos2 θ + 6 cos3 θ + 8 cos4 θ + 4 cos5 θ −
4
−π/3
=
π/3
√
π/3
1
160π + 327 3
,
=
=
240θ + 195 sin θ + 150 sin 2θ + 55 sin 3θ + 15 sin 4θ + 3 sin 5θ
60
60
−π/3
Mx =
π/3
−π/3
1+2 cos θ
r3 sin θ dr dθ =
2
π/3
−π/3
1
(sin θ)(1 + 2 cos θ)4 − 4 sin θ
4
dθ
15
2
3
4
sin θ dθ
=
2 sin θ cos θ + 6 sin θ cos θ + 8 sin θ cos θ + 4 sin θ cos θ −
4
−π/3
π/3
π/3
1
= 0.
35 cos θ − 30 cos 2θ − 15 cos 3θ − 5 cos 4θ − cos 5θ
20
−π/3
=
Therefore the centroid of the lamina is at the point
(x, y) =
√
981 3 + 480π
√
, 0 ≈ (2.2377522671212835, 0).
900 3 − 40π
13.5.31: The polar moment of inertia of the lamina is
I0 =
2π
a
r
0
n+3
0
rn+4
dr dθ = 2π ·
n+4
a
=
0
2πan+4
.
n+4
13.5.32: The polar moment of inertia of the lamina is
I0 =
π
a
4
π
r sin θ dr dθ =
0
0
0
1 5
a sin θ dθ =
5
1
− a5 cos θ
5
π
=
0
2 5
a .
5
13.5.33: The polar moment of inertia of the lamina is
I0 =
π/2
−π/2
0
2 cos θ
π/2
1
3
kr dr dθ =
4k cos θ dθ = k 12θ + 8 sin 2θ + sin 4θ
= πk.
8
2
−π/2
−π/2
3
π/2
4
1720
13.5.34: See the figure that accompanies the solution of Problem 29. The polar moment of inertia of the
lamina is
5π/6
I0 =
2 sin θ
r4 sin θ dr dθ =
1
π/6
5π/6
1
(32 sin6 θ − sin θ) dθ
5
π/6
√
5π/6
4π + 3 3
1
≈ 5.9208410123552683.
=
=
60θ + 6 cos θ − 45 sin 2θ + 9 sin 4θ − sin 6θ
30
3
π/6
13.5.35: The polar moment of inertia of the lamina is
I0 =
π/4
√
cos 2θ
−π/4
r5 dr dθ =
0
π/4
−π/4
π/4
1
1
1
cos3 2θ dθ =
= .
9 sin 2θ + sin 6θ
6
144
9
−π/4
13.5.36: In Problem 21 we found that the mass of the lamina is m = a3 . Next,
Iy =
a
0
a
x2 (x + y) dy dx =
0
0
a
1 2 2
a x + ax3
2
dx =
1 2 3 1 4
a x + ax
6
4
a
=
0
5 5
a ,
12
√
and Ix = Iy by symmetry. Therefore x̂ = ŷ = 16 a 15 .
13.5.37: In Problem 23 we found that the mass of the lamina is m =
Iy =
2
21
2
−2
√
4
4
y 3 dy dx =
4
3
2
2
√
5.
13.5.38: In Problem 24 we found that the mass of the lamina is m =
Iy =
3
2x+3
x4 dy dx =
x2
−1
Ix =
3
−1
x2
3
−1
2x+3
and
2
2048
1
1 9
x
.
=
64 − x8 dx = 64x −
4
36
9
−2
−2
x2
105 and ŷ =
Next,
2
8 3
1
1 7
512
x −
x
=
8x2 − x6 dx =
2
3
14
21
−2
−2
x2 y dy dx =
x2
−2
Ix =
Therefore x̂ =
2
128
5 .
(3x4 + 2x5 − x6 ) dx =
96
5 .
Next,
3 5 1 6 1 7
x + x − x
5
3
7
3
=
−1
8032
105
8 5 1 8
2
3
4
x y dy dx =
9x + 18x + 12x + x − x
dx
3
3
−1
2 2
3
9
12 5 4 6
1 9
x + x −
x
= 3x + x4 +
2
5
9
27
3
3
=
−1
84256
.
135
Therefore
√
x̂ =
1757
≈ 1.9960278014413988
21
and
1721
√
2633
ŷ =
≈ 5.7014184936299995.
9
and
13.5.39: In the solution of Problem 27 we found that the mass of the lamina is m = 13 πa3 . Next,
Iy =
π
0
Ix =
a
r4 cos2 t dr dθ =
0
π
0
π
0
a
r4 sin2 θ dr dθ =
0
Therefore x̂ = ŷ =
0
1
10 a
π
π
1 5
1
1 5
a cos2 θ dθ =
a (2θ + sin 2θ) =
πa5
5
20
10
0
π
1 5
1 5 2
1
a sin θdθ =
a (2θ − sin 2θ) =
πa5 .
5
20
10
0
and
√
30 .
13.5.40: In the solution of Problem 33 we found that the mass of the lamina is m = πk. Next,
Iy =
π/2
2 cos θ
3
2
π/2
4k cos6 θ dθ
kr cos θ dr dθ =
−π/2
0
−π/2
=
Ix =
π/2
−π/2
2 cos θ
kr3 sin2 θ dr dθ =
0
π/2
1
5
k 60θ + 45 sin 2θ + 9 sin 4θ + sin 6θ
= πk;
48
4
−π/2
π/2
4k cos4 θ sin2 θ dθ
−π/2
π/2
1
1
=
k 12θ + 3 sin 2θ − 3 sin 4θ − sin 6θ
= πk.
48
4
−π/2
Therefore x̂ =
13.5.41:
1
2
√
5 and ŷ = 12 .
The quarter of the circular disk x2 + y 2 a2 that lies in the first quadrant has the polar-
coordinates description 0 r a, 0 θ 1
2 π.
If we assume that it has uniform density δ = 1, then its
mass is m = 14 πa2 . Next,
My =
π/2
0
a
r2 cos θ dr dθ =
0
0
π/2
π/2
1 2
1 3
1
a cos θ dθ =
a sin θ
= a3 ,
3
3
3
0
and Mx = My by symmetry. Therefore the centroid is located at the point
(x, y) =
4a 4a
,
3π 3π
≈ [0.4244131815783876]a, [0.4244131815783876]a .
13.5.42: Given: The quarter of the circular disk x2 + y 2 a2 that lies in the first quadrant, with centroid
(x, y). Note that x = y by symmetry. By the first theorem of Pappus,
2πy ·
which we solve for y =
1 2
2
πa = πa3 ,
4
3
4a
.
3π
1722
13.5.43: The arc x2 + y 2 = r2 can be parametrized in this way:
x(t) = r cos θ,
y(t) = r sin θ,
0θ
π
.
2
We may also assume that it has unit density. The arc length element is
[x (θ)]2 + [y (θ)]2 dθ = r dθ,
ds =
and hence its mass is
π/2
m=
r dθ =
0
1
πr.
2
Its moment around the y-axis is
My =
π/2
2
π/2
2
r cos θ dθ = r sin θ
0
= r2 .
0
Therefore, because y = x by symmetry,
(x, y) =
2r 2r
,
π π
.
13.5.44: By the second theorem of Pappus,
2πy ·
and hence x = y =
13.5.45:
πr
= 2πr2 ,
2
2r
.
π
We may assume that the triangle has unit density. The hypotenuse of this right triangle has
equation
x
y = f (x) = h 1 −
,
r
and hence the mass and moments of the triangle are
r
0
r
0
f (x)
r
x dy dx =
0
Mx =
r
1 dy dx =
0
My =
f (x)
m=
0
r
0
f (x)
r
h 2
x
1
dx = hx −
x
h 1−
= hr,
r
2r
2
0
r
x
h 3
1
h 2
hx 1 −
= hr2 ,
dx =
x −
x
r
2
3r
6
0
1 2 h2
h2
h −
x + 2 x2 dx
2
r
2r
0
2
r
h
h2 2
h2
1
x−
x + 2 x3 = h2 r.
=
2
2r
6r
6
0
r
y dy dx =
0
0
1723
and
Therefore the centroid of the triangle is
C = (x, y) =
The midpoint of the hypotenuse is the point M
1
1
2 r, 2 h
1 1
r, h .
3 3
and the line from the origin to M has equation
y = hx/r, so it is clear that C lies on this line and is two-thirds of the way from the origin to M .
r 1
1
13.5.46: Here we simply observe that 2π · rh = πr2 h.
3 2
3
r
13.5.47: Here we simply observe that 2π · L = πrL.
2
13.5.48: First we compute the mass and moments of the rectangular part of the lamina (we assume that
it has unit density):
r2
h
m=
r2
1 dy dx =
0
0
My =
r2
h dx = hr2 ,
0
h
r2
x dy dx =
0
0
Mx =
r2
hx dx =
h
0
r2
y dy dx =
0
0
0
1 2
hr ,
2 2
and
1 2
1
h dx = h2 r2 .
2
2
Next we compute the mass and moments of the triangular part of the lamina. First note that the equation
of the diagonal side of the lamina is
y=
h
(r1 − x) = f (x).
r1 − r2
Thus
r1
f (x)
m=
r1
1 dy dx =
0
r2
r1
My =
r2
f (x)
r1
x dy dx =
0
r2
r1
Mx =
r2
f (x)
r1
y dy dx =
r2
0
r2
r
2hr1 x − hx2 1
h(r1 − x)
1
dx =
= h(r1 − r2 );
r1 − r2
2(r1 − r2 ) r2
2
r
3hr1 x2 − 2hx3 1
hr1 x − hx2
1
dx =
= h(r1 − r2 )(r1 + 2r2 );
r1 − r2
6(r1 − r2 )
6
r2
h2 (r1 − x)2
dx =
2(r1 − r2 )2
3h2 (r1 )2 x − 3h2 r1 x2 + h2 x3
6(r1 − r2 )2
r1
=
r2
1 2
h (r1 − r2 ).
6
Now moments, like masses, are additive. Thus the mass and moments of the entire lamina can be obtained
by addition; they are
m=
1
h(r1 + r2 ),
2
My =
1
h(r12 + r1 r2 + r22 ),
6
and Mx =
Therefore the centroid of the lamina is at
(x, y) =
r12 + r1 r2 + r22 h(r1 + 2r2 )
,
3(r1 + r2 )
3(r1 + r2 )
1724
.
1 2
h (r1 + 2r2 ).
6
Finally, using the first theorem of Pappus, the volume generated by rotating the trapezoid around the
y-axis—the volume of the conical frustum—is
V = 2πx ·
13.5.49:
1
1
h(r1 + r2 ) = πh(r12 + r1 r2 + r22 ).
2
3
The diagonal side of the trapezoid has centroid at its midpoint, so that x =
1
2 (r1
+ r2 ). By
the second theorem of Pappus, the curved surface area of the conical frustum—which is generated by the
diagonal side—is
A = 2πx · L = πL(r1 + r2 ).
13.5.50: The vertical line segment L in the xy-plane connecting the two points (r, 0) and (r, h) generates
the curved side of the cylinder of radius r and height h when L is rotated around the y-axis. The centroid
of L has x-coordinate x = r, so by the second theorem of Pappus the curved surface area of the cylinder is
A = 2πx · h = 2πrh.
This result also follows from the result in Problem 49 by letting r1 = r2 = r, in which case L = h, so that
A = π(r1 + r2 )L = 2πrh.
13.5.51: First we compute the mass and moments of the rectangle (we assume that it has unit density):
a
m=
a
b dx = 2ab;
0
−a
b
a
bx dx =
0
−a
b
a
y dy dx =
−a
a
x dy dx =
−a
Mx =
a
1 dy dx =
−a
My =
b
0
−a
1 2
bx
2
a
= 0;
−a
1 2
b dx = ab2 .
2
Next we compute the mass and moments of the semicircle. Note that its curved boundary has equation
√
y = b + a2 − x2 = f (x). Because we have assumed unit density, its mass is m = 12 πa2 . By symmetry,
My = 0. And
2 1 1
2
2
b+ a −x
Mx =
y dy dx =
− b2 dx
2
−a b
−a 2
a
a
a
1 2
1 3
2
2 1/2
2
2
2 1/2
1 2
(a − x )
dx + 2 a x − x
=
b(a − x ) + (a − x ) dx = b
2
6
−a
−a
−a
=b·
a
f (x)
a
πa2
1
3πa2 b + 4a3
+ a3 − a3 =
.
2
3
6
1725
Moments, like masses, are additive. Thus we find the mass and moments for the entire lamina by adding
those of the rectangle and semicircle:
4ab + πa2
,
2
m=
Therefore x = 0 and
y=
My = 0,
and Mx =
6ab2 + 3πa2 b + 4a3
.
6
2
6ab2 + 3πa2 b + 4a3
6b2 + 3πab + 4a2
·
.
=
2
6
4ab + πa
12b + 3πa
Finally, we apply the first theorem of Pappus to find the volume of the solid generated by rotation of the
lamina around the x-axis:
V = 2πy ·
6b2 + 3πab + 4a2
4ab + πa2
πa
=π·
· (4ab + πa2 ) =
· (6b2 + 3πab + 4a2 ).
2
12b + 3πa
3
To check the answer, note that it has the correct domensions of volume and that in the extreme case b = 0,
it becomes
4
3
3 πa ,
the correct formula for the volume of a sphere of radius a.
13.5.52: The area of the region is
h
√
2py
A=
1 dx dy =
0
0
h
2 3/2 y
2py dy =
2p
3
0
My =
h
√
2py
h
x dx dy =
0
Mx =
py dy =
0
h
0
√
2py
h
y 3/2
y dx dy =
0
0
=
0
2 3/2 2
h
2p = rh
3
3
r2
for p). The moments are
2h
(the last equality follows from the substitution of
h
0
Therefore the centroid of the region is
(x, y) =
r2 h
h2 p
=
2
4
and
2
2
2p dy = h5/2 2p = rh2 .
5
5
3 3
r, h .
8 5
By the first theorem of Pappus, the volume of the paraboloid generated by rotating this region around the
y-axis is therefore
V = 2πx · A = 2π ·
2
1
3
r · rh = πr2 h.
8
3
2
13.5.53: The plate is a rectangle of area ab, area density δ, and mass m, and hence m = abδ. The moment
of inertia with respect to the y-axis is
Iy =
a/2
b/2
2
a/2
δx dy dx =
−a/2
−b/2
2
δbx dx =
−a/2
1726
1
δbx3
3
a/2
=
−a/2
1 3
1
δa b =
ma2 .
12
12
1
2
12 mb .
By symmetry or by a very similar computation Ix =
So, by the comment following Eq. (6) of the
text,
1
m(a2 + b2 ).
12
I0 = Ix + Iy =
13.5.54: Note that the moments Mx and My are both zero because (x, y) = (0, 0). Therefore the moment
of inertia of the region R with respect to the line perpendicular to the xy-plane at the point (x0 , y0 ) is
I=
(x − x0 )2 + (y − y0 )2 · δ dA
R
2
2
(x + y ) · δ dA − 2
=
R
R
= I0 − 2x0 δ
(x20 + y02 ) · δ dA
(x0 x + y0 y) · δ dA +
y dA + (x20 + y02 )
x dA − 2y0 δ
R
R
R
δ dA
R
= I0 − 2x0 δMy − 2y0 δMx + (x20 + y02 )m = I0 + m(x20 + y02 ).
13.5.55: Suppose that the plane lamina L is the union of the two nonoverlapping laminae R and S. Let
IL , IR , and IS denote the polar moments of inertia of L, R, and S, respectively. Let δ(x, y) denote the
density of L at the point (x, y). Then
2
IL =
2
2
(x + y )δ(x, y) dA =
L
2
(x + y )δ(x, y) dA +
R
(x2 + y 2 )δ(x, y) dA = IR + IS .
S
Next, let I1 denote the polar moment of inertia of the lower rectangle in Fig. 14.5.25 and let I2 denote the
polar moment of inertia of the upper rectangle. Then
I1 =
1
−1
I2 =
4
−4
=k
3
0
4
3
(x2 + y 2 ) · k dy dx = k
(x2 + y 2 ) · k dy dx = k
1 3 37
x +
x
3
3
4
−4
=k·
1
1
(3x2 + 9) dx = k x3 + 9x
= 20k;
−1
4
−1
x2 y +
−4
128 296
+
3
3
=
1 3
y
3
4
4 37
dx = k
x2 +
dx
3
−4
3
424
k.
3
Thus, by the first result in this solution, the polar moment of inertia of the T-shaped lamina is
I0 = I1 + I2 =
484
k.
3
13.5.56: Let k denote the [constant] density of the racquet. The area of the racquet is
π/4
A=
θ=−π/4
1 2
r dθ =
2
π/4
−π/4
π/4
1
1
1
cos 2θ dθ =
sin 2θ
= ,
2
4
2
−π/4
1727
and hence its mass is m = 12 k. Next we compute the moment of inertia I of the racquet with respect to the
line x = −1:
π/4
√
cos 2θ
I =2
0
kr(1 + r cos θ)2 dr dθ = k
0
π/4
0
4
1
r2 + r3 cos θ + r4 cos2 θ
3
2
√cos 2θ
dθ = J1 + J2 + J3
0
where
J1 = k
cos 2θ dθ = k ·
0
J2 = k
π/4
π/4
0
1
sin 2θ
2
π/4
=
0
1
k,
2
1
k
cos2 θ cos2 2θ dθ =
12θ + 9 sin 2θ + 3 sin 4θ + sin 6θ
2
96
and
4k
J3 =
3
π/4
π/4
=
0
k
(3π + 8),
96
(cos 2θ)3/2 cos θ dθ.
0
To evaluate J3 using Mathematica, the command
Integrate[ (4∗k/3)∗(Cos[2∗t])∧(3/2)∗Cos[t], t]
produces the antiderivative
√
4k
3 arcsin( 2 sin t) √
1
1
√
·
sin t + sin 3t
+ cos 2t
3
4
8
8 2
Then we evaluate the antiderivative at the limits of integration:
(% /.
t → Pi/4) - (% /.
t → 0)
kπ
√
4 2
To evaluate J3 by hand, note that
4k
J3 =
3
The substitution u = arcsin
sin u =
Moreover, cos u du =
π/4
0
(1 − 2 sin2 θ)3/2 cos θ dθ.
√
2 sin θ then yields
√
2 sin θ;
sin2 u = 2 sin2 θ;
1 − 2 sin2 θ = 1 − sin2 u = cos2 u.
√
2 cos θ dθ, and therefore
1
cos θ dθ = √ cos u du.
2
1728
These substitutions yield
4k
J3 =
3
=
π/2
u=0
√ π/2 2
1 + cos 2u
2k 2
cos u
(cos u) · √ du =
du
3
2
2
0
3
√
2k 2 1 π/2
1 + cos 4u
·
1 + 2 cos 2u +
du
3
4 0
2
√
√
√
π/2
3
1
πk 2
k 2 3 π
k 2
·
u + sin 2u + sin 4u
· · =
.
=
=
6
2
8
6
2 2
8
0
Therefore
I = J1 + J2 + J3 =
√ k 56 + 3π + 12π 2 ,
96
and hence the radius of gyration of the racquet with respect to the line x = −1 is
x̂ =
Iy
=
m
√
56 + 3π + 12π 2
≈ 1.5728117948615531.
48
The coordinates of the “sweet spot” are thus approximately (0.573, 0); this point is marked with a bullet in
the following sketch of the racquet.
y
0.5
-1
0.5
1
x
-0.5
13.5.57: The mass and moments are
π
2 sin θ
0
0
π
0
Mx =
π
r sin θ dr dθ =
m=
My =
2
0
2 sin θ
r3 sin θ cos θ dr dθ =
0
π
π
4 sin5 θ cos θ dθ =
0
2 sin θ
3
2
π
r sin θ dr dθ =
0
π
1
8
4
sin θ dθ =
= π;
12θ − 8 sin 2θ + sin 4θ
3
12
0
0
Therefore the centroid is at (x, y) =
0
2
sin6 θ
3
π
= 0;
0
π
1
5
4 sin θ dθ =
= π.
60θ − 45 sin 2θ + 9 sin 4θ − sin 6θ
48
4
0
6
5
0,
.
4
1729
13.5.58: The mass and moments are
π 2 sin θ
m=
r3 sin θ dr dθ =
0
My =
π
0
Mx =
0
π
4 sin5 θ dθ =
0
2 sin θ
r4 sin θ cos θ dr dθ =
0
π
4
2
π
− 150 cos θ + 25 cos 3θ − 3 cos 5θ
π
32
32
sin6 θ cos θ dθ ==
sin7 θ
5
35
0
2 sin θ
1
60
π
=
0
64
;
15
= 0;
0
π
32
sin7 θ dθ
5
0
π
1
1024
=
=
− 1225 cos θ + 245 cos 3θ − 49 cos 5θ + 5 cos 7θ
.
350
175
0
r sin θ dr dθ =
0
0
48
Therefore the centroid is at the point (x, y) = 0,
.
35
13.5.59: The mass and moments are
π/2 2 cos θ
m=
r2 cos θ dr dθ =
0
My =
π/2
0
Mx =
0
0
2 cos θ
0
π/2
0
π/2
r3 cos2 θ dr dθ =
π/2
8
1
1
4
cos θ dθ =
= π;
12θ + 8 sin 2θ + sin 4θ
3
12
2
0
π/2
4 cos6 θ dθ =
0
2 cos θ
0
r3 sin θ cos θ dr dθ =
π/2
Therefore the centroid is located at (x, y) =
13.5.60: The mass and moments are
π/2 2 cos θ
m=
r5 cos2 θ sin2 θ dr dθ =
=
=
0
π/2
0
−
2
cos6 θ
3
π/2
=
0
2
.
3
5 4
,
.
4 3π
π/2
0
32
cos8 θ sin2 θ dθ
3
2 cos θ
r6 cos3 θ sin2 θ dr dθ =
0
π/2
0
128
cos10 θ sin2 θ dθ
7
π/2
3
1
π;
=
2520θ + 1440 sin 2θ − 225 sin 4θ − 400 sin 6θ − 195 sin 8θ − 48 sin 10θ − 5 sin 12θ
6720
16
0
Mx =
π/2
7
1
π;
=
840θ + 420 sin 2θ − 120 sin 4θ − 130 sin 6θ − 45 sin 8θ − 6 sin 10θ
2880
48
0
My =
4 sin θ cos5 θ dθ =
0
0
π/2
1
5
= π;
60θ + 45 sin 2θ + 9 sin 4θ + sin 6θ
48
8
0
π/2
0
0
2 cos θ
r6 cos2 θ sin3 θ dr dθ =
0
π/2
128
cos9 θ sin3 θ dθ
7
1
=
−1080 cos 2θ − 405 cos 4θ + 20 cos 6θ + 90 cos 8θ + 36 cos 10θ + 5 cos 12θ
6720
Hence the centroid is located at the point (x, y) =
9 512
,
7 245π
1730
π/2
=
0
32
.
105
≈ (1.2857142857, 0.6652027009).
Section 13.6
Note: For some triple integrals, there are six possible orders of integration of the corresponding iterated
integrals; for many triple integrals, there are two. To save space, in this section we do not show all possibilities,
but only the one that seems most natural.
13.6.1: The value of the triple integral is
1
3
2
I=
1
3
(x + y + z) dx dy dz =
z=0
y=0
1
x=0
z=0
3
1
(2 + 2y + 2z) dy dz =
=
z=0
y=0
y=0
2y + y 2 + 2yz
3
z=0
z=0
y=0
dy dz
x=0
1
dz =
1
(15 + 6z) dz = 15z + 3z 2 = 18.
0
y=0
13.6.2: The value of the triple integral is
π π π
xy sin z dx dy dz =
J=
2
1 2
x + xy + xz
2
0
π
1 2
1 4
π y sin z dy dz =
π sin z dz
2
0
0
0 4
π
1
1
= − π 4 cos z = π 4 ≈ 48.7045455170012186.
4
2
0
x=0
π
2
π
13.6.3: The value of the triple integral is
6
2
3
K=
6
xyz dx dy dz =
z=−2
y=0
x=−1
z=−2
y=0
6
1 2
x yz
2
z=−2
6
2
dy dz =
4yz dy dz
z=−2
x=−1
2
2
2y z
=
3
y=0
6
dz =
8z dz = 4z
2
z=−2
y=0
6
= 128.
z=−2
13.6.4: The value of the triple integral is
6
2
3
I=
6
2
(x + y + z) dx dy dz =
z=−2
y=0
6
x=−1
z=−2
2
6
(4 + 4y + 4z) dy dz =
=
z=−2
y=0
2
2
2y + 4y + 4yz
y=0
z=−2
6
= 16z + 4z 2
1 2
x + xy + xz
2
3
dy dz
x=−1
6
dz =
(16 + 8z) dz
z=−2
y=0
= 256.
z=−2
13.6.5: The value of the triple integral is
1
1−x
1−x−y
J=
2
1
1−x x dz dy dx =
x=0
1
=
x=0
y=0
z=0
1
x y − x y − x2 y 2
2
2
3
1−x−y
x z
x=0
1−x
dx =
y=0
2
0
1
y=0
1−x
dy dx =
x=0
z=0
1 2
1
x − x3 + x4
2
2
1731
1
dx =
(x2 − x3 − x2 y) dy dx
y=0
1 3 1 4
1 5
x − x +
x
6
4
10
1
=
0
1
.
60
13.6.6: The value of the triple integral is
3 (6−2x)/3 6−2x−3y
K=
(2x + 3y) dz dy dx =
0
0
3
0
(6−2x)/3
=
0
3
0
0
2
4 3 2 4
x + x
3
9
= 12x −
6−2x−3y
0
dy dx
0
(12x − 4x2 + 18y − 12xy − 9y 2 ) dy dx
2
(6−2x)/3
2xz + 3yz
0
2
12xy − 4x y + 9y − 6xy − 3y
=
3
3
(6−2x)/3
dx =
0
0
3
3
8 3
2
12 − 4x + x
dx
9
= 18.
0
13.6.7: The value of the triple integral is
0 2 1−x2
xyz dz dy dx =
I=
1−x2
1
2
xyz
dy dx
x=−1 y=0 z=0
x=−1 y=0 2
z=0
2
0 0 2 1
1 2 1 3 2 1 5 2
1 5
3
xy − x y + x y dy dx =
xy − x y + x y
dx
=
2
2
2
4
x=−1 y=0
x=−1 4
y=0
0
2
0
1
1
2
4
6
(x − 2x + x ) dx =
=− .
=
3x − 3x + x
6
6
x=−1
x=−1
0
3
5
13.6.8: The value of the triple integral is
1 2 4−y2
J=
(2y + z) dz dy dx =
4−y2
1 2
dy dx
2yz + z
2
x=−1 y=−2 z=0
−1 −2
0
2
1 2
1
1
4
1
1 5
y
dx
=
8 + 8y − 4y 2 − 2y 3 + y 4 dy dx =
8y + 4y 2 − y 3 − y 4 +
2
3
2
10
−1 −2
−1
−2
1
512
256
=
dx =
≈ 34.1333333333333333.
15
−1 15
1
13.6.9: The value of this triple integral is
1 3 2−x2
K=
(x + y) dz dy dx =
−1
1
x2
0
3
=
−1
0
2
1
3
0
(2x − 2x + 2y − 2x y) dy dx =
1
dy dx
x2
2
2−x2
xy + xz
−1
3
1
−1
3
3
13.6.10: The value of this triple integral is
1 2 8−y2
z dz dy dx =
I=
−1
2
=
−1
−2
y2
8
32y − y 3
3
1
−1
2
1
dx =
−2
−1
2
2
−2
1 2
z
2
3
8−y2
1
dx
0
= 12.
1
2
−1
−2
(32 − 8y 2 ) dy dx
512
256
dx =
≈ 170.66666666666666667.
3
3
1732
3
−1
dy dx =
y2
2 2
2xy − 2x y + y − x y
3
=
(9 + 6x − 9x − 6x ) dx = 9x + 3x − 3x − x4
2
−1
2
2
13.6.11: The solid resembles the tetrahedron shown in Fig. 13.6.4 of the text. You will see it if you execute
the Mathematica command
ParametricPlot3D[ {{x, 0, 6 - 2∗x - 3∗y}, {x, y, 6 - 2∗x - 3∗y}},
{x, 0, 3}, {y, 0, 2}, PlotRange → {0, 3}, {0, 2}, {0, 6}},
AspectRatio → 1.0, ViewPoint → {3.5, -1.6, 3.0},
LightSources → {{{1., 0., 1.}, RGBColor[1., 0., 0.]},
{{1., 1., 1.}, RGBColor[0., 1., 0.]}, {{0., 1., 1.}, RGBColor[1., 0., 1.]}} ];
If you change the Viewpoint parameters, you will see that we plotted only the diagonal face of the tetrahedron
and the face that lies in the xz-plane. The volume of this tetrahedron is given by
3
(6−2x)/3
6−2x−3y
V =
3
(6−2x)/3
1 dz dy dx =
x=0
3
y=0
z=0
=
6y − 2xy −
x=0
3 2
y
2
x=0
(6−2x)/3
(6 − 2x − 3y) dy dx
y=0
3
2
2
6 − 4x + x2 dx = 6x − 2x2 + x3 = 6.
3
9
x=0
0
dx =
y=0
3
13.6.12: The volume is
2
4
y
V =
2
1 dz dy dx =
−2
x2
0
4
2
y dy dx =
−2
x2
−2
1 2
y
2
4
dx
x2
2
128
1 4
1 5
=
x
= 25.6.
=
8− x
dx = 8x −
2
10
5
−2
−2
2
13.6.13: The view of the figure on the left below was generated with the Mathematica command
ParametricPlot3D[ {{u, 0, v}, {u, 4 - u∧2, v}, {u, v, 4 - v}},
{u, -2, 2}, {v, 0, 4}, AspectRatio → Automatic, Shading → False,
PlotRange → {{-2.2, 2.2}, {-0.2, 4.2}, { -0.2, 4.2}}, ViewPoint → {4, 4, 5} ];
1733
4
0
-2
y
x
0
y
2
x
0
2
4
2
4
z
-2
0
2
4
2
2
z
Change the ViewPoint command parameters to {4, -4, 7} to see the figure from another angle (this view
is shown above, on the right). The volume of the solid is
2
4−x2
4−y
2
4−x2
1 dz dy dx =
V =
x=−2
y=0
z=0
x=−2
(4 − y) dy dx =
2
4y − y
2
x=−2
y=0
4−x2
dx
y=0
2
128
1
1 5
x
= 25.6.
=
8 − x4 dx = 8x −
2
10
5
−2
−2
=
2
13.6.14: The volume is
1
1−x
x2 +y 2
1
1−x
1 dz dy dx =
V =
0
0
0
0
=
0
13.6.15:
(x2 + y 2 ) dy dx =
0
1
0
1
x2 y +
1 3
y
3
1−x
dx
0
1
1
1
1
2
3
2
3
4
(1 − 3x + 6x − 4x ) dx =
= .
2x − 3x + 4x − 2x
3
6
6
0
It’s not easy to get a clear three-dimensional plot of this solid. We got fair results with the
Mathematica command
ParametricPlot3D[ {{u, u∧2, v}, {u*u, u, v}, {u*Sqrt[10], v, 10 - u∗u - v∗v}},
{u, 0, 1}, {v, 0, 10}, AspectRatio → 0.8,
PlotRange → {{0, Sqrt[10]}, {0, Sqrt[10]}, {0, 10}}, ViewPoint → {6, -4, 12} ];
If you have the time and interest, experiment with changing the AspectRatio and ViewPoint parameters
1734
until you get a clear view of the solid. Its volume is
1 1 √x 10−x2 −y2
1 dz dy dx =
V =
y=x2
x=0
1
=
x=0
=
1
(30y − 3x2 y − y 3 )
3
y=√x
(10 − x2 − y 2 ) dy dx
1
1
10x1/2 − x3/2 − 10x2 − x5/2 + x4 + x6 dx
3
3
1
dx =
0
y=x2
x
y=x2
x=0
z=0
√
20 3/2
2 5/2 10 3 2 7/2 1 5
1 7
x −
x −
x − x + x +
x
3
15
3
7
5
21
13.6.16: The volume is
−1 2 V =
−3
8−z 2
−3
−1
13.6.17: The volume is
2 4 V =
z=x2
4−z
0
332
≈ 3.1619047619047619.
105
(8 − 2z ) dz dy =
−1
−3
2
8z − z 3
3
2
−3
128
64
dy =
≈ 42.6666666666666667.
3
3
2
4
1 dy dz dx =
y=0
=
2
−2
=
x=−2
2
1 dx dz dy =
z2
−2
−1
1
z=x2
x=−2
2
(4 − z) dz dx =
x=−2
dy
−2
1
4z − z 2
2
4
dx
z=x2
2
1 4
1 5
256
4 3
2
x
≈ 17.0666666666666667.
=
8 − 4x + x
dx = 8x − x +
=
2
3
10
15
x=−2
−2
2
13.6.18: The volume is
1 1−y2 V =
y=−1
z=y 2 −1
1
1−z
z=y 2
1
=
y=0
=
x=0
z=y 2 −1
y=−1
1
1
(1 − z) dz dy =
8
≈ 2.6666666666666667.
3
√
z=y 2
y=0
y
1
y=−1
=
−1
13.6.19: The volume is
1 √y 2−y−z
V =
1 dx dz dy =
y=0
1−y 2
1 dx dz dy =
x=0
2
=
(2 − 2y ) dy = 2y − y 3
3
y=−1
2
1
(2 − y − z) dz dy =
1
y=0
1
z − z2
2
1−y2
dy
y 2 −1
1
2z − yz − z 2
2
√ y
dy
z=y 2
1
(4y 1/2 − y − 2y 3/2 − 4y 2 + 2y 3 + y 4 ) dy
2
1
11
1
≈ 0.3666666666666667.
80y 3/2 − 15y 2 − 24y 5/2 − 40y 3 + 15y 4 + 6y 5 =
60
30
0
13.6.20: The volume is
2 2−x 4−x2 −z2
V =
1 dy dz dx =
0
=
0
0
2
0
0
2
0
2−x
(4 − x2 − z 2 ) dz dx =
0
2
4z − x2 z −
2
1
1
16
(16 − 12x2 + 4x3 ) dx =
≈ 5.3333333333333333.
16x − 4x3 + x4 =
3
3
3
0
1735
1 3
z
3
2−x
dx
0
13.6.21: Because the solid has constant density δ = 1, its mass is
2 4 y
2 4
m=
1 dz dy dx =
y dy dx =
y=x2
x=−2
z=0
y=x2
x=−2
2
x=−2
1 2
y
2
4
dx
y=x2
2
128
1
1 5
x
.
=
8 − x4 dx = 8x −
2
10
5
x=−2
−2
2
2
=
Its moments are
Myz =
y
2
4
x dz dy dx =
x=−2
=
y=x2
2
x=−2
1 2
xy
2
xy dy dx =
z=0
y=x2
x=−2
4
dx
y=x2
2
1 6
1
x
= 0,
8x − x5 dx = 4x2 −
2
12
x=−2
−2
2
2
Mxz =
4
y
2
4
y dz dy dz =
x=−2
y=x2
2
=
x=−2
2
Mxy =
z=0
64 1 6
− x
3
3
4
y=x2
x=−2
dx =
64
1 7
x−
x
3
21
y
2
1
2
y=x2
2
x=−2
z=0
4
x=−2
y 2 dy dx =
y=x2
1 7
32
x−
x
=
3
42
2
=
−2
1
2
2
x=−2
1 2
y dy dx =
2
2
=
−2
4
z dz dy dx =
x=−2
=
4
y=x2
1 3
y
3
2
x=−2
1 3
y
3
4
dx
y=x2
512
,
7
1 2
z
2
y
dy dx
z=0
4
2
dx =
y=x2
x=−2
32 1 6
− x
3
6
dx
256
.
7
20 10
,
Therefore the centroid of the solid is located at the point (x, y, z) = 0,
.
7 7
13.6.22: Because the hemispherical solid has unit density δ = 1, its mass is m =
2
3
3 πR .
By symmetry,
x = y = 0. It remains only to compute the moment Mxy . Let D denote the circular disk x2 + y 2 R2 ,
z = 0. Then
√R2 −x2 −y2
Mxy =
z dz
2π
R
=
θ=0
dA =
z=0
D
r=0
1 2
(R − r2 ) · r dr dθ =
2
D
2π
θ=0
1 2
z
2
√R2 −x2 −y2
dA =
z=0
1 2 2 1 4
R r − r
4
8
D
1 2
(R − x2 − y 2 ) dA
2
R
dθ = 2π ·
r=0
1 4
1
R = πR4 .
8
4
3
Therefore the centroid is located at the point (x, y, z) = 0, 0, R .
8
13.6.23: Because the solid has unit density δ = 1, its mass and moments are
4
2 4 4−z
2 4
2 1
m=
1 dy dz dx =
(4 − z) dz dx =
dx
4z − z 2
2
x=−2 z=x2 y=0
x=−2 z=x2
x=−2
z=x2
1736
2
1 4
1 5
256
4 3
2
x
;
=
=
8 − 4x + x
dx = 8x − x +
2
3
10
15
−2
−2
2
2
Myz =
4
4−z
2
4
x dy dz dx =
z=x2
x=−2
y=0
z=x2
x=−2
2
(4x − xz) dz dx =
4xz −
x=−2
1 2
xz
2
4
dx
z=x2
2
1 5
1 6
3
2
4
x
=
= 0;
8x − 4x + x
dx = 4x − x +
2
12
−2
=2
2
2
Mxz =
4
4−z
2
4
y dy dz dx =
z=x2
x=−2
2
=
−2
y=0
32
1
− 8x2 + 2x4 − x6
3
6
2
Mxy =
4
z=x2
x=−2
dx =
4−z
z=x2
2
=
−2
y=0
2
x=−2
32
1
− 2x4 + x6
3
3
dx =
4
z=x2
(4z − z 2 ) dz dx =
2
8z − 2z 2 +
x=−2
32
8
2
1 7
x − x3 + x5 −
x
3
3
5
42
z dy dz dx =
x=−2
1
(4 − z)2 dz dx =
2
2
−2
2z 2 −
x=−2
32
2
1 7
x − x5 +
x
3
5
21
2
=
−2
4
dx
z=x2
2048
;
105
=
2
1 4
z
6
1 3
z
3
4
dx
z=x2
1024
.
35
Therefore the centroid of the solid is located at the point (x, y, z) =
8 12
0, ,
.
7 7
13.6.24: Because the solid has unit density δ = 1, its mass is
1
1
1−x2
1
1 dz dy dx =
m=
−1
−1
−1
0
1
1
2
(1 − x ) dy dx =
(2 − 2x ) dx = 2x − x3
3
−1
−1
2
2
1
=
−1
8
.
3
By symmetry x = y = 0, and
Mxy =
1
1
1−x2
1
1
z dz dy dx =
−1
−1
−1
0
−1
1
(1 − x2 )2 dy dx =
2
1
−1
(1 − x2 )2 dx
1
2
1
16
= x − x3 + x5
.
=
3
5
15
−1
Therefore the centroid of the solid is located at the point (x, y, z) =
2
0, 0,
.
5
13.6.25: Because the solid has unit density δ = 1, its mass and moments are given by
π/2
cos x 1−z
π/2
1 dy dz dx =
m=
x=−π/2
Myz = 0
y=0
x=−π/2
1
z − z2
2
cos x
dx
z=0
π/2
π/2
1
8−π
1
;
(2 cos x − cos2 x) dx =
=
cos x − cos2 x dx =
−2x + 8 sin x − sin 2x
2
4
4
−π/2
0
0
=
z=0
π/2
(by symmetry);
1737
π/2
cos x 1−z
Mxz =
π/2
cos x
1
(1 − z)2 dz dx
2
y dy dz dx =
x=−π/2
z=0
π/2
y=0
x=−π/2
1
1
1
z − z2 + z3
2
2
6
=
x=−π/2
cos x
z=0
π/2
dx =
−π/2
z=0
1
1
1
2
3
cos x − cos x + cos x dx
2
2
6
π/2
1
44 − 9π
;
=
−18x + 45 sin x − 9 sin 2x + sin 3x
72
36
−π/2
=
π/2
cos x 1−z
Mxy =
π/2
cos x
z dy dz dx =
x=−π/2
π/2
=
−π/2
z=0
y=0
x=−π/2
(z − z 2 ) dz dx =
z=0
π/2
x=−π/2
1 2 1 3
z − z
2
3
cos x
dx
z=0
π/2
1
1
1
9π − 16
2
3
cos x − cos x dx =
.
=
18x − 18 sin x + 9 sin 2x − 2 sin 3x
2
3
72
36
−π/2
Therefore the centroid of the solid is located at the point
44 − 9π 9π − 16
,
(x, y, z) = 0,
≈ (0, 0.359643831963, 0.280712336074).
72 − 9π 72 − 9π
13.6.26: (See Problem 12.) The moment of inertia of the solid (with density δ = 1) with respect to the
z-axis is
Iz =
2
−2
4
x2
y
(x2 + y 2 ) dz dy dx =
0
2
4
(x2 y + y 3 ) dy dx =
x2
−2
2
−2
1 2 2 1 4
x y + y
2
4
4
dx
x2
2
1
1
1 7
1 9
15872
8
x −
x
≈ 251.936507936508.
=
648x2 − x6 − x8 dx = 64x + x3 −
2
4
3
14
36
63
−2
−2
=
2
13.6.27: (See Problem 24.) The moment of inertia of the solid (with density δ = 1) with respect to the
y-axis is
1
1
1−x2
Iy =
2
2
1
1
(x + z ) dz dy dx =
x=−1
y=−1
1
1
=
x=−1
−1
x=−1
z=0
1
(1 − x6 ) dy dx =
3
1
−1
y=−1
1
x z + z3
3
2
1−x2
dy dx
z=0
1
2 7
2
8
2
(1 − x6 ) dx =
x−
x
= .
3
3
21
7
−1
13.6.28: Given: The solid cylinder x2 + y 2 R2 , 0 z H; we assume constant density δ rather than
constant density 1. Let D denote the base of the cylinder—the circular disk x2 + y 2 R2 , z = 0. The
moment of inertia of the cylinder with respect to the z-axis is
H
2
2
δ(x + y ) dz dA = δH
(x2 + y 2 ) dA = δH
Iz =
z=0
D
= 2πδH
1 4
r
4
D
R
=
0
1
πδR4 H.
2
1738
2π
θ=0
R
r=0
r3 dr dθ
Because the mass of the cylinder is M = δπR2 H, the answer can also be expressed in the form
2
1
R
,
Iz = M R2 = M · √
2
2
demonstrating that the answer has the correct dimensions—the product of mass and square of distance.
For a physical interpretation of the last equation, the cylinder behaves for purposes of angular acceleration
√
around the z-axis as if all its mass were concentrated at distance 12 R 2 from the z-axis.
13.6.29: With unit density δ = 1, the moment of inertia of the given tetrahedron with respect to the z-axis
is
1
1−z
1−y−z
Iz =
2
2
1
1−z
(x + y ) dx dy dz =
z=0
y=0
1
1−z
x=0
=
z=0
y=0
1
=
z=0
1
=
0
z=0
y=0
1 3
x + xy 2
3
1−y−z
dy dz
x=0
4
1
1
− y + 2y 2 − y 3 − z + 2yz − 2y 2 z + z 2 − yz 2 − z 3
3
3
3
1
2
1
1
− y 4 + y 3 (1 − z) − y 2 (1 − z)2 + y(1 − z)3
3
3
2
3
dy dz
1−z
dz
y=0
1
1
1
1
(1 − 4z + 6z 2 − 4z 3 + z 4 ) dz =
.
5z − 10z 2 + 10z 3 − 5z 4 + z 5 =
6
30
30
0
13.6.30: Problem 54 of Section 13.5 provides an alternative method of solving this problem, one that is
frequently simpler. But we will solve it by direct methods. The moment of inertia of the given solid cube of
unit density with respect to the z-axis is
1/2
4
1/2
Iz =
2
2
1/2
4
(x + y ) dx dy dz =
z=−1/2
y=3
1/2
4
x=−1/2
1
+ y2
12
=
z=−1/2
z=−1/2
y=3
1/2
dy dz =
−1/2
y=3
1
1
y + y3
12
3
1 3
x + xy 2
3
4
1/2
dz =
−1/2
3
1/2
dy dz
−1/2
149
149
dz =
≈ 12.41666667.
12
12
13.6.31: It should be clear that x = y = 0 by symmetry, but there is a slight suggestion in the wording
of the problem that we should prove this more rigorously. Hence we will compute the mass and all three
moments. Assuming unit density δ = 1, we have
m=
√
x=− h
=
√
h
√
y=− h−x2
√
h
√
x=− h
The substitution x =
√
h−x2
1 dz dy dx =
z=x2 +y 2
1
hy − x y − y 3
3
2
√
h
h
√h−x2
√
x=− h
√
y=− h−x2
dx =
√
h
√
x=− h
√
h−x2
(h
√
y=− h−x2
− x2 − y 2 ) dy dx
4
(h − x2 )3/2 dx.
3
√
√
h sin θ, dx = h cos θ dθ then yields
π/2
m=
θ=−π/2
π/2
1 2
1
4 2
4
h cos θ dθ =
h 12θ + 8 sin 2θ + sin 4θ
= πh2 .
3
24
2
−π/2
1739
Next,
Myz =
√
h
√
x=− h
=
√
h−x2
1
hxy − x y − xy 3
3
4
= − (h − x2 )5/2
15
Mxz =
√
h
√
√
x=− h
√
√
dx =
√
y=− h−x2
√
− h
(hx
√
y=− h−x2
h
− x3 − xy 2 ) dy dx
4
x(h − x2 )3/2 dx
3
√
− h
h−x2
= 0;
y dz dy dx =
√h−x2
√
y=− h−x2
dx =
z=x2 +y 2
h−x2
(hy
√
y=− h−x2
√
− h
√
h
h
√
− x2 y − y 3 ) dy dx
√
h
z dz dy dx =
√
y=− h−x2
h
√
x=− h
z=x2 +y 2
h−x2
√
h
1 2 1 2 2 1 4
hy − x y − y
2
2
4
h
√
√h−x2
√
h−x2
√ h
√
y=− h−x2
√
x=− h
Mxy =
h
√
x=− h
=
√
√
x=− h
z=x2 +y 2
3
√
x=− h
√
h
h
x dz dy dx =
√
y=− h−x2
h
√
√
x=− h
0 dx = 0;
√
h−x2
√
y=− h−x2
1 2 1 2
h − (x + y 2 )2
2
2
dy dx.
Now rewrite the last integral in polar coordinates. Then
2π
√
h
Mxy =
θ=0
r=0
1 2
(h − r4 ) · r dr dθ =
2
2π
θ=0
1 2 2
1 6
h r −
r
4
12
√ h
dθ = 2π ·
r=0
1 3 1 3
h = πh .
6
3
Thus the centroid of the parabolic segment is located at the point
2
(x, y, z) = 0, 0, h ,
3
and therefore the centroid is on the axis of symmetry of the segment, two-thirds of the way from the vertex
(0, 0, 0) to the base.
13.6.32: We assume that the cone has unit density δ = 1, that its vertex is at the origin, that its axis of
symmetry lies on the nonnegative z-axis, and that it has radius R and height H. Then the equation of its
base is z = H and the equation of its curved side is
z=
H 2
x + y2 .
R
Let D denote the circular disk x2 + y 2 R2 in the xy-plane. Then the mass of the cone is
H
H 2
2
1
dz
dA
=
x
+
y
m=
H
−
dA.
√
R
z=(H/R) x2 +y 2
D
D
Rewrite the double integral as an iterated integral in polar coordinates. Thus
2π
m=
θ=0
R
H 2
H 3
1
H
r −
r
= πR2 H.
H − r · r dr dθ = 2π
R
2
3R
3
r=0
0
R
1740
Next, we could argue that the centroid lies on the axis of the cone by symmetry, but there is an implication
in the statement of the problem that this should be proved rigorously. Hence we compute all three moments.
First,
H
Myz =
D
z=(H/R)
√
x dz
dA =
x2 +y 2
H Hx − x x2 + y 2 dA.
R
D
Rewrite the double integral as an iterated integral in polar coordinates (don’t forget to replace x with
r cos θ). Thus
R
2π 1
H 4
H 2
3
Hr cos θ −
r cos θ
=
dθ
Hr cos θ − r cos θ · r dr dθ =
R
4R
θ=0 r=0
θ=0 3
r=0
2π
2π
1
1 3
3
R H cos θ dθ =
=
= 0.
R H sin θ
12
12
0
0
Myz
2π
R
Also,
H
Mxz =
D
z=(H/R)
√
y dz
dA =
x2 +y 2
D
H Hy − y x2 + y 2
R
dA.
As before, rewrite the double integral as an iterated integral in polar coordinates (and don’t forget the
substitution of r sin θ for y). So
R
2π R 2π H 2
1
H 4
3
Hr sin θ − r sin θ r dθ =
dθ
Hr sin θ −
r sin θ
Mxz =
R
4R
θ=0 r=0
θ=0 3
r=0
2π
2π
1 3
1
=
R H sin θ dθ = − R3 H cos θ
= 0.
12
12
0
0
Finally,
Mxy =
D
H
z=(H/R)
√
z dz
dA =
x2 +y 2
D
1 2
H2 2
2
H −
(x + y ) dA.
2
2R2
As before, rewrite the double integral in polar form. Thus
2
R
2π R 1 2
H 2
H2 2
H2 4
1
H −
r
Mxy =
r
−
r
= πR2 H 2 .
·
r
dr
dθ
=
2π
·
2
2
2
2R
4
8R
4
θ=0 r=0
0
Therefore the centroid of the uniform solid right circular cone is located at the point
3
(x, y, z) = 0, 0, H ,
4
on the axis of the cone and three-fourths of the way from the vertex to the base.
13.6.33: Place the cube in the first octant with three of its faces in the coordinate planes, one vertex at
(0, 0, 0), and the opposite vertex at (a, a, a). With density δ = 1, its moment of inertia with respect to the
z-axis is
a
a
Iz =
z=0
y=0
a
a a
1 3
1 3
2
2
x + xy
a + ay
(x + y ) dx dy dz =
dy dz =
dy dz
3
3
x=0
0
0
0
0
0
a
a
a
1 3
2 4
1
2
2
a y + ay 3 dz =
a dz = a · a4 = a5 .
=
3
3
3
3
3
0
0
0
a
2
2
a
a
1741
Because the mass of the cube is m = a3 , we see that Iz = 23 ma2 , which is dimensionally correct.
13.6.34: The density at P (x, y, z) is δ(x, y, z) = k(x2 + y 2 + z 2 ) where k is a positive constant. It will
not change the answer to assume that k = 1. Then the mass and moments of the cube are
a a
a a a
1 3
2
2
2
2
2
a + ax + ay
(x + y + z ) dz dy dx =
dy dx
m=
3
0
0
0
0
0
a
a
a 1 3
1 a 4
1
(2a + 3a2 x2 ) dx =
=
a y + 3ax2 y + ay 3 dx =
2a4 x + a2 x3 = a5 ;
3 0
3
0 3
0
0
Myz =
a
0
a
=
0
1 3
a x + ax3 + axy 2 dy dx
3
0
0
0
0
a
a
a
1 3
1
1 4 2 3 2 4
7 6
3
3
4
2 3
a ;
dx =
(2a x + 3a x ) dx =
=
a xy + 3ax y + axy
a x + a x
3
3
3
4
12
0
0
0
a
Mxz = Mxy =
a
(x3 + xy 2 + xz 2 ) dz dy dx =
7 6
a
12
a
a
by symmetry.
Therefore the centroid of this cube is located at the point (x, y, z) =
7
7
7
a,
a,
a .
12 12 12
13.6.35: With density δ(x, y, z) = k(x2 + y 2 + z 2 ) at the point (x, y, z) (k is a positive constant), the
moment of inertia of the cube of Problem 34 with respect to the z-axis is
a a a
Iz =
k(x2 + y 2 + z 2 )(x2 + y 2 ) dz dy dx
0
0
a
0
a
=k
0
0
a
a
x4 z + 2x2 y 2 z + y 4 z +
=k
0
a
1 3 2
1
a x + ax4 + a3 y 2 + 2ax2 y 2 + ay 4
3
3
dy dx
0
dy dx
a
1 3 2
1
2
1
a x y + ax4 y + a3 y 3 + ax2 y 3 + ay 5 dx
3
9
3
5
0
0
a
a
14 6
14 6
1 4 3 1 2 5
38 7
4 2
2 4
a +a x +a x
a x+ a x + a x
ka .
=
=k
dx = k
45
45
3
5
45
0
0
a
0
1 2 3 1 2 3
x z + y z
3
3
=k
13.6.36: The density of the cube at the point (x, y, z) is δ(x, y, z) = kz where k is a positive constant.
Its mass and moments are
1 1
m=
Myz
Mxz
1
1
1
1
1
1
k dy dx = 1 · 1 · k = k;
2
2
2
0
0
0
0
0
1
1 1 1
1 1
1
1
1
kx dx dy =
k dy = 1 · k =
=
kxz dz dx dy =
4
0
0
0
0
0 2
0 4
1
1 1 1
1 1
1
1
1
ky dy dx =
k dx = 1 · k =
=
kyz dz dy dx =
2
4
4
0
0
0
0
0
0
kz dz dy dx =
1742
1
k;
4
1
k;
4
Mxy =
1
0
1
0
1
kz 2 dz dy dx =
0
1
0
Therefore its centroid is located at the point
1
0
1
1
1
k dy dx = 1 · 1 · k = k.
3
3
3
1 1 2
, ,
.
2 2 3
13.6.37: The moment of inertia of the cube of Problem 36 with respect to the z-axis is
1 1 1
1 1
1
k(x2 + y 2 ) dy dx
k(x2 + y 2 )z dz dy dx =
Iz =
2
0
0
0
0
0
1
1
1
1
1 3 1
1 2
1 2 1
1
1
kx y + ky 3 dx =
kx + k dx = k ·
x + x = k.
=
2
6
2
6
6
6
3
0
0
0
0
13.6.38: Assume that the sphere is centered at the origin, so that it consists of the points (x, y, z) for
which x2 + y 2 + z 2 a2 . Let D denote the circular disk x2 + y 2 a2 in the xy-plane. With constant
density δ, the moment of inertia of the sphere with respect to the z-axis is
√a2 −x2 −y2
2
2
Iz =
δ(x
+
y
)
dz
dA
=
2δ(x2 + y 2 ) a2 − x2 − y 2 dA.
√
a2 −x2 −y 2
z=−
D
D
Now rewrite the last double integral as an iterated integral in polar coordinates. The result:
a
2π a
2π 6r4 − 2a2 r2 − 4a4
3 2
2 1/2
2
2 1/2
Iz =
· (a − r )
2δr (a − r )
dr dθ =
dθ
δ·
15
θ=0 r=0
θ=0
r=0
2π
4 5
4 5
8
δa dθ = 2π ·
δa =
πδa5 .
=
15
15
15
θ=0
Because the mass of the sphere is m = 43 πδa3 , we see that Iz =
2
2
5 ma ,
which is dimensionally correct and
of plausible magnitude.
13.6.39: With constant density δ = 1, the mass and moments are
1
√
1−z 2
√
1−z 2
m=
1
√
1−z 2
1 dx dy dz =
z=0
1
=
y=0
z=0
x=0
(1 − z 2 ) dz = z −
z=0
1
√
1−z 2
√
1 3
z
3
1
=
0
y=0
2
;
3
1−z 2
Myz =
1
√
1−z 2
x dx dy dz =
z=0
1
=
z=0
1
y=0
z=0
x=0
y=0
1
(1 − z 2 ) dy dz
2
5z − 2z 3
3
1
(1 − z 2 )3/2 dz =
· (1 − z 2 )1/2 +
arcsin z
2
16
16
√
1−z 2
√
1−z 2
Mxz =
1
√
1−z 2
y dx dy dz =
z=0
1 − z 2 dy dz
y=0
z=0
x=0
1743
y=0
1
=
0
3
π;
32
y(1 − z 2 )1/2 dy dz
1
5z − 2z 3
1
3
3
2 3/2
2 1/2
(1 − z )
· (1 − z ) +
arcsin z =
π;
dz =
2
16
16
32
0
1
=
z=0
√
1
1−z 2
√
1−z 2
Mxy =
√
1−z 2
1
z dx dy dz =
z=0
1
y=0
(z − z 3 ) dz =
=
z=0
x=0
z=0
1 2 1 4
z − z
2
4
1
z(1 − z 2 )1/2 dy dz
y=0
1
.
4
=
0
Therefore the centroid of the solid is located at the point
9
9
3
π,
π,
.
64
64
8
13.6.40: Assuming constant density δ = 1, the moment of inertia of the solid of Problem 39 with respect
to the z-axis is
√
1
1−z 2
√
1−z 2
Iz =
z=0
y=0
√
1
1−z 2
y=0
1
=
z=0
1
=
0
(x2 + y 2 ) dx dy dz =
1
1
(1 − z 2 )3/2 + y 2 (1 − z 2 )1/2
3
1
1
y(1 − z 2 )3/2 + y 3 (1 − z 2 )1/2
3
3
√
z=0
x=0
=
z=0
1−z 2
y=0
1 3
x + xy 2
3
√1−z2
dy dz
x=0
dy dz
√1−z2
1
dz =
z=0
y=0
1
y(1 + y 2 − z 2 )(1 − z 2 )1/2
3
√1−z2
dz
y=0
1
2
2
16
(1 − z 2 )2 dz =
.
15z − 10z 3 + 3z 5 =
3
45
45
0
13.6.41: The given solid projects onto the circular disk D with radius 2 and center (0, 0) in the xy-plane.
Hence the volume of the solid is
12−x2 −2y 2
V =
1 dz
z=2x2 +y 2
D
(12 − 3x2 − 3y 2 ) dA.
dA =
D
Rewrite the last double integral as an iterated integral in polar coordinates. Thus
2π
V =
θ=0
2
3
(12 − 3r ) · r dr dθ = 2π · 6r − r4
4
r=0
2
2
2
0
= 24π ≈ 75.3982236861550377.
13.6.42: Note first that the two surfaces intersect in a curve that projects vertically onto the ellipse
x−1
2
2
+ z2 = 1
in the xz-plane. Hence the volume of the solid is
1
V =
z=−1
√
1+2 1−z 2
√
x=1−2 1−z 2
2x+3
1
1 dy dx dz =
y=x2 +4z 2
z=−1
1744
√
1+2 −z 2
√
x=1−2 1−z 2
(3 + 2x − x2 − 4z 2 ) dx dz
1
=
z=−1
1+2√1−z2
√
x=1−2 1−z 2
dz
2 1 3
1 + 2 1 − z2
−
3 1 + 2 1 − z 2 − 4z 2 1 + 2 1 − z 2 + 1 + 2 1 − z 2
3
−1
=
1
3x + x − x3 − 4xz 2
3
2
1
2 1 3 2
2
2
2
2
+ 4z 1 − 2 1 − z
− 1−2 1−z
1−2 1−z
−3 1−2 1−z
+
dz
3
1
=
−1
1
4
32
2 3/2
3
2
(1 − z )
(5z − 2z ) 1 − z + 4 arcsin z
dz =
= 4π ≈ 12.566370614359.
3
3
−1
13.6.43: Following the Suggestion, the volume is
1
√z2 −4y2
z/2
V =
z=0
y=−z/2
x=−
√
1
z/2
1 dx dy dz =
z 2 −4y 2
z=0
2 z 2 − 4y 2 dy dz.
y=−z/2
Let y = 12 z sin u, dy = 12 z cos u du. This substitution yields
1
π/2
V =
2
2
1
z cos u du dz =
z=0
=
0
1
u=−π/2
z=0
1 2
z (2u + sin 2u)
4
π/2
dz
u=−π/2
1
1 3
1 2
1
πz dz =
πz
= π ≈ 0.5235987755982989.
2
6
6
0
Methods of single-variable calculus also succeed here. A horizontal cross section of the solid at z = h
(0 < h 1) is an ellipse with equation x2 + 4y 2 = h2 . This ellipse has major semiaxis of length h and minor
1
2 h.
semiaxis of length
Therefore its area is
1
2
2 πh .
So, by the method of parallel cross sections (see Eq. (3)
of Section 6.2), the volume of the solid is
0
1
1
1 3
1 2
1
πh dh =
πh
= π.
2
6
6
0
13.6.44: First interchange the roles of z and x. Thus we are to find the volume of the region bounded by
the paraboloid z = 2x2 + y 2 and the parabolic cylinder z = 2 − y 2 . The two surfaces meet in a curve that
projects vertically onto the circle x2 + y 2 = 1, z = 0. Let D be the disk bounded by that circle. Then the
volume of the solid bounded by the two surfaces is
2−y 2
V =
1 dz
D
dA =
z=2x2 +y 2
(2 − 2x2 − 2y 2 ) dA.
D
Rewrite the last double integral as an iterated integral in polar coordinates. Thus
2π
V =
θ=0
1
1
1
(2 − 2r2 ) · r dr dθ = 2π r2 − r4 = 2π · = π ≈ 3.14159265358979323846.
2
2
r=0
0
1
1745
13.6.45: If the pyramid (tetrahedron) of Example 2 has density z at the point (x, y, z), then its mass and
moments are
2
(6−3x)/2 6−3x−2y
2
(6−3x)/2
z dz dy dx =
m=
x=0
y=0
2
0
2
0
2
0
(6−3x)/2
dx
0
1
3
9 2
3
2
3
2
(6 − 3x) − 9x(6 − 3x) + x(6 − 3x) + x (6 − 3x) dx
9(6 − 3x) − (6 − 3x) +
2
12
4
4
=
=
0
9
2
18y − 18xy + x2 y − 6y 2 + 3xy 2 + y 3
2
3
=
0
z=0
1
(6 − 3x − 2y)2 dy dx
2
2
9 4
27 2 9 3
27 2 9 3
x − x
x + x −
x
= 9;
18 − 27x +
dx = 18x −
2
4
2
2
16
0
2
(6−3x)/2 6−3x−2y
Myz =
2
(6−3x)/2
1
x(6 − 3x − 2y)2 dy dx
2
xz dz dy dx =
x=0
2
y=0
(6−3x)/2
=
0
0
2
0
2
0
2
0
(6−3x)/2
dx
0
1
3 2
9 3
3
2
3
2
2
x(6 − 3x) − 9x (6 − 3x) + x (6 − 3x) + x (6 − 3x) dx
9x(6 − 2x) − x(6 − 3x) +
2
12
4
4
=
=
0
9
18x − 18x2 + x3 − 12xy + 6x2 y + 2xy 2 dy dx
2
9
2
xy(x − 2)2 + 3xy 2 (x − 2) + xy 3
2
3
=
0
z=0
2
27 3 9 4
27 4
9 5
18
x − x
x −
x
;
=
18x − 27x2 +
dx = 9x2 − 9x3 +
2
4
8
20
5
0
2
(6−3x)/2 6−3x−2y
Mxz =
2
(6−3x)/2
1
y(6 − 3x − 2y)2 dy dx
2
yz dz dy dx =
x=0
2
y=0
(6−3x)/2
=
0
0
2
0
2
2
0
=
0
9
18y − 18xy + x2 y − 12y 2 + 6xy 2 + 2y 3 dy dx
2
2
27
81 2 27 3 27 4
− 27x +
x −
x +
x
2
4
4
32
2
(6−3x)/2 6−3x−2y
Mxy =
x=0
2
y=0
=
0
(6−3x)/2
dx
0
1
1
9
1
9 2
9
2
3
4
2
3
2
(6 − 3x) − (6 − 3x) +
(6 − 3x) − x(6 − 3x) + x(6 − 3x) +
x (6 − 3x) dx
4
2
32
4
4
16
=
0
9
1
9y − 9xy + x2 y 2 − 4y 3 + 2xy 3 + y 4
4
2
2
=
0
z=0
0
(6−3x)/2
z=0
z 2 dz dy dx =
27
27 2 27 3 27 4
27 5
x−
x +
x −
x +
x
dx =
2
2
4
16
160
0
2
(6−3x)/2
0
2
=
0
27
;
5
1
(6 − 3x − 2y)3 dy dx
3
8
72 − 108x + 54x2 − 9x3 − 72y + 72xy − 18x2 y + 24y 2 − 12xy 2 − y 3 dy dx
3
1746
2
=
0
2
72y − 108xy + 54x y − 9x y − 36y + 36xy − 9x y + 8y − 4xy − y 4
3
2
3
2
2
2 2
0
Therefore the centroid of the pyramid is located at the point
13.6.46: The mass and moments are
√ 2
2
y+2
2
−1
2
y+2
1 dx dz dy =
y2
−1
=
z−y
m=2
3
(6−3x)/2
dx
y=0
2
27 4
27 4 27 5
108
x
x +
x
.
=
54 − 108x + 81x2 − 27x3 +
dx = 54x − 54x2 + 27x3 −
8
4
40
5
0
2
=
=
3
y2
−1
0
2 3 12
, ,
.
5 5 5
2
2 z − y dz dy =
2
−1
4
(z − y 2 )3/2
3
y+2
dy
y2
4
(2 + y − y 2 )3/2 dy
3
2
2y − 1
81
81
1
(−43 + 78y + 24y 2 − 16y 3 )(2 + y − y 2 )1/2 +
arcsin
π;
=
48
32
3
32
−1
Myz = 2
2
y+2
√z−y2
2
y+2
x dx dz dy = 2
y2
−1
y2
−1
0
1
(z − y 2 ) dz dy = 2
2
2
−1
1 2 1 2
z − y z
4
2
y+2
dy
y2
2
1
1
1
1
1 5
81
3
1
y
;
=
= 2int2−1 1 + y − y 2 − y 3 + y 4 dy = 2 y + y 2 − y 3 − y 4 +
4
2
4
2
4
8
20
20
−1
Mxz = 2
2
2
−1
2
y+2
y2
−1
0
2y(z − y 2 )1/2 dz dy =
2
−1
4
y(z − y 2 )3/2
3
y+2
dy
y2
4
y(2 + y − y 2 )3/2 dy
3
2
2y − 1
61
63 2 11 3 1 5
81
81
727
2 1/2
−
y+
y +
y − y · (2 + y − y ) +
arcsin
π;
=
−
640 320
80
40
5
128
3
64
−1
4
=
3
2
y2
√z−y2
2
y2
z dx dz dy =
−1
2
−1
2
=
−1
y+2
=
√z−y2
y2
−1
y+2
y dx dz dy =
=
Mxy = 2
−1
0
4 2
4 2
8 4
z −
y z−
y
5
15
15
2z(z − y 2 )1/2 dz dy
y+2
2 1/2
· (z − y )
y+2
dy
y2
4
(12 + 12y + y 2 − y 3 − 2y 4 )(2 + y − y 2 )1/2 dy
15
2
2y − 1
427 2 23 3 1 4 1 5
567
4 7075 2303
−
y−
y −
y + y + y · (2 + y − y 2 )1/2 +
arcsin
= −
15 768
384
96
48
6
3
128
3
−1
=
567
π.
128
1747
Therefore the centroid of the parabolic segment is located at the point
13.6.47: First we compute
2 (6−3x)/2 6−3x−2y
2
8 1 7
, ,
.
5π 2 4
(6−3x)/2
1
(6 − 3x − 2y)2 dy dx
2
x=0 y=0
z=0
0
0
2 (6−3x)/2 9
=
18 − 18x + x2 − 12y + 6xy + 2y 2 dy dx
2
0
0
(6−3x)/2
2
2
9
=
dx
18y − 18xy + x2 y − 6y 2 + 3xy 2 + y 3
2
3
0
0
2
2
9 4
27 2 9 3
27 2 9 3
x − x
x + x −
x
= 9.
=
18 − 27x +
dx = 18x −
2
4
2
2
16
0
0
z dz dy dx =
The volume of the pyramid is 6, and hence the average value of the density function δ(x, y, z) = z on the
pyramid is δ =
9
6
= 32 .
13.6.48: The volume of the cube is 1, and hence the average value of f (x, y, z) = x2 + y 2 + z 2 on the cube
is
1
1
1
f=
0
0
1
=
0
0
(x2 + y 2 + z 2 ) dz dy dx
0
1
1
+ x2 + y 2
3
1
dy dx =
0
2
+ x2
3
dx =
2
1
x + x3
3
3
13.6.49: The centroid of the cube of Problem 48 is, by symmetry, its midpoint
1
= 1.
0
1
1 1
2, 2, 2
. Because the cube
has volume 1, the average value of
2 2 2
1
1
1
+ y−
+ z−
g(x, y, z) = x −
2
2
2
on the cube is
1 g=
1
1
1
1
3
1
1
z − xz + x2 z − yz + y 2 z − z 2 + z 3
g(x, y, z) dz dy dx =
4
2
3
x=0 y=0 z=0
0
0
1
1 1
1
7
7
1 2 1 3
2
2
2
−x+x −y+y
y − xy + x y − y + y
dx
=
dy dx =
12
12
2
3
0
0
0
0
1
1
5
5
1
1
1
=
− x + x2 dx =
x − x2 + x3 = .
12
12
2
3
4
0
0
1
dy dx
0
13.6.50: If the cube of Problem 48 has density δ(x, y, z) = x + y + z at the point (x, y, z), then—because
the volume of the cube is 1—the average value of the density function on the cube is
1 1 1
1 1
1
+ x + y dy dx
δ=
(x + y + z) dz dy dx =
2
0
0
0
0
0
1
1
1
1
1
3
=
y + xy + y 2 dx =
(1 + x) dx = .
2
2
2
0
0
0
1748
13.6.51: First we compute
2
(6−3x)/2
6−3x−2y
J=
x=0
2
y=0
0
(6−3x)/2
0
2
=
0
2
0
=
2
0
(6−3x)/2
x2 z + y 2 z +
0
1 3
z
3
6−3x−2y
dy dx
0
14 3
y
72 − 108x + 60x2 − 12x3 − 72y + 72xy − 20x2 y + 30y 2 − 15xy 2 −
dy dx
3
72y − 108xy + 60x2 y − 12x3 y − 36y 2 + 36xy 2 − 10x2 y 2 + 10y 3 − 5xy 3 −
=
z=0
=
(x2 + y 2 + z 2 ) dz dy dx =
135
441 2 171 3 207 4
− 135x +
x −
x +
x
2
4
4
32
dx =
7 4
y
6
(6−3x)/2
dx
0
135
135 2 147 3 171 4 207 5
x−
x +
x −
x +
x
2
2
4
16
160
2
0
147
.
5
Because the pyramid has volume V = 6, the average squared distance of its points from its centroid is
d=
147
49
J
=
=
= 4.9.
V
30
10
13.6.52: Suppose that the tetrahedron T has vertices at (0, 0, 0), (a, 0, 0), (0, b, 0), and (0, 0, c) where
a, b, and c are positive constants. Suppose also that its density is δ = 1. We plan first to find the centroid
of T . Its mass is simply m = 16 abc. The equation of its bounding diagonal plane is
x y
so that z = f (x, y) = c 1 − −
.
a
b
x y z
+ + = 1,
a
b
c
and therefore
a
b−bx/a
f (x,y)
Myz =
a
b−bx/a
x dz dy dx =
x=0
y=0
0
z=0
0
cx −
c 2 c x − xy dy dx
a
b
b−bx/a
a 2
2abcxy − 2bcx2 y − acxy 2
a bcx − 2abcx2 + bcx3
dx =
dx
2ab
2a2
0
0
0
2 2
a
6a bcx − 8abcx3 + 3bcx4
1 2
a bc.
=
=
24a2
24
0
a
=
and z = 14 c. It follows that the centroid of the pyramid of
Example 2, with unit density, is located at the point 12 , 34 , 32 . Let h(x, y, z) be the squared distance of the
Therefore x =
1
4 a.
By symmetry, y =
1
4b
point (x, y, z) of the pyramid from its centroid:
2 2 2
1
3
3
h(x, y, z) = x −
+ y−
+ z−
.
2
4
2
Then we compute
2
(6−3x)/2
J=
6−3x−2y
h(x, y, z) dz dy dx
0
0
0
1749
2
(6−3x)/2
=
0
0
2
(6−3x)/2
=
0
0
49
3
3
1
z − xz + x2 z − yz + y 2 z − z 2 + z 3
16
2
2
3
2
=
0
dy dx
0
291 1107
99 2
409
−
x+
x − 12x3 −
y
8
16
2
8
+
6−3x−2
121
14 3
xy − 20x2 y + 27y 2 − 15xy 2 −
y
2
3
dy dx
1107
99 2
409 2
291
y−
xy +
x y − 12x3 y −
y
8
16
2
16
121 2
7
xy − 10x2 y 2 + 9y 3 − 5xy 3 − y 4
4
6
2
441 1125
4833 2 585 3 207 4
−
x+
x −
x +
x
=
dx
16
16
64
16
32
0
(6−3x)/2
+
=
441
1125 2 1611 3 585 4 207 5
x−
x +
x −
x +
x
16
32
64
64
160
2
=
0
dx
0
441
.
40
Because the pyramid has volume V = 6, the average value of h(x, y, z) on the pyramid is
h=
147
J
=
= 1.8375.
V
80
13.6.53: Using Mathematica for each single-integration step, we find that the average distance of points of
the cube of Problem 48 from the origin is
1
1
1
d=
x=0
y=0
1
1
=
0
0
x2 + y 2 + z 2 dz dy dx
z=0
1
1 2
1 2
2
2
2
2
2
2
z x + y + z + (x + y ) ln z + x + y + z
dy dx
2
2
0
1
1
1 2
2
2
2
2
2
2
2
2
2
+ (x + y ) ln 1 + x + y + 1
=
x + y + 1 − (x + y ) ln
x +y
dy dx
2
2
2
0
0
1 1
1 2
y
1
2
2
y x + y + 1 − x arctan
x2 + y 2
− y(3x2 + y 2 ) ln
=
3
3
6
x x2 + y 2 + 1
0
1
1
1
1
1
2
2
2
2
2
2
2
+ y(3x + y ) ln 1 + x + y + 1 + (3x + 1) ln y + x + y + 1
dx
6
6
0
1 1
1
1
1
√
x2 + 2 − x3 arctan
x2 + 1
=
− (3x2 + 1) ln
3
3
3
x x2 + 2
0
1
2
2
+ (3x + 1) ln 1 + x + 2
dx
3
x
1
1
1
1 4
√
x x2 + 2 + arcsinh √
x arctan
−
=
4
3
12
2
x x2 + 2
1750
1
1
x
1
2
2
2
2
√
x + 1 + x(x + 1) ln 1 + x + 2
− x(x + 1) ln
3
3
x2 + 2
0
√
√
1
=
6 3 − π + 12 ln 2 + 3
≈ 0.960591956455052959425108.
24
1
− arctan
6
Section 13.7
13.7.1: The volume is
2π V =
θ=0
2
r=0
4
2
1
r dz dr dθ = 2π
(4r − r ) dr = 2π 2r − r4
4
2
z=r
r=0
3
2
2
= 8π.
0
13.7.2: We assume unit density. Then the mass of the solid is m = 8π by the result in the solution of
Problem 1. It is clear by symmetry that x = y = 0, but we will also demonstrate this rigorously:
2π 2 4
2π 2
2
Myz =
r cos θ dz dr dθ =
(4r2 cos θ − r4 cos θ) dr dθ
0
2π =
0
r2
0
4 3
1
r cos θ − r5 cos θ
3
5
0
2
0
2π
dθ =
0
0
64
cos θ dθ =
15
64
sin θ
15
2π
= 0.
0
Replacement of cos θ with sin θ in the first integral will clearly lead to the result Mxz = 0. There remains
only this:
Mxy =
2π
2
4
2π
2
rz dz dr dθ =
0
r2
0
0
0
2π =
0
1 2
rz
2
4r2 −
4
2π
dr dθ =
0
r2
1 6
r
12
2
2π
dθ =
0
0
0
2
1 5
8r − r
dr dθ
2
64
32
dθ =
π.
3
3
Therefore the centroid of the solid is located at 0, 0, 83 . Compare this answer with that obtained using the
result in Problem 31 of Section 13.6.
13.7.3: Place the center of the sphere at the origin. Then its volume is
2π
a
V =
θ=0
r=0
√
a
2 2
4
2 3/2
2
2
r dz dr dθ = 2π
2r a − r dr = 2π · − (a − r )
= πa3 .
3
3
r=0
0
a2 −r 2
√
z=− a2 −r 2
a
13.7.4: The moment of inertia of a solid sphere of density δ, radius a, and center (0, 0, 0) with respect to
the z-axis is
Iz =
0
2π
0
a
√
a2 −r 2
√
− a2 −r 2
3
δr dz dr dθ = 2πδ
0
a
2r3 (a2 − r2 )1/2 dr
a
2
8
2
πδa5 = ma2
= 2πδ · − (2a4 + a2 r2 − 3r4 )(a2 − r2 )1/2 =
15
15
5
0
1751
where m = 43 πδa3 is the mass of the sphere.
13.7.5: The volume is
2π
V =
θ=0
√
1
1
2
2r(4 − r2 )1/2 dr = 2π · − (4 − r2 )3/2
3
r=0
0
√
√
16
4
− 2 3 = π 8 − 3 3 ≈ 11.7447292674805137.
= 2π
3
3
4−r 2
√
z=− 4−r 2
r=0
1
r dz dr dθ = 2π
13.7.6: Assuming unit density, the result in the solution of Problem 5 shows that the mass of the solid is
m=
√ 2 π 8−3 3 .
3
By symmetry, the centroid lies on the z-axis, so we need only compute the moment with respect to the
xy-plane:
Mxy =
2π
1
√
4−r 2
1
1
7
7
1
(4r − r3 ) dr = 2π · r2 − r4 = 2π · = π.
2
8
8
4
0
1
rz dz dr dθ = 2π
0
0
0
0
Therefore the centroid is located at the point (x, y, z) where x = y = 0 and
z=
√ 21 3
7π
√ =
·
8 + 3 3 ≈ 0.9362135164758083.
4 2π 8 − 3 3
296
13.7.7: The mass of the cylinder is
2π
a
h
m=
a
rz dz dr dθ = 2π
θ=0
r=0
z=0
r=0
a
1 2 2
1 2
1
rh dr = 2π ·
r h
= πa2 h2 .
2
4
2
0
13.7.8: We saw in the solution of Problem 7 that the mass of the cylinder is m =
1
2 2
2 πa h .
By symmetry,
its centroid lies on the z-axis. So we need only compute its moment with respect to the xy-plane:
Mxy =
2π
a
h
2
a
rz dz dr dθ = 2π
0
0
0
0
a
1 2 3
1 3
1
rh dr = 2π ·
r h
= πa2 h3 .
3
6
3
0
2
Therefore its centroid is located at the point 0, 0, h .
3
13.7.9: The moment of inertia of the cylinder of Problem 7 with respect to the z-axis is
2π
a
h
Iz =
θ=0
r=0
a
= 2π
r=0
r3 z dz dr dθ = 2π
z=0
a
r=0
1 3 2
r z
2
h
dr
z=0
a
1 4 2
1 3 2
1
1
r h dr = 2π ·
r h
= 2π · a4 h2 = πa4 h2 .
2
8
8
4
r=0
1752
13.7.10: The cylinder x2 + y 2 − 2x = 0 meets the xy-plane in the circle with polar equation r = 2 cos θ
and the entire circle is swept out as θ runs through the values from − 12 π to
1
2 π.
(We will integrate from
0 to π/2 and double the result.) Hence the volume of the region within both the cylinder and the sphere
r2 + z 2 = 4 is
π/2
2 cos θ
V =2
0
π/2
0
√
4−r 2
√
− 4−r 2
0
=2
2
− (4 − r2 )3/2
3
π/2
r dz dr dθ = 2
0
2 cos θ
π/2
dθ = 2
0
0
0
2 cos θ
2r(4 − r2 )1/2 dr dθ
16
(1 − sin3 θ) dθ
3
π/2
32
1
1 − (1 − cos2 θ) sin θ dθ =
θ + cos θ − cos3 θ
3
3
0
0
1
32 π
16
−1+
(3π − 4) ≈ 9.6440497080344528.
=
=
3 2
3
9
=
32
3
π/2
This problem has a pitfall. Here are the details of the simplification in the second line of this solution:
2
− (4 − r2 )3/2
3
2 cos θ
=
16 2
16 2
− (4 − 4 cos2 θ) 4 − 4 cos2 θ =
− (4 sin2 θ) 4 sin2 θ
3
3
3
3
=
16 1 − (sin2 θ) sin2 θ .
3
0
If 0 θ 12 π, then sin θ 0, and hence the last expression can be replaced with
16 1 − sin3 θ .
3
But if − 12 π θ 0, then sin θ 0, so that
16 16 1 − (sin2 θ) sin2 θ =
1 + sin3 θ .
3
3
If this important detail is overlooked by a student who integrates from − 12 π to
the incorrect answer
16
3 π
1
2 π,
he or she will obtain
for the volume of the solid.
13.7.11: The volume is
2π
3
9−r 2
V =
θ=0
r=0
z=0
3
9 2 1 4
r − r
r dz dr dθ = 2π
(9r − r ) dr = 2π ·
2
4
r=0
3
3
=
0
81
π.
2
By symmetry, x = y = 0. The moment of the solid with respect to the xy-plane is
2π
3
9−r 2
3
1
r(9 − r2 )2 dr
2
θ=0 r=0 z=0
r=0
3
3 81
81 2 9 4
1
1 6
243
= 2π
r − 9r3 + r5 dr = 2π ·
r − r +
r
π.
=
2
2
4
4
12
2
r=0
0
Mxy =
rz dz dr dθ = 2π
1753
Therefore the z-coordinate of the centroid is z = 3. Suggestion:
Compare this answer with the answer
obtained by using the result in Problem 31 of Section 13.6.
13.7.12: The paraboloids meet in the circle x2 + y 2 = 4, z = 4. Therefore the volume between them is
2
2π 2 12−2r2
2
3 4
3
2
V =
r dz dr dθ = 2π
(12r − 3r ) dr = 2π · 6r − r
= 2π · 12 = 24π.
4
0
0
0
r2
0
We are to assume that the solid has unit density, so its mass is m = 24π as well. By symmetry, the centroid
lies on the z-axis. The moment of the solid with respect to the xy-plane is
2
2π 2 12−2r2
2
3
1
Mxy =
rz dz dr dθ = 2π
72r − 24r3 + r5 dr = 2π · 36r2 − 6r4 + r6
2
4
0
0
0
r2
0
= 2π · 64 = 128π.
16
Therefore the centroid of the solid is located at the point 0, 0,
.
3
13.7.13: The curve formed by the intersection of the paraboloids lies on the cylinder x2 + y 2 = 4, and
hence the solid projects vertically onto the disk D with boundary x2 + y 2 = 4 in the xy-plane. Therefore
the volume of the solid is
2π 2 12−r2 −r2 sin2 θ
r dz dr dθ =
V =
θ=0
z=r 2 +r 2 cos2 θ
r=0
2π
θ=0
2π
=
0
2
(12r − 3r3 ) dr dθ =
r=0
2π
θ=0
6r2 −
3 4
r
4
2
dθ
r=0
12 dθ = 2π · 12 = 24π ≈ 75.3982236861550377.
13.7.14: The paraboloid z = r2 and the plane z = 2r cos θ intersect in the cylinder r = 2 cos θ, which
projects vertically onto the circle with the same polar equation in the xy-plane. Note that this circle is swept
out as θ varies from − 12 π to
π/2
2 cos θ
Hence the volume of the solid between the paraboloid and the plane is
2r cos θ
V =
π/2
r dz dr dθ =
−π/2
π/2
=
−π/2
=
1
2 π.
0
r2
2 3
1
r cos θ − r4
3
4
−π/2
2 cos θ
π/2
dθ =
0
−π/2
0
2 cos θ
(2r2 cos θ − r3 ) dr dθ
π/2
4
1
4
cos θ dθ =
12θ + 8 sin 2θ + sin 4θ
3
24
−π/2
1
π ≈ 1.57079632679489661923.
2
13.7.15: The spherical surface r2 + z 2 = 2 and the paraboloid z = r2 meet in a horizontal circle that
projects vertically onto the circle x2 + y 2 = 1 in the xy-plane. Hence the volume between the two surfaces is
1
2π 1 √2−r2
2π 1
1 4
1
2 1/2
3
2 3/2
V =
r dz dr dθ =
r(2 − r ) − r dr dθ = 2π · − (2 − r ) − r
3
4
θ=0 r=0 z=r 2
0
0
0
2√
7
1 √
= 2π ·
2 −
= π 8 2 − 7 ≈ 2.2586524883563962.
3
12
6
1754
13.7.16: Choose a coordinate system in which the points of the cylinder are described by
0 r a,
0 θ 2π,
0zh
and denote the (constant) density of the homogeneous cylinder by δ. Then the mass of the cylinder will be
m = πδa2 h. So its moment of inertia with respect to its axis of symmetry—the z-axis—is
Iz =
2π
0
a
0
h
0
1 4
r h
= 2πδ ·
4
δr3 dz dr dθ =
0
a
dθ =
0
2π
r
δr3 h dr dθ
0
1
1
1
πδa4 h = a2 · πδa2 h = ma2 .
2
2
2
13.7.17: Set up a coordinate system in which the points of the cylinder are described by
0 r a,
0 θ 2π,
0 z h.
Because the cylinder has constant density δ, its mass is m = πδa2 h. One diameter of its base coincides with
the x-axis, so we will find the moment of inertia I of the cylinder with respect to that axis. The square of
the distance of the point (x, y, z) from the x-axis is y 2 + z 2 = z 2 + r2 sin2 θ. Therefore
2π
a
h
2π
a
1 3
h r + hr3 sin2 θ
3
θ=0 r=0 z=0
0
0
a
2π 2π 1 3 2 1 4 2
1 2 3 1 4
h r + hr sin θ dθ = δ
a h + a h sin2 θ dθ
=δ
6
4
6
4
0
0
0
I = Ix =
=δ
δ(z 2 + r2 sin2 θ) · r dz dr dθ = δ
6a4 hθ + 8a2 h3 θ − 3a4 h sin 2θ
48
2π
=
0
dr dθ
1
1
δπa2 h(3a2 + 4h2 ) =
m(3a2 + 4h2 )
12
12
where m is the mass of the cylinder.
13.7.18: By symmetry, the centroid lies on the axis of the cylinder midway between its two bases. This is
so obvious that the intent of this problem must be to verify this by actually computing the integrals. Assume
that the cylinder has unit density and is described in cylindrical coordinates by
0 r a,
0 θ 2π,
0 z h.
Then its mass and moments are
2π
a
a
1 2
a h = πa2 h;
2
0
0
0
0
2π a h
2π a
=
r2 cos θ dz dr dθ =
r2 h cos θ dr dθ =
m=
Myz
h
r dz dr dθ = 2π
0
0
0
rh dr = 2π ·
0
1755
0
0
2π
1 3
a h cos θ dθ
3
1 3
a h sin θ
=
3
Mxz =
2π
a
2π
= 0;
0
h
2
2π
a
r sin θ dz dr dθ =
0
0
0
1
= − a3 h cos θ
3
Mxy =
2π
a
0
0
0
1 3
a h sin θ dθ
3
2π
= 0;
0
h
a
rz dz dr dθ = 2π
0
2π
r h sin θ dr dθ =
0
2
0
0
a
1 2 2
1 2
1
rh dr = 2π ·
r h
= πa2 h2 .
2
4
2
0
Therefore the centroid of the cylinder is located at the point
1
0, 0, h ,
2
on its axis of symmetry and midway between its two bases.
13.7.19: The volume is
2π
1
1
V =
r dz dr dθ = 2π
θ=0
r=0
0
z=r
1
1 2 1 3
r − r
(r − r ) dr = 2π ·
2
3
2
1
=
0
1
π.
3
13.7.20: Suppose that the cone has base radius R and height H. Set up a coordinate system in which its
vertex is at the origin, its axis lies on the nonnegative z-axis, its base (at the top) is part of the horizontal
plane z = H, and its curved side has cylindrical description
z=
H
r,
R
0 r R,
0 θ 2π.
Assume that the cone has constant unit density. Then its mass and moments are
2π
R
r dz dr dθ = 2π
0
0
Mxz
0
Hr/R
H 2
H 3
r −
r
2
3R
R
H
Hr − r2 dr
R
R
1 2
1
R H = πR2 H;
6
3
0
2π R H
2π R H 3
2
2
=
r cos θ dz dr dθ =
Hr cos θ − r cos θ dr dθ
R
0
0
Hr/R
0
0
= 2π
Myz
H
m=
= 2π ·
2π
1 2
1 2
R H cos θ dθ =
R H sin θ
=
dθ =
= 0;
12
12
0
0
0
0
2π R H
2π R H 3
2
2
=
r sin θ dz dr dθ =
Hr sin θ − r sin θ dr dθ
R
0
0
Hr/R
0
0
2π
2π
=
0
H 3
H 4
r cos θ −
r cos θ
3
4R
H 3
H 4
r sin θ −
r sin θ
3
4R
R
R
2π
2π
dθ =
0
0
1756
2π
1 2
1
R H sin θ dθ = − R2 H cos θ
= 0;
12
12
0
Mxy =
2π
R
H
R
rz dz dr dθ = 2π
0
0
0
Hr/R
H2 2
H2 4
r −
r
= 2π ·
4
8R2
R
0
H2
H2 3
r−
r
2
2R2
dr
1
1
= 2π R2 H 2 = πR2 H 2 .
8
4
Therefore the centroid is located at the point
3
0, 0, H ,
4
on the axis of the cone three-quarters of the way from its vertex to its base. Compare this with the solution
of Problem 32 in Section 13.6.
13.7.21: Without loss of generality we may assume that the hemispherical solid has density δ = 1. Choose
a coordinate system in which the solid is bounded above by the spherical surface ρ = a and below by the
xy-plane. Then its mass and moments are
2π
π/2
a
ρ2 sin φ dρ dφ dθ = 2π
m=
θ=0
φ=0
ρ=0
2π
π/2
a
2π
π/2
a
π/2
φ=0
π/2
1 3
2
2
a sin φ dφ = πa3 − cos φ
= πa3 ;
3
3
3
0
2π
π/2
2π
π/2
π/2
1 4 2
a sin φ cos θ dφ dθ
θ=0 φ=0 ρ=0
θ=0 φ=0 4
2π
2π π/2
π/2
1 2 2
1 2 2
a sin φ dφ cos θ dθ =
a sin φ dφ · sin θ
= 0;
=
θ=0
φ=0 4
φ=0 4
θ=0
ρ3 sin2 φ cos θ dρ dφ dθ =
Myz =
1 4 2
a sin φ sin θ dφ dθ
θ=0 φ=0 ρ=0
θ=0 φ=0 4
2π
2π π/2
π/2
1 2 2
1 2 2
a sin φ dφ sin θ dθ =
a sin φ dφ · − cos θ
= 0;
=
θ=0
φ=0 4
φ=0 4
θ=0
Mxz =
ρ sin φ sin θ dρ dφ dθ =
2π
π/2
a
ρ3 sin φ cos φ dρ dφ dθ = 2π
Myx =
θ=0
φ=0
ρ=0
=
2
3
1 4 1
πa
sin2 φ
2
2
φ=0
π/2
=
0
1 4
a sin φ cos φ dφ
4
1 4
πa .
4
3
Therefore the centroid of the hemispherical solid is located at the point 0, 0, a .
8
13.7.22: We are given the information that the hemispherical solid has density δ = kz at the point (x, y, z).
Then its mass and moments are
2π
π/2
a
m=
θ=0
φ=0
kρ3 sin φ cos φ dρ dφ dθ = 2π
ρ=0
π/2
φ=0
1757
1 4
ka sin φ cos φ dφ
4
π/2
1
1
4
2
= kπa sin φ
= kπa4 ;
4
4
0
2π
π/2
a
Myz =
2
4
2π
kρ sin φ cos φ cos θ dφ dθ =
θ=0
φ=0
ρ=0
1
ka5 sin θ
15
=
2π
π/2
θ=0
dθ
φ=0
= 0;
θ=0
a
2
4
2π
kρ sin φ cos φ sin θ dφ dθ =
φ=0
π/2
2π
Myz =
θ=0
1
ka5 sin3 φ cos θ
15
ρ=0
θ=0
1
ka5 sin3 φ sin θ
15
π/2
dθ
φ=0
2π
1
= − ka5 cos θ
= 0;
15
θ=0
2π
π/2
a
Mxy =
4
2
2π
π/2
kρ sin φ cos φ dρ dφ dθ =
θ=0
φ=0
= 2π · −
ρ=0
θ=0
1
ka5 cos3 φ
15
π/2
=
φ=0
φ=0
1 5
ka sin φ cos2 φ dφ dθ
5
2
kπa5 .
15
Therefore the centroid of this hemispherical solid is located at the point
13.7.23:
0, 0,
8
a .
15
The plane z = 1 has the spherical-coordinates equation ρ = sec φ; the cone with cylindrical-
coordinates equation r = z has spherical-coordinates equation φ =
1
4 π.
Hence the volume bounded by the
plane and the cone is
2π
π/4
sec φ
V =
θ=0
φ=0
ρ2 sin φ dρ dφ dθ = 2π
ρ=0
π/4
φ=0
π/4
1
1
1
sec2 φ tan φ dφ = 2π ·
sec2 φ
= π.
3
6
3
φ=0
13.7.24: Without loss of generality the cone has density δ = 1. Assume that its vertex is at the origin
and that its axis lies on the nonnegative z-axis. Assume that its curved side has spherical-coordinates
equation φ = α (where 0 < α <
1
2 π)
and that its base lies on the plane z = a > 0; thus its base has
spherical-coordinates equation ρ = a sec φ. Then its mass and moments are
2π
α
a sec φ
m=
2
α
ρ sin φ dρ dφ dθ = 2π
0
= 2π
2π
0
0
1 3 2
1
a sec α − a3
6
6
α
a sec φ
0
=
1 3
a sec2 φ tan φ dφ
3
1 3
πa tan2 α;
3
2π
α
1 4
a sec2 φ tan2 φ cos θ dφ dθ
4
0
0
0
0
0
2π
α
2π α
1
1
sec2 φ tan2 φ dφ cos θ dθ = a4
sec2 φ tan2 φ dφ · sin θ
= 0;
= a4
4
4
0
0
0
0
Myz =
ρ3 sin2 φ cos θ dρ dφ dθ =
1758
Mxz =
α
a sec φ
2
3
α
a sec φ
ρ sin φ dρ dφ sin θ dθ =
0
Myz =
2π
0
2π
α
0
0
a sec φ
3
α
= 2π
0
0
1 4 2
1
a sec α − a4
8
8
0
=
2
0
1 4 3
a sec φ sin φ dφ
4
ρ sin φ cos φ dρ dφ dθ = 2π
0
0
2π
ρ sin φ dρ dφ · − cos θ
= 0;
3
1 4
πa tan2 α.
4
Therefore the centroid of the cone is on its axis of symmetry and three-quarters of the way from the vertex
to the base, because it is located at the point
3
(x, y, z) = 0, 0, a .
4
13.7.25:
Assume unit density. Then the mass and the volume are numerically the same; they and the
moments are
2π
π/4
a
m=V =
0
=
0
0
π/4
1 3
a − cos φ
3
0
ρ2 sin φ dρ dφ dθ = 2π ·
√ 2 2
1 1√
πa 1 −
2 = π 2 − 2 a3 ;
3
2
3
Myz =
2π
π/4
a
2
3
π/4
a
ρ sin φ cos θ dρ dφ dθ =
0
0
Mxz =
2π
0
π/4
2π
a
ρ3 sin2 φ sin θ dρ dφ dθ =
0
0
0
0
Mxy =
0
π/4
0
0
π/4
a
0
0
0
a
2π
ρ sin φ dρ dφ · sin θ
= 0;
3
2
0
2π
3
2
ρ sin φ dρ dφ · − cos θ
= 0;
0
π/4
1 4 1
1
2
sin φ
ρ sin φ cos φ dρ dφ dθ = 2π · a ·
= πa4 .
4
2
8
0
3
So the z-coordinate of the centroid is
√ √ 3 2+ 2 a
3 3πa4
3a
√ √ =
=
2 + 2 a;
z=
= 16
16
8π 2 − 2 a3
8 2− 2
clearly x = y = 0. For a plausibility check, note that z ≈ (0.6401650429449553)a.
13.7.26: Assume that the solid of Problem 25 has constant density δ. Then its moment of inertia with
respect to the z-axis is
Iz =
2π
π/4
a
4
3
0
0
0
1
= 2πδa ·
cos 3φ − 9 cos φ
60
5
π/4
0
π/4
1 5 3
a sin φ dφ
5
0
√
√ 2 5
2 5
1
a −
a
πδ 8 − 5 2 a5 .
= 2πδ
=
15
12
30
δρ sin φ dρ dφ dθ = 2πδ
1759
13.7.27:
Set up a coordinate system so that the center of the sphere is at the point with Cartesian
coordinates (a, 0, 0). Then its Cartesian equation is
(x − a)2 + y 2 + z 2 = a2 ;
x2 − 2ax + y 2 + z 2 = 0;
and thus it has spherical-coordinates equation ρ = 2a sin φ cos θ (but note that θ ranges from − 12 π to
1
2 π ).
We plan to find its moment of inertia with respect to the z-axis, and the square of the distance of a point of
the sphere from the z-axis is
x2 + y 2 = ρ2 sin2 φ cos2 θ + ρ2 sin2 φ sin2 θ = ρ2 sin2 φ.
Moreover, if the sphere has mass m and constant density δ, then we also have m = 43 πa3 δ. Finally,
Iz =
=
=
π/2
−π/2
π
0
64 5
δa
5
0
128 5
δa ·
5
2a sin φ cos θ
δρ2 sin2 φ dρ dφ dθ = 2δ
0
π/2
0
π/2 π
sin8 φ cos5 θ dφ dθ =
0
0
π/2
sin8 φ dφ ·
π/2
128 5
δa
5
0
π
1
(2a sin φ cos θ)5 sin3 φ dφ dθ
5
0
π/2 π/2
sin8 φ cos5 θ dφ dθ
0
cos5 θ dθ .
0
Then Formula (113) from the long table of integrals (see the endpapers) yields
Iz =
28
128 5 1 3 5 7 π 2 4
4
7
7
δa · · · · · · · =
πδa5 = πδa3 · a2 = ma2 .
5
2 4 6 8 2 3 5
15
3
5
5
13.7.28: We will find its moment of inertia with respect to the z-axis (which contains a diameter) using
density δ = ρ2 at the point (ρ, φ, θ):
Iz =
2π
0
π
0
2a
π
127 7 3
δa sin φ dφ
7
0
π
127 7
1016 7
=
πa cos 3φ − 9 cos φ =
πa .
42
21
0
δρ6 sin3 φ dρ dφ dθ = 2πδ
a
13.7.29: The surface with spherical-coordinates equation ρ = 2a sin φ is generated as follows. Draw the
circle in the xz-plane with center (a, 0) and radius a. Rotate this circle around the z-axis. This generates
the surface with the given equation. It is called a pinched torus—a doughnut with an infinitesimal hole. Its
volume is
2π
π
2a sin φ
V =
θ=0
=
φ=0
16 3
πa · 2
3
ρ2 sin φ dρ dφ dθ = 2π
ρ=0
π/2
φ=0
π
φ=0
sin4 φ dφ =
1
(2a sin φ)3 sin φ dφ
3
32 3 1 3 π
πa · · · = 2π 2 a3 .
3
2 4 2
1760
We evaluated the last integral with the aid of Formula (113) from the long table of integrals (see the
endpapers). The volume of the pinched torus is also easy to evaluate using the first theorem of Pappus
(Section 13.5).
13.7.30: Draw the cardioid with polar equation r = 1 + sin θ, then replace the y-axis with the z-axis. Such
a cardioid is shown in the following figure.
z
2
1
-1
x
1
To generate the surface with spherical-coordinates equation ρ = 1 + cos φ, rotate this cardioid around the
x-axis. The resulting surface resembles an inverted apple. The volume that it bounds is
π
2π π 1+cos φ
1
2
(1 + cos φ)3 sin φ dφ
ρ sin φ dρ dφ dθ = 2π
V =
0
0
0
0 3
π
1
8
4
4
= 2π − (1 + cos φ)
= 2π · = π.
12
3
3
0
13.7.31: Assuming constant density δ, we have
2π a √4a2 −r2
δ(r2 sin2 θ + z 2 ) · r dz dr dθ
Ix = 2
0
0
2π
0
a
= 2δ
0
0
2π
= 2δ
0
1
r(4a2 − r2 )3/2 + r3 (4a2 − r2 )1/2 sin2 θ
3
dr dθ
a
1
dθ
(4a2 − r2 )3/2 (8a2 + 3r2 ) cos 2θ − 16a2 − r2
30
r=0
√
√
1 5
a 128 − 51 3 + 33 3 − 64 cos 2θ dθ
30
0
2π
√
√ √ 2 1 5 128 − 5 3 · 2θ + 33 3 − 64 sin 2θ
δa
128 − 51 3 δπa5 .
=
=
30
15
0
2π
= 2δ
13.7.32: Assuming constant density δ, we have
2π π/6 2a cos φ
δρ4 sin3 φ dρ dφ dθ = 2πδ
Iz =
0
0
0
0
1761
π/6
1
(2a cos φ)5 sin3 φ dφ
5
64
πδa5
=
5
π/6
64
πδa5
cos φ sin φ dφ =
5
3
5
0
π/6
0
(cos5 φ − cos7 φ) sin φ dφ
4
3
π/6
3
3
64
1
64
1
1 1
5 1
8
6
5 1
πδa
cos φ − cos φ
πδa
·
=
=
− ·
− +
5
8
6
5
8
4
6
4
8 6
0
67
64
67
πδa5 ·
=
πδa5 .
5
6144
480
=
13.7.33: If the density at (x, y, z) of the ice-cream cone is z, then its mass and moments are
2π
π/6
2a cos φ
ρ3 sin φ cos φ dρ dφ dθ = 2π
m=
0
0
0
π/6
4a4 sin φ cos5 φ dφ
0
π/6
37 4
4
πa ;
=
= − πa4 cos6 φ
3
48
0
Myz =
2π
π/6
2a cos φ
4
2
ρ sin φ cos φ dρ dφ cos θ dθ
0
0
π/6
0
2a cos φ
=
0
Mxz = 0
2
0
(by a similar computation);
Mxy =
0
2π
ρ sin φ cos φ dρ dφ · sin θ
= 0;
4
2π
π/6
2a cos φ
4
2
π/6
ρ sin φ cos φ dρ dφ dθ = 2π
0
0
0
0
4
= 2π · − a5 cos8 φ
5
π/6
=
0
32 5
a cos7 φ sin φ dφ
5
35 5
πa .
32
105
a .
Hence the centroid is located at the point 0, 0,
74
13.7.34: The moment of inertia of the ice-cream cone of Problem 33 with respect to the z-axis is
Iz =
2π
0
π/6
0
2a cos φ
0
2
2
(ρ cos φ) · (ρ sin φ) · ρ sin φ dρ dφ dθ = 2π
=
1
πa6 cos 10φ + 5 cos 8φ + 5 cos 6φ − 20 cos 4φ − 70 cos 2φ
240
π/6
0
π/6
=
0
32 6
a cos7 φ sin3 φ dφ
3
47
πa6 .
240
13.7.35: The similar star with uniform density k has mass m2 = 43 kπa3 . The other star has mass
m1 =
2π
0
0
= 2π
0
π
π
0
a
k 1−
ρ 2 a
2
· ρ sin φ dρ dφ dθ = 2π
0
π
π
2
4
8
ka3 sin φ dφ = − kπa3 cos φ
kπa3 .
=
15
15
15
0
1762
5ka2 ρ3 sin φ − 3kρ5 sin φ
15a2
a
dφ
0
Finally,
m1
2
= .
m2
5
13.7.36: The moment of inertia of the first star of Problem 35 with respect to its diameter that lies on the
z-axis is
Iz =
2π
0
π
0
a
0
a
π
ρ 2 7ka2 ρ5 sin3 φ − 5kρ7 sin3 φ
k 1−
dφ
· ρ4 sin3 φ dρ dφ dθ = 2π
a
35a2
0
0
π
1
16
2
ka5 sin3 φ dφ =
πka5 cos 3φ − 9 cos φ =
πka5 .
35
105
105
0
π
= 2π
0
13.7.37: The given triple integral takes the following form:
2π
0
π
0
a
0
ρ2 exp −ρ3 sin φ dρ dφ dθ = 2π
= 2π
0
π
π
0
1
− exp −ρ3 sin φ
3
a
dφ
0
1
4 1 − exp −a3 sin φ dφ = π 1 − exp −a3 .
3
3
Clearly the value of the integral approaches
4
π as a → +∞.
3
13.7.38: If the given integral is evaluated over the ball B of Problem 37, the result is
2π
0
π
0
a
3
ρ exp −ρ
0
2
sin φ dρ dφ dθ = 2π
0
π
1
− (ρ2 + 1) exp −ρ2 sin φ
2
a
dφ
0
π
2 2
2
1
2
= 2π
sin φ dφ = π (a + 1) exp −a − 1 cos φ
1 − (a + 1) exp −a
0 2
0
= 2π 1 − (a2 + 1) exp −a2 .
π
Now let a → +∞ to see that the value of the integral given in the statement of this problem is 2π.
13.7.39: Let V =
4
3
3 πa ,
the volume of the ball. The average distance of points of such a ball from its
center is then
d=
1
V
0
2π
0
π
0
a
ρ3 sin φ dρ dφ dθ =
2π
V
0
π
π
3
1 4
πa4
πa4
a sin φ dφ =
= a.
=
− cos φ
4
2V
V
4
0
Note that the answer is both plausible and dimensionally correct.
13.7.40: The key to solving such problems is to keep the integrand as simple as possible. To do so we
set up a coordinate system in which the ball is centered at the point with Cartesian coordinates (0, 0, a).
Thus the bounding spherical surface has spherical-coordinates equation ρ = 2a cos φ where 0 φ 12 π and
1763
0 θ 2π. Let V = 43 πa3 , the volume of the ball. The average distance of points of the ball from its south
pole (at the origin) is then
d=
=
1
V
2π
V
2π
0
π/2
0
π/2
2a cos φ
ρ3 sin φ dρ dφ dθ
0
4a4 cos4 φ sin φ dφ =
0
π/2
2π
2π 4 4
6
4
· a = a.
=
− a4 cos5 φ
V
5
V 5
5
0
As in the solution of Problem 39, note that the answer is both plausible and dimensionally correct.
13.7.41: A Mathematica solution:
m = Integrate[ Integrate[ delta∗a∧2∗Sin[ phi ], { phi, 0, Pi } ],
{ theta, 0, 2∗Pi } ]
4πa2 δ
r = a∗Sin[ phi ];
I0 = Integrate[ Integrate[ r∧2∗delta∗a∧2∗Sin[ phi ], { phi, 0, Pi } ],
{ theta, 0, 2∗Pi } ]
8 4
πa δ
3
I0/m
2 2
a
3
Second solution, by hand: The spherical surface S of radius a is described by ρ = a, 0 φ π,
0 θ 2π. Hence its moment of inertia with respect to the z-axis is
Iz =
δ(x2 + y 2 ) dA =
θ=0
S
= 2πδa
2π
4
0
π
δa4 sin3 φ dφ dθ
φ=0
π
2
(1 − cos φ) sin φ dφ = 2πδa
4
1
cos3 φ − cos φ
3
π
=
0
13.7.42: A Mathematica solution:
m = Integrate[ Integrate[ Integrate[ delta∗rho∧2∗Sin[ phi ],
{ rho, a, b } ], { phi, 0, Pi } ], { theta, 0, 2∗Pi } ]
4
πδ(b3 − a3 )
3
r = rho∗Sin[ phi ];
1764
2 2
2
a · 4πδa2 = ma2 .
3
3
I0 = Integrate[ Integrate[ Integrate[ delta∗r∧2∗delta∗rho∧2∗Sin[ phi ],
{ rho, a, b } ], { phi, 0, Pi } ], { theta, 0, 2∗Pi } ]
8
πδ(b5 − a5 )
15
Simplify[ I0/m ]
2 b 5 − a5
·
5 b 3 − a3
Second solution, by hand: Let r =
x2 + y 2 denote the usual radial polar coordinate. Then the
moment of inertia of the solid with respect to the z-axis is
2π
π
b
Iz =
π
b
δr ρ sin φ dρ dφ dθ = 2πδ
θ=0
= 2πδ
φ=0
0
ρ=a
b5 − a
5
The mass of the shell S is m =
5
π
sin3 φ dφ = 2πδ
0
4
3
3 πδ(b
b5 − a
5
ρ4 sin3 φ dρ dφ
a
5
·
8
4
=
πδ(b5 − a5 ).
3
15
− a3 ), and therefore
2
b 5 − a5
4
2(b5 − a5 )
2
πδ(b3 − a3 ) ·
=
m
·
= mc2
3
3
3
3
3
5(b − a )
5
b −a
5
Iz =
where c2 =
2 2
b 5 − a5
.
b 3 − a3
13.7.43: A Mathematica solution:
z = Sqrt[ b∧2 − a∧2 ];
m = 2∗Integrate[ Integrate[ delta∗r∗z, { r, a, b } ], { theta, 0, 2∗Pi } ]
4
πδ(b2 − a2 )3/2
3
I0 = 2∗Integrate[ Integrate[ delta∗r∧3∗z, { r, a, b } ], { theta, 0, 2∗Pi } ]
4
πδ(2b4 + b2 a2 − 3a4 ) b2 − a2
15
Simplify[ I0/m ]
1
(3a2 + 2b2 )
5
Second solution, by hand: Choose a coordinate system so that the z-axis is the axis of symmetry of the
sphere-with-hole. The central cross section of the solid in the xz-plane is bounded by the circle with polar
(or cylindrical coordinates) equation r2 + z 2 = b2 . Hence the mass of the solid is
2π
b
√
b2 −r 2
m = 2
δr dz dr dθ = 4πδ
θ=0
r=a
a
z=0
1765
b
r(b2 − r2 )1/2 dr
= 4πδ
b
1
− (b2 − r2 )3/2
3
4
πδ(b2 − a2 )3/2 .
3
=
a
The moment of inertia of this solid with respect to the z-axis is
2π
b
√
b2 −r 2
Iz = 2
3
δr dz dr dθ = 4πδ
θ=0
r=a
b
r3 (b2 − r2 )1/2 dr.
a
z=0
Integration by parts with u = r2 , dv = (b2 − r2 )1/2 dr, so that
du = 2r dr
1
v = − (b2 − r2 )3/2 ,
3
and
then yields
Iz = 4πδ
= 4πδ
1
− r2 (b2 − r2 )3/2
3
b
a
2
+
3
b
r(b2 − r2 )3/2 dr
a
b 2
1 2
1 2 2
2 3/2
2 5/2
a (b − a )
+
− (b − r )
3
3
5
a
4
4
2 2
2 3/2
2
2
2
2 3/2
πδ 5a (b − a ) + 2(b − a )(b − a )
πδ(b2 − a2 )3/2 (5a2 + 2b2 − 2a2 )
=
=
15
15
=
1
1
4
πδ(b2 − a2 )3/2 · (3a2 + 2b2 ) = m(3a2 + 2b2 ).
3
5
5
13.7.44: As we examine Fig. 13.7.15(b), with the positive x-axis to the east and the positive y-axis to the
north, we see that one-sixteenth of the solid is bounded below by the region R in the xy-plane described in
polar (cylindrical) coordinates by
0 r 1,
0 θ π
4
and above by part of the cylindrical surface with equation x2 + z 2 = 1 (not y 2 + z 2 = 1; over R, that
surface is higher). We get the total volume V of the solid of Problem 44 by multiplying by 16:
π/4
1
√
1−r 2 cos2 θ
π/4
1
r dz dr dθ = 16
V = 16
θ=0
π/4
= 16
0
r=0
r
0
z=0
(1 − r2 cos2 θ)3/2
−
3 cos2 θ
3
1
dθ =
r=0
16
3
0
0
π/4
1 − r2 cos2 θ dr dθ
1
(1 − cos2 θ)3/2
−
2
cos θ
cos2 θ
dθ
1 − (1 − cos2 θ) sin θ
dθ
cos2 θ
0
0
16 π4 1 + cos2 θ sin θ − sin θ
16 π/4
=
dθ =
(sec2 θ + sin θ − sec θ tan θ) dθ
3 0
cos2 θ
3 0
π/4
√ 16
16
3√
=
tan θ − cos θ − sec θ
=
2
= 8 2 − 2 ≈ 4.68629150101523961.
3−
3
3
2
=
16
3
π/4
16
1 − sin θ
dθ =
2
cos θ
3
π/4
0
1766
Replacement of (1 − cos2 θ)3/2 with sin3 θ in the third line is permitted because sin θ is nonnegative on R.
(Thus we have avoided the “pitfall” indicated in the solution of Problem 10.) The plausibility of the answer
is enhanced by the observation that a sphere of radius 1 will fit snugly within the boundary of the region
bounded by all three cylinders, and the volume of such a sphere is approximately 4.18879020478639098.
Second solution: Mathematica presents the last antiderivative in such a way as to produce a seemingly
improper integral, so care must be taken in the computer algebra solution. We write t for θ here, as usual;
recall also that the symbol “ % ” refers to the last output.
16∗Integrate[ r, { z, 0, Sqrt[ 1 − (r∗Cos[t])∧2 ] } ]
16r 1 − r2 cos2 t
Integrate[ %, r ]
1√ 2 − r2 − r2 cos 2t ·
2 r2 − sec2 t
8
3
(% /. r → 1) − (% /. r → 0)
√
1√ 16
sec2 t + 8 1 − cos 2t ·
2 1 − sec2 t
3
3
Integrate[ %, t ]
√
1√
16
tan t −
1 − cos 2t ·
2 (16 cot t + 8 tan t)
3
3
Limit[ %, t → Pi/4 ] − Limit[ %, t → 0 ]
√
√
32
16
−8 2 +
= 16 − 8 2
3
3
N[ %, 40 ]
4.68629150101523960958649020632241537144
13.7.45:
First we let g(φ, θ) = 6 + 3 cos 3θ sin 5φ, so that the boundary of the bumpy sphere B has
spherical equation ρ = g(φ, θ). Then the volume V of B is simply
2π
π
g(φ,θ)
V =
θ=0
φ=0
ρ2 sin φ dρ dφ dθ.
ρ=0
To evaluate V with the aid of Mathematica, we write r in place of ρ, f instead of φ, and t for θ, as usual.
Then, to find the volume V of B step-by-step, we proceed as follows:
Integrate[ r∗r∗Sin[f], r ]
1 3
ρ sin φ
3
1767
% /. r → g[f,t]
1
(sin φ)(6 + 3 cos 3θ sin 5φ)3
3
Integrate[ %, f ]
3
[−2247168 cos φ − 19712 cos 9φ + 16128 cos 11φ − 177408 cos(φ − 6θ)
78848
− 9856 cos(9φ − 6θ) + 8064 cos(11φ − 6θ) − 177408 cos(φ + 6θ)
− 9856 cos(9φ + 6θ) + 8064 cos(11φ + 6θ) + 2772 sin(4φ − 9θ)
− 1848 sin(6φ − 9θ) − 264 sin(14φ − 9θ) + 231 sin(16φ − 9θ)
+ 185724 sin(4φ − 3θ) − 123816 sin(6φ − 3θ) − 792 sin(14φ − 3θ)
+ 693 sin(16φ − 3θ) + 185724 sin(4φ + 3θ) − 123816 sin(6φ + 3θ)
− 792 sin(14φ + 3θ) + 693 sin(16φ + 3θ) + 2772 sin(4φ + 9θ) − 1848 sin(6φ + 9θ)
− 264 sin(14φ + 9θ) + 231 sin(16φ + 9θ) ]
(% /. f → Pi) - (% /. f → 0) // FullSimplify
12 (157 + 25 cos 6θ)
11
Integrate[ %, t ]
1884θ + 50 sin 6θ
11
(% /. t → 2∗Pi) - (% /. t → 0)
3768π
11
N[ %, 60 ]
1076.138283520576447502476388017924266069590311789503624849396
Thus V =
3768
11 π
≈ 1076.13828352.
13.7.46: First we let g(φ, θ) = 6 + 3 cos 3θ sin 5φ. To find the moment of the bumpy sphere with respect
to the yz-plane, we computed
2π
π
g(φ,θ)
Myz =
θ=0
φ=0
ρ3 sin2 φ cos θ dρ dφ dθ
ρ=0
by executing the following Mathematica commands:
Integrate[ r∧3∗(Sin[f])∧2∗Cos[t], { r, 0, g[f,t] } ];
1768
Integrate[ %, { f, 0, Pi } ];
Integrate[ %, { t, 0, 2∗Pi } ]
0
Therefore x = 0. Similarly, we found that
2π
π
g(φ,θ)
Mxz =
θ=0
and
2π
φ=0
ρ=0
π
g(φ,θ)
Mxy =
θ=0
ρ3 sin2 φ sin θ dρ dφ dθ = 0
φ=0
ρ3 sin φ cos θ dρ dφ dθ = 0.
ρ=0
Thus the centroid of the bumpy sphere is, indeed, located at its center (0, 0, 0). (We suppressed the
“intermediate output” in this solution because it is quite long, as in the solution of Problem 45.)
13.7.47: The following figure makes some of the equations we use easy to derive.
0,0,c
Α
w
P
c
Ρ
Φ
0,0,0
A mass element δ dV located at the point P (ρ, φ, θ) of the ball exerts a force on the mass m at (0, 0, c)
that has vertical component
dF = −
Gmδ cos α
dV.
w2
Note the following:
M=
4
πδa3 ;
3
w2 = ρ2 + c2 − 2ρc cos φ;
2w dw = 2ρc sin φ dφ;
1769
w cos α + ρ cos φ = c;
ρ cos φ =
ρ 2 + c2 − w 2
;
2c
ρ sin φ dφ =
w
dw.
c
Hence the total force exerted by the ball on m is
F =−
2π
θ=0
= −2π
ρ=0
a
0
a
φ=0
π
Gmδ cos α 2
· ρ sin φ dφ dρ dθ = −
w2
c+ρ
= −2π
ρ=0
w=c−ρ
= −πGmδ
πGmδ
=− 2
c
πGmδ
=− 2
c
=−
a
0
0
c+ρ
c−ρ
a
0
a
2π
0
Gmδ
(c − ρ cos φ) · ρ2 sin φ dφ dρ = −2π
w3
0
a
π
a
0
a
0
π
0
c+ρ
c−ρ
Gmδw cos α 2
· ρ sin φ dφ dρ dθ
w3
Gmδ
w3
ρ 2 + c2 − w 2
ρw
dw dρ
c−
·
2c
c
w
Gmδ 2c2 − ρ2 − c2 + w2
· ρ · dw dρ
·
w3
2c
c
1 c2 − ρ 2 + w 2
πGmδ
·
· ρ dw dρ = − 2
w2
c2
c
ρ 2 − c2
+w
ρ·
w
c+ρ
w=c−ρ
πGmδ
dρ = − 2
c
0
a
0
a
ρ·
πGmδ
ρ(ρ − c + ρ + c + c + ρ − c + ρ) dρ = − 2
c
c+ρ
c−ρ
c2 − ρ 2
+ 1 · ρ dw dρ
w2
ρ 2 − c2
c2 − ρ2
+c+ρ+
−c+ρ
ρ+c
c−ρ
0
a
dρ
a
4 3
πGmδ
ρ
4ρ dρ = − 2 ·
c
3
0
2
πGmδ 4 3
4
Gm
GM m
· a = − πδa3 · 2 = − 2 .
c2
3
3
c
c
Magnificent! You can even extend this result to show that if the density of the ball varies only as a function
of ρ, then the same conclusion follows: The ball acts, for purposes of gravitational attraction of an external
mass m, as if all its mass M were concentrated at its center. And note one additional item of interest: This
is one of the rare spherical triple integrals not that is best evaluated in the order dρ dφ dθ.
13.7.48: The following figure again makes some of the equations we use easy to derive.
1770
0,0,c
Α
w
P
c
Ρ
Φ
0,0,0
A mass element δ dV located at the point P (ρ, φ, θ) of the ball exerts a force on the mass m at (0, 0, c)
that has vertical component
dF =
Gmδ cos α
dV.
w2
Note the following:
M=
4
πδa3 ;
3
w2 = ρ2 + c2 − 2ρc cos φ;
2w dw = 2ρc sin φ dφ;
w cos α + ρ cos φ = c;
ρ 2 + c2 − w 2
;
2c
w
ρ sin φ dφ =
dw.
c
ρ cos φ =
Hence the total force exerted by the ball on m is
2π b π
2π b π
Gmδ cos α 2
Gmδw cos α 2
F =
·
ρ
sin
φ
dφ
dρ
dθ
=
· ρ sin φ dφ dρ dθ
2
w
w3
θ=0 ρ=a φ=0
0
a
0
b
π
0
Gmδ
(c − ρ cos φ) · ρ2 sin φ dφ dρ = 2π
w3
c+ρ
= 2π
a
b
= 2π
ρ=a
w=ρ−c
b
c+ρ
= πGmδ
a
ρ−c
b
a
c+ρ
ρ−c
Gmδ
w3
ρ 2 + c2 − w 2
ρw
dw dρ
c−
·
2c
c
w
Gmδ 2c2 − ρ2 − c2 + w2
· ρ · dw dρ
·
3
w
2c
c
1 c2 − ρ 2 + w 2
πGmδ
·
· ρ dw dρ =
2
2
w
c
c2
1771
a
b
c+ρ
ρ−c
c2 − ρ 2
+
1
· ρ dw dρ
w2
πGmδ
=
c2
πGmδ
=
c2
b
a
b
a
ρ 2 − c2
+w
ρ·
w
c+ρ
w=ρ−c
πGmδ
dρ =
c2
b
ρ·
a
πGmδ
ρ(ρ − c + ρ + c − ρ − c + c − ρ) dρ =
c2
ρ 2 − c2
ρ 2 − c2
+c+ρ−
+c−ρ
ρ+c
ρ−c
dρ
b
0 dρ = 0.
a
The key difference between this derivation and that in the solution of Problem 47 is that the lower limit of
integration on w is here ρ − c rather than c − ρ. The reason for the change is that here we have c ρ, so
that when φ = 0 we have w2 = (ρ − c)2 and thus w = ρ − c.
13.7.49: A Mathematica solution:
r = 6370∗1000;
d1 = 11000;
k = 0.371;
d2 = 5000;
m1 = (4/3)∗Pi∗d1∗x∧3;
m2 = (4/3)∗Pi∗d2∗(r∧3 - x∧3);
m = m1 + m2;
I1 = (2/5)∗m1∗x∧2;
By Problem 42, the moment of inertia of the mantle with respect to the polar axis (through the poles of the
planet) is
I2 = (2/5)∗m2∗(r∧5 - x∧5)/(r∧3 - x∧3);
Therefore we proceed to solve for x as follows.
I0 = I1 + I2;
eq1 = I0 == k∗m∗r∧2;
soln = NSolve[ eq1, x ];
We suppress the output, but there are five solutions. Only two are real and positive,
x1 ≈ 2.76447 × 106
and
x2 ≈ 5.87447 × 106 .
We are given the information that the mantle is “a few thousand kilometers thick,” and x2 does not satisfy
this condition, as it implies that the mantle is less than 496 km thick. We conclude that the radius of the
core is x1 /1000 km, and hence that the thickness of the mantle is (r − x1 )/1000 ≈ 3605.53 km.
1772
Section 13.8
13.8.1: The surface area element is
dS =
√
1 + 12 + 32 dA = 11 dA.
So the area in question is
2
A=4
3√1−(x/2)2 √
11 dy dx =
x=0
2
0
y=0
√ 6 11 4 − x2 dx
x 2
√
√
√ = 6π 11 ≈ 62.5169044565658738.
= 3x 11 4 − x2 + 12 11 arcsin
2
0
13.8.2: The surface area element is
dS =
1 + 22 + 22 dA = 3 dA.
So the area in question is
1
√
x
A=
1
3 dy dx =
0
1/2
(3x
x2
0
2
3/2
− 3x ) dx = 2x
3
1
−x
= 1.
0
13.8.3: The paraboloid meets the plane in the circle with equation x2 + y 2 = 4, z = 5. Let D denote the
circular disk x2 + y 2 4 in the xy-plane. The surface area element is
dS =
1 + 4x2 + 4y 2 dA,
so the surface area in question is
2
2
A=
1 + 4x + 4y dA =
= 2π
√
17 17 − 1
12
=
dS =
0
0
1
1+
4r2
r=0
1
dr dθ = 2π ·
(1 + 4r2 )3/2
12
2
0
1 √
π 17 17 − 1 ≈ 36.1769031974114084.
6
13.8.4: The surface area element is
so the area is
1 x
A=
1 + x2 dy dx =
2
r
θ=0
D
2π
1 + x2 dA,
x(1 + x2 )1/2 dx =
0
13.8.5: The surface area element is
dS =
1
(1 + x2 )3/2
3
2 + 4y 2 dA,
1773
1
=
0
1 √
2 2 − 1 ≈ 0.609475708249.
3
so the area is
2
1
(2 + 4y 2 )1/2 dx dy =
A=
y=0
x=0
2
(2 + 4y 2 )1/2 dy =
y=0
√ 2
1
1
y(2 + 4y 2 )1/2 + arcsinh y 2
2
2
0
√ √ √
√
1
1 = 3 2 + arcsinh 2 2 = 3 2 + ln 3 + 2 2 ≈ 5.1240142741388282.
2
2
2 + 4y 2 dA, so the surface area is
13.8.6: The surface area element is dS =
1
y
A=
0
(2 + 4y 2 )1/2 dx dy =
0
1
y(2 + 4y 2 )1/2 dy =
0
1
(2 + 4y 2 )3/2
12
1
0
√ 1 √
3 6 − 2 ≈ 0.9890426199607321.
=
6
√
14 dA, so the surface area is
13.8.7: The surface area element is dS =
3
(6−2x)/3
A=
x=0
√
14 dy dx =
3
√
√
√
1
1 2√
(6 − 2x) 14 dx = 2x 14 − x 14
= 3 14 .
3
3
0
3
0
y=0
Alternatively, the vectors u = −3, 2, 0 and v = −3, 0, 6 span two adjacent sides of the triangular
surface. So its area is half the magnitude of their cross product:
√
1
1√
1
|u × v| = |12, 18, 6 | =
504 = 3 14 ≈ 11.22497216032182415675.
2
2
2
A=
13.8.8: The surface area element is dS =
2π
A=
0
0
√
2
2π
1
1 2√
r 14
r 14 dr dθ = 2π ·
2
2 1/2
r(1 + r )
A=
θ=0
r=0
2π
0
0
2
0
1
(1 + r2 )3/2
dr dθ = 2π ·
3
r(1 + 4r2 )1/2 dr dθ = 2π ·
√
= 2π 14 ≈ 23.5095267170779957.
1 + x2 + y 2 dA, so the surface area is
13.8.10: The surface area element is dS =
√2
√
13.8.9: The surface area element is dS =
√
14 dA. Hence the surface area is
1
=
0
2 √
π 2 2 − 1 ≈ 3.8294488151512928.
3
1 + 4x2 + 4y 2 dA, so the surface area is
1
(1 + 4r2 )3/2
12
2
=
0
1 √
π 17 17 − 1 ≈ 36.1769031974114084.
6
13.8.11: The paraboloid meets the xy-plane in the circle with equation x2 + y 2 = 16. The surface area
element is
dS =
1 + 4x2 + 4y 2 dA,
1774
so the area is
2π
4
A=
2 1/2
r(1 + 4r )
0
0
1
(1 + 4r2 )3/2
dr dθ = 2π ·
12
4
=
0
1 √
π 65 65 − 1 ≈ 273.866639786258.
6
13.8.12: Let z(r, θ) = br. Then the surface area element is
dS =
r2 + (rzr )2 + (zθ )2 dA = r 1 + b2 dA,
so the surface area is
2π
A=
a
2 1/2
r(1 + b )
0
0
1 2
r (1 + b2 )1/2
dr dθ = 2π ·
2
a
= πa
a2 + a2 b2 = πa a2 + h2 = πaL.
0
Note that when you use Eq. (10), you should see dr dθ where you are accustomed to see r dr dθ.
13.8.13: Let r(θ, z) = a cos θ, a sin θ, z . Then
i
rθ × rz = −a sin θ
0
j
a cos θ
0
and hence
|rθ × rz | =
k 0 = a cos θ, a sin θ, 0 1
a2 cos2 θ + a2 sin2 θ = a.
Therefore, by Eq. (8), the area of the zone is
2π
h
A=
θ=0
13.8.14: Let z(r, θ) =
h
a dz dθ = 2π · az
z=0
= 2πah.
z=0
√
a2 − r2 . Then the surface area element in cylindrical coordinates is
dS =
ar
r2 + (rzr )2 + (zθ )2 dA = √
dA.
2
a − r2
Thus by Eq. (10), the area of the zone is
A=
0
2π
√
a2 −b2
√
a2 −c2
√a2 −b2
ar
2
2 1/2
√
dr dθ = 2π · −a(a − r )
= 2π(ac − ab) = 2πa(c − b) = 2πah.
√
a2 − r 2
a2 −c2
13.8.15: Let z(x, y) =
√
a2 − x2 . Then the surface area element in Cartesian coordinates is
a
dS = √
dA.
a2 − x2
1775
Let D be the disk in which the vertical cylinder meets the xy-plane. Then the area of the part of the
horizontal cylinder—top and bottom—that lies within the vertical cylinder is
2π a
a
2a
√
√
dA =
· r dr dθ
A=2
2 − x2
2 − r2 cos2 θ
a
a
θ=0 r=0
D
π/2
a
π/2
−2a(a2 − r2 cos2 θ)1/2
2
2
2
tan
θ
−
sec
θ
dθ
=
4
2a
(sec
θ
−
sec
θ
tan
θ)
dθ
=
8a
cos2 θ
θ=0
θ=0
r=0
0
= 8a2 + 8a2 ·
lim − tan θ − sec θ
= 8a2 + 8a2 · 0 = 8a2 .
2π
=
θ→(π/2)
The change in the limits of integration in the second line was necessary because we needed the simplification
(1 − cos2 θ)1/2 = sin θ, which is not valid if π < θ < 2π; moreover, we needed to avoid the discontinuity of
the following improper integral at θ = π/2.
13.8.16: Let z(r, θ) =
√
a2 − r2 . Then the surface area element in cylindrical coordinates is
dS =
ar
r2 + (rzr )2 + (zθ )2 dA = √
dA.
2
a − r2
Hence, by Eq. (10), the area of the part of the sphere (top and bottom) that lies within the cylinder is
a sin θ
π
ar
2
2 1/2
√
A=2
dr dθ =
dθ
− 2a(a − r )
a2 − r 2
0
0
0
0
π/2
π/2
π
− 1 = 2a2 (π − 2) ≈ (2.283185307180)a2.
2a2 (1 − cos θ) dθ = 4a2 θ − sin θ
= 4a2
=2
2
0
0
π
a sin θ
The change of limits of integration in the second line is necessary because the important simplification
(a2 − a2 sin2 θ)1/2 = a cos θ is not valid on the interval
1
2π
< θ < π. The student who fails to notice this
2
will probably obtain the incorrect answer 2πa .
13.8.17: The surface y = f (x, z) is parametrized b
r(x, z) = x, f (x, z), z for (x, z) in the region R in the xz-plane. Then
i
rx × rz = 1
0
j
fx
fz
k
0 = fx , −1, fz .
1
Therefore the area of the surface y = f (x, z) lying “over” the region R is
2 2
∂f
∂f
A=
1+
+
dx dz.
∂x
∂z
R
1776
The surface x = f (y, z) is parametrized by
r(y, z) = f (y, z), y, z for (y, z) in the region R in the xz-plane. Then
i
ry × rz = fy
fz
k
0 = 1, −fy , −fz .
1
j
1
0
Therefore the area of the surface x = f (y, z) lying “over” the region R is
A=
1+
R
∂f
∂y
2
+
∂f
∂z
2
dy dz.
13.8.18: The equations in (6) take the form
x = a sin φ cos θ,
y = a sin φ sin θ,
z = a cos φ,
and hence the spherical surface corresponding to the region R in the φθ-plane is parametrized by
r(φ, θ) = a sin φ cos θ, a sin φ sin θ, a cos φ .
Thus
i
rφ × rθ = a cos φ cos θ
−a sin φ sin θ
Therefore
|rφ × rθ | =
−a sin φ = a2 sin2 φ cos θ, a2 sin2 φ sin θ, a2 sin φ cos φ .
0
j
k
a cos φ sin θ
a sin φ cos θ
a4 sin4 φ cos2 θ + a4 sin4 φ sin2 θ + a4 sin2 φ cos2 φ = a2 sin φ.
Thus, by Eq. (8), the surface area of the part of the sphere corresponding to R is
A=
a2 sin φ dφ dθ.
R
13.8.19: By Problem 50 of Section 13.2,
θ2
φ2
A=
θ1
a2 sin φ dφ dθ = a2 sin φ̂ ∆φ ∆θ
φ1
1777
for some φ̂ in (φ1 , φ2 ). Therefore
ρ2
∆V =
(ρ2 sin φ̂ ∆φ ∆θ) dρ =
ρ1
=
1 3
(ρ − ρ31 ) sin φ̂ ∆φ ∆θ
3 2
1 ρ32 − ρ31
1
·
sin φ̂ ∆φ ∆θ ∆ρ = · 3ρ̂2 sin φ̂ ∆φ ∆θ ∆ρ = ρ̂2 sin φ̂ ∆ρ ∆φ ∆θ
3 ρ2 − ρ1
3
for some ρ̂ in (ρ1 , ρ2 ), and this is Eq. (8) of Section 13.7.
13.8.20: See the solution of Problem 29 of Section 14.7. The surface area of this pinched torus can be
computed (without the second theorem of Pappus) as follows. First parametrize it via
r(φ, θ) = 2a sin2 φ cos θ, 2a sin2 φ sin θ, 2a sin φ cos φ where 0 φ π and 0 θ 2π. (To verify that this parametrization is correct, execute the Mathematica
command
ParametricPlot3D[ { 2∗Sin[u]∗Sin[u]∗Cos[v], 2∗Sin[u]∗Sin[u]∗Sin[v], 2∗Sin[u]∗Cos[u] },
{u, 0, Pi}, {v, 0, 2∗Pi}, AspectRatio → Automatic ];
to see a copy of the pinched torus for the case a = 1.) Next,
rφ = 4a sin φ cos φ cos θ, 4a sin φ cos φ sin θ, 2a cos2 φ − 2a sin2 φ and
rθ = −2a sin2 φ sin θ, 2a sin2 φ cos θ, 0 .
After some lengthy computations, you will find that |rφ × rθ | = 4a2 sin2 φ. Hence the surface area of the
pinched torus is
2π
π
2
2
π
4a sin φ dφ dθ = 2πa · 2φ − sin 2φ
A=
θ=0
2
φ=0
= 4π 2 a2 .
0
13.8.21: Given:
x = f (z) cos θ,
y = f (z) sin θ,
z=z
where 0 θ 2π and a z b. The surface thereby generated is thereby parametrized by
r(θ, z) = f (z) cos θ, f (z) sin θ, z ,
and thus
rθ = −f (z) sin θ, f (z) cos θ, 0 rz = f (z) cos θ, f (z) sin θ, 1 .
1778
and
Therefore
i
rθ × rz = −f (z) sin θ
f (z) cos θ
and hence
|rθ × rz | =
j
f (z) cos θ
f (z) sin θ
k 0 = f (z) cos θ, f (z) sin θ, −f (z) · f (z) 1
2
2
2
[f (z)] + [f (z) · f (z)] = f (z) 1 + [f (z)] .
Thus, by Eq. (8), the area of the surface of revolution is
2π
2
f (z) 1 + [f (z)] dz dθ =
b
A=
θ=0
z=a
2
2πf (z) 1 + [f (z)] dz.
b
z=a
Compare this with Eq. (8) in Section 6.4.
13.8.22: Part (a):
2π
A=
0
0
π
π
a2 sin φ dφ dθ = 2πa2 · − cos φ = 4πa2 .
0
Part (b): The area is
2π
π/6
A=
0
0
π/6 √ a2 sin φ dφ dθ = 2πa2 · − cos φ
= 2 − 3 πa2 ,
0
a little less than 6.6987% of the total surface area of the sphere.
13.8.23: In the result in Problem 21, take f (z) = r (the constant radius of the cylinder). Then
2
f (z) 1 + f (z) = r,
so the curved surface area of the cylinder is
2π
13.8.24: Let z(x, y) =
h
r dz dθ = 2π · rz
A=
θ=0
h
z=0
= 2πrh.
z=0
√
r2 − x2 . Then
dS =
r
1 + (zx )2 + (zy )2 dA = √
dA.
2
r − x2
Then Eq. (9) yields the curved surface area of the cylinder:
h
r ry
r
√
√
dy dx = 4
dx
r2 − x2
r2 − x2 y=0
x=0 y=0
x=0
r
r
x
rh
πrh
√
= 2πrh.
dx = 4 · rh arctan √
=4·
=4
2
r2 − x2
r2 − x2
0
x=0
r
h
A=4
1779
Why the factor 4? We had to double twice; once because the integral gives only the area of the top of
the cylinder, once again because we integrated only over the interval 0 x r rather than the interval
−r x r.
13.8.25: Part (a): The Mathematica command
Plot3D[ x∧2 + y∧2, {x, -1, 1}, {y, -1, 1} ];
produced the view of the surface that is shown next.
1
y
0
1
2-1
z
1
0
-1
0
x
The surface area element is dS =
1
x=−1
1
1 + 4x2 + 4y 2 dA, so the area of the surface is
1 + 4x2 + 4y 2 dy dx
1
A=
1
y=−1
1
1 1
2
2
2
2
2
y 1 + 4x + 4y + (4x + 1) ln 2y + 1 + 4x + 4y
dx
2
4
−1
=
−1
1
1
2
2
2
2
2
=
4x + 5 − (4x + 1) ln −2 + 4x + 5 + (4x + 1) ln 2 + 4x + 5
dx
4
4
−1
2x
4x
2
7
1
x 4x2 + 5 + arcsinh √
=
− arctan √
3
6
6
5
4x2 + 5
1
1
1
x(4x2 + 3) ln −2 + 4x2 + 5 +
x(4x2 + 3) ln 2 + 4x2 + 5
+
12
12
−1
√ 4
2 5
1
7
7
− arctan
= 4 + arcsinh
+ ln 5 ≈ 7.44625672301236346326.
3
5
3
3
6
Part (b):
1
By symmetry, we integrate over the quarter of the square that lies in the first quadrant and
multiply by 4. Thus the area is
1
A=4
x=0
1−x 1 + 4x2 + 4y 2 dy dx
y=0
1780
1
=
1−x
2
2
2
2
2
dx
2y 1 + 4x + 4y + (4x + 1) ln 2y + 1 + 4x + 4y
0
1
=
0
0
1 + 4x2
2(1 − x) 1 + 4(1 − x)2 + 4x2 − (1 + 4x2 ) ln
2
2
+ (1 + 4x ) ln 2(1 − x) + 1 + 4(1 − x) + x
dx
2
√
√
1
6
5 2
2
2
(4x − 2x − 1) 8x − 8x + 5 +
arcsinh
[2x − 1]
=
3
6
3
1
− arctan
6
1
+ arctan
6
√
72 + 288x2 − 25(4x + 1) 8x2 − 8x + 5
184x2 − 200x − 29
√
−72 − 288x2 + 25(4x + 1) 8x2 − 8x + 5
184x2 − 200x − 29
1
1
1
x(4x2 + 3) ln
1 + 4x2 + x(4x2 + 3) ln 2 − 2x + 8x2 − 8x + 5
3
3
0
√
√
√ √ √
5 2
1
72 − 25 5
1
72 − 25 5
6
2 5
+
arcsinh
− arctan
− arctan
=
3
3
3
6
71
6
29
−
1
+ arctan
6
√
√
25 5 − 72
1
25 5 − 72
+ arctan
≈ 3.0046254342814410.
29
6
71
Of course it was Mathematica that computed and evaluated the antiderivatives in this solution.
13.8.26: Part (a): The surface area element is
dS = 1 + (zx )2 + (zy )2 dA =
1+
x2
√
x2
y2
+ 2
dA = 2 dA.
2
2
+y
x +y
Hence the area of the surface is
1
x=−1
1
√
√
2 dy dx = 4 2 ≈ 5.6568542494923802.
y=−1
Part (b): By symmetry, we integrate over the quarter of the square that lies in the first quadrant, then
multiply by 4. The area is thus
1 1−x √
√
1 √
4
2 dy dx = · 4 2 = 2 2 ≈ 2.8284271247461901.
2
x=0 y=0
In both cases we integrated the constant function by multiplying its value by the area of the domain of the
integral.
13.8.27: Part (a): The following graph of the surface was generated by the Mathematica command
1781
Plot3D[ 1 + x∗y, {x, -1, 1}, {y, -1, 1} ];
1
y
0
1
2-1
z
1
0
-1
0
x
The surface area element is dS =
1
x=−1
1 + x2 + y 2 dA. Hence the surface area is
1 + x2 + y 2 dy dx =
1
1
1
−1
y=−1
1
1 1
dx
y 1 + x2 + y 2 + (1 + x2 ) ln y + 1 + x2 + y 2
2
2
−1
1
1
2 + x2 − (1 + x2 ) ln −1 + 2 + x2 + (1 + x2 ) ln 1 + 2 + x2
dx
2
2
−1
√ x
2
x 2
2
4
x 2 + x2 + arcsinh
− arctan √
=
3
3
2
3
2 + x2
1
=
1
1
1
x(3 + x2 ) ln −1 + 2 + x2 + x(3 + x2 ) ln 1 + 2 + x2
6
6
−1
√
√
√ 4 √ 2
8
4 4 3
2
− π + arcsinh
− ln −1 + 3 + ln 1 + 3 ≈ 5.123157101094.
=
3
9
3
2
3
3
−
Part (b): Using symmetry, we integrate over the quarter of the square in the first quadrant and multiply
the answer by 4. Thus the surface area is
1
4
x=0
1−x 1 + x2 + y 2 dy dx =
1
2y
1
1 + x2 + y 2 + 2(1 + x2 ) ln y + 1 + x2 + y 2
dx
0
y=0
1
=
0
0
1 + x2
2(1 − x) 2x2 − 2x + 2 − 2(1 + x2 ) ln
+ 2(1 + x2 ) ln 1 − x + 2x2 − 2x + 2
dx
√
√
2 − 2x + 2
2x
−
1
1
2
2x
7
4
(4x − 2x2 − 1) 2x2 − 2x + 2 +
arcsinh √
=
+ arctan
3
6
3
x+1
3
−
2
1
2
x(3 + x2 ) ln
1 + x2 + x(3 + x2 ) ln 1 − x + 2x2 − 2x + 2
3
3
0
1782
√ √ √
√
√ 7 2
4
4
3
2
2 2
+
arcsinh
+ arctan
− arctan
=
2 ≈ 2.302310960471.
3
3
3
3
2
3
13.8.28: The surface area element is
dS =
1+
x2
y2
2
+
dA = dA.
2
2
2
2
4−x −y
4−x −y
4 − x2 − y 2
Part (a): The surface area is double the integral of dS because half of the surface is above the xy-plane
and half is below. Thus the area is
1 1
1
1
4
y
dy dx =
dx
4 arctan 4 − x2 − y 2
4 − x2 − y 2
x=−1 y=−1
−1
−1
1
1
=
8 arctan √
dx
3 − x2
−1
x
3 − 2x
1
= 8 arcsin √
+ 8 arctan √
+ 8x arctan √
3
3 − x2
3 − x2
1
3 + 2x
− 8 arctan √
3 − x2
−1
√ √ √ 5 2
3
2
+ 32 arctan
− 16 arctan
≈ 8.8205695749204929.
= 16 arcsin
3
2
2
Part (b): Using symmetry, we integrate dS over the quarter of the square in the first quadrant, multiply
by 4, then multiply by 2 as well because half of the surface is above the xy-plane and half is below. Thus
the area is
1 8
x=0
1−x
y=0
2
1
y
1−x
dy dx =
dx
16 arctan 4 − x2 − y 2
4 − x2 − y 2
0
0
1
1−x
=
16 arctan √
dx
3 + 2x − 2x2
0
√
2x − 1
1−x
√
= 8 2 arcsin
+ 16x arctan √
7
3 + 2x − 2x2
1
5 − 3x
5x + 1
√
− 16 arctan √
− 16 arctan
3 + 2x − 2x2
3 3 + 2x − 2x2
0
√
1
1
2
5
√
= 16 2 arcsin √
+ 16 arctan
− 32 arctan √
+ 16 arctan √
7
3 3
3
3
≈ 4.183189651006409398670043719362732266.
13.8.29: We readily verify that x, y, and z satisfy the equation
x2
y2
z
+
=
2
2
a
b
c
1783
of an elliptic paraboloid. For a typical graph, we executed the Mathematica commmands
a = 2; b = 1; c = 3;
x = a∗u∗Cos[v]; y = b∗u∗Sin[v]; z = c∗u∧2;
ParametricPlot3D[ { x, y, z }, { u, 0, 1 }, { v, 0, 2∗Pi } ];
and the resulting graph is next.
3
z
2
1
0
1
-2
-1
0 y
0
1
x
2
-1
13.8.30: We readily verify that x, y, and z satisfy the equation
x2
y2
z2
+ 2 + 2 = 1
2
a
b
c
of an ellipsoid. For a typical example, we executed the Mathematica commands
a = 2; b = 1; c = 1;
x = a∗Sin[u]∗Cos[v]; y = b∗Sin[u]∗Sin[v]; z = c∗Cos[u];
ParametricPlot3D[ { x, y, z }, { u, 0, Pi }, { v, 0, 2∗Pi } ];
and the result is shown next.
1
z
0
-1
1
-2
-1
0 y
0
1
x
2
1784
-1
13.8.31: We readily verify that x, y, and z satisfy the equation
z2
x2
y2
−
−
= 1
c2
a2
b2
of a hyperboloid of two sheets. To see the upper half of a typical example, we executed the following
Mathematica commands:
a = 2; b = 1; c = 4;
x = a∗Sinh[u]∗Cos[v]; y = b∗Sinh[u]∗Sin[v]; z = c∗Cosh[u];
ParametricPlot3D[ { x, y, z }, { u, 0, 1 }, { v, 0, 2∗Pi } ];
and the result is next.
6
z 5
4
3
1
-2
-1
0 y
0
x
1
2
-1
13.8.32: It’s easy to verify that x, y, and z satislfy the equation
x2
y2
z2
+
−
= 1
a2
b2
c2
of a hyperboloid of one sheet. To graph a typical example, we executed the Mathematica commands
a = 2; b = 1; c = 2;
x = a∗Cosh[u]∗Cos[v]; y = b∗Cosh[u]∗Sin[v]; z = c∗Sinh[u];
ParametricPlot3D[ { x, y, z }, { u, -1, 1 }, { v, 0, 2∗Pi } ];
and the result is shown next.
1785
y
1
0
x
0
-3
-1
3
2
z 0
-2
13.8.33: The ellipsoid is parametrized via
r(φ, θ) = 4 sin φ cos θ, 3 sin φ sin θ, 2cosφ .
Then
rφ = 4 cos φ cos θ, 3 cos φ sin θ, −2 sin φ and
rθ = −4 sin φ sin θ, 3 sin φ cos θ, 0 .
It now follows that
rφ × rθ = 6 sin2 φ cos θ, 8 sin2 φ sin θ, 12 sin φ cos φ ,
so that
|rφ × rθ | = 36 sin4 φ cos2 θ + 144 sin2 φ cos2 φ + 64 sin4 φ sin2 θ .
We used the NIntegrate command in Mathematica to approximate the surface area of the ellipsoid; it is
2π π
(36 sin4 φ cos2 θ + 144 sin2 φ cos2 φ + 64 sin4 φ sin2 θ)1/2 dφ dθ ≈ 111.545774984838.
θ=0
φ=0
13.8.34: Part (a): Begin with the ellipse
x−b
a
2
+
2
z
= 1,
c
and note that this equation is satisfied if
x−b
= cos ψ
a
and
z
= sin ψ.
c
With the aid of a figure much like Fig. 13.8.13, we see that the ellipsoidal torus generated by rotation of this
ellipse around the z-axis is parametrized by
x = (b + a cos ψ) cos θ,
y = (b + a cos ψ) sin θ,
1786
z = c sin ψ
where 0 θ 2π and 0 ψ 2π.
Part (b): Parametrize the surface of revolution via r(ψ, θ) where the components of r are those given in
Part (a), but using a = 2, b = 3, and c = 1. Then
rψ = −2 sin ψ cos θ, −2 sin ψ sin θ, cos ψ ;
rθ = −(3 + 2 cos ψ) sin θ, (3 + 2 cos ψ) cos θ, 0 ;
rψ × rθ = −(3 + 2 cos ψ) cos ψ cos θ, −(3 + 2 cos ψ) cos ψ sin θ, −2(3 sin ψ + sin 2ψ) ;
|rψ × rθ | =
(5 − 3 cos 2ψ)(3 + 2 cos ψ)2
2
1/2
.
It follows that the surface area of the ellipsoid is
2π
2π
A=
θ=0
ψ=0
(5 − 3 cos 2ψ)(3 + 2 cos ψ)2
2
1/2
2π
dψ dθ = 2π
0
(5 − 3 cos 2ψ)(3 + 2 cos ψ)2
2
1/2
dψ.
The command
2∗Pi∗NIntegrate[ Sqrt[ (1/2)∗(5 - 3∗Cos[2*t])*(3 + 2∗Cos[t])∧2 ],
{t, 0, 2∗Pi}, WorkingPrecision → 32 ]
in Mathematica returns the approximation A ≈ 182.622946526146.
Part (c): Let
(x − 3)2
f (x) = 1 −
4
1/2
.
The graph of f is the top half of the ellipse. Hence the length of the ellipse is
L=2
5
1+
2
f (x)
5
dx =
1
1
3x2 − 18x + 11
x2 − 6x + 5
1/2
dx.
Note that this integral is improper at each endpoint of the interval [1, 5]. A Mathematica command similar
to the one in Part (b) yields the approximation L ≈ 9.688448220548.
Section 13.9
13.9.1: It is easy to solve the given equations for
x=
u+v
,
2
y=
1787
u−v
.
2
Hence
1
∂(x, y) 2
=
∂(u, v) 1
2
1
2
1
−
2
1
=− .
2
13.9.2: It is easy to solve the given equations for
x=
Thus
u + 2v
,
7
y=
1
∂(x, y) 7
=
∂(u, v) 3
−
7
v − 3u
.
7
2
7
1
7
1
= .
7
13.9.3: When we solve the equations u = xy and v = y/x for x and y, we find that there are two solutions:
√
√
u
u
x=
, y = uv
, y = − uv .
and
x=−
v
v
It doesn’t matter which we choose; the value of the Jacobian will be the same. (Why?) So we choose the
first solution. Then
1
∂(x, y) 2u1/2 v 1/2
=
∂(u, v) v 1/2
2u1/2
u1/2
− 3/2
2v
u1/2
2v 1/2
= 1 .
2v
13.9.4: When we solve the equations u = 2(x2 + y 2 ), v = 2(x2 − y 2 ) for x and y, we get four solutions—all
possible combinations of
√
u+v
x=±
,
2
√
u−v
y=±
.
2
We choose the solution for which x and y are both nonnegative. Then
1
√ 1
√
4 u+v
4 u+v ∂(x, y) 1
=
.
=− √ 2
∂(u, v) 8 u − v2
1
1
√
− √
4 u−v
4 u−v
13.9.5: When we solve the equations u = x + 2y 2 , v = x − 2y 2 for x and y, we get two solutions:
√
u+v
u−v
x=
,
y=±
.
2
2
We choose the solution for which y is nonnegative. Then
1
1
2
2
∂(x, y) =
∂(u, v) 1
1
− √
√
4 u−v
4 u−v
1788
1
.
=− √
4 u−v
13.9.6: Given
u=
2x
,
x2 + y 2
note first that
u2 + v 2 =
v=−
2y
,
x2 + y 2
(1)
4x2 + 4y 2
4
= 2
,
(x2 + y 2 )2
x + y2
so that
x2 + y 2 =
u2
4
.
+ v2
(2)
Therefore, using the equations in (1), then Eq. (2), we have
x=
Then
1
2u
u(x2 + y 2 ) = 2
2
u + v2
1
2v
y = − v(x2 + y 2 ) = − 2
.
2
u + v2
and
2(v 2 − u2 )
2
(u + v 2 )2
∂(x, y) =
∂(u, v) 4uv
2
(u + v 2 )2
4
=
(u2 + v 2 )2 .
2(v 2 − u2 ) (u2 + v 2 )2
−
4uv
2
(u + v 2 )2
13.9.7: First we solve the equations u = x + y and v = 2x − 3y for
x=
3u + v
,
5
y=
2u − v
.
5
Substitution in the equation x + y = 1 then yields
1=
3u + v 2u − v
5u
+
=
= u.
5
5
5
Similarly, x + y = 2 yields u = 2, 2x − 3y = 2 yields v = 2, and 2x − 3y = 5 yields v = 5. Moreover,
3
1 5 ∂(x, y) 5
1
=
=− .
∂(u, v) 5
2
1 − 5
5
Therefore
5
2
1 dy dx =
R
v=2
u=1
1
3
1
du dv = 3 · 1 · = .
5
5
5
Note: Because R is a parallelogram with adjacent sides represented by the two vectors a = 35 , 25 , 0 and
b = 35 , − 53 , 0 , we have the following alternative method of finding the area A of R:
i
j
k 3
2
3
0 = 0, 0, −
a×b=
,
5
5
5
3
3
−
0
5
5
1789
and therefore A = |a × b| =
3
.
5
13.9.8: Given u = xy and v =
y
, we have
x
y
uv = xy · = y 2
x
and thus we choose
x=
u
v
and
u
x
= xy · = x2 ,
v
y
and
y=
1
2u1/2 v 1/2
∂(x, y) =
∂(u, v) v 1/2
2u1/2
Then
√
uv .
u1/2
− 3/2
2v
u1/2
2v 1/2
(1)
= 1 .
2v
Also, if y = x, then substitution of the equations in (1) yields
(uv)1/2 =
u1/2
;
v 1/2
uv =
u
;
v
v 2 = 1.
So we choose v = 1. (This choice implies that if we have a similar choice with u, we must choose u > 0
because of the equations in (1).) Similarly, y = 2x yields v = 2, xy = 1 yields u = 1, and xy = 2 yields
u = 2. Hence the area of the region of Fig. 13.9.7 is
2
2 2
1
1
1
du dv =
dv = ln 2 ≈ 0.3465735902799727.
1 dx dy =
A=
2
v=1 u=1 2v
1 2v
R
13.9.9: If u = xy and v = xy 3 , then
uy 2 = xy 3 = v,
Then
x=
y2 =
so that
v
;
u
y=
v 1/2
.
u1/2
u1/2
u
u3/2
= u · 1/2 = 1/2 .
y
v
v
(We do not need the solution in which x and y are negative.) Then
3u1/2
u3/2 − 3/2 1/2
2v
∂(x, y) 2v
= 3 − 1 = 1 .
=
4v 4v
∂(u, v) 2v
1
v 1/2
− 3/2
2u
2u1/2 v 1/2
We also find by substitution that xy = 2 corresponds to u = 2, xy = 4 corresponds to u = 4, xy 3 = 3
corresponds to v = 3, and xy 3 = 6 corresponds to v = 6. Hence the area of the region shown in Fig. 13.9.8
is
6
4
1 dx dy =
A=
D
v=3
u=2
1
du dv =
2v
1790
3
6
1
dv = ln 2 ≈ 0.6931471805599453.
v
13.9.10: If y = ux2 and x = vy 2 , then
y = uv 2 y 4 ;
y3 =
1
;
uv 2
1
y=
u1/3 v 2/3
.
Then it follows that
x = vy 2 =
Next,
v
1
= 2/3 1/3 .
u2/3 v 4/3
u v
2
−
5/3
1/3
∂(x, y) 3u v
=
∂(u, v) 1
−
3u4/3 v 2/3
−
−
1
= 2 2.
3u v
1
3u2/3 v 4/3
2
3u1/3 v 5/3
Next, y = x2 corresponds to u = 1, y = 2x2 corresponds to u = 2, x = y 2 corresponds to v = 1, and
x = 4y 2 corresponds to v = 4. Therefore the area of the region shown in Fig. 13.9.9 is
A=
4
2
1 dx dy =
R
v=1
u=1
1
du dv =
3u2 v 2
4
v=1
1
−
3uv 2
2
dv =
1
u=1
4
4
1
1
1
dv = −
= .
2
6v
6v 1
8
13.9.11: Given: the region R bounded by the curves y = x3 , y = 2x3 , x = y 3 , and x = 4y 3 . Choose u
and v so that y = ux3 and x = vy 3 . Then
y = uv 3 y 9 ;
y=
1
u1/8 v 3/8
;
y8 =
1
;
uv 3
x = vy 3 =
v
u3/8 v 9/8
=
1
u3/8 v 1/8
.
Then the curve y = x3 can be written as
1
1
= 9/8 3/8 ;
u1/8 v 3/8
u v
u1/8 v 3/8 = u9/8 v 3/8 ;
u = 1.
Similarly, the curve y = 2x3 corresponds to u = 2, the curve x = y 3 corresponds to v = 1, and the curve
x = 4y 3 corresponds to v = 4. Next,
3
− 11/8 1/8
8u
v
∂(x, y)
= ∂(u, v)
1
−
8u9/8v 3/8
−
−
1791
1
8u3/8 v 9/8
3
8u1/8v 11/8
1
.
=
8u3/2 v 3/2
Hence the area of R is
4
2
1 dx dy =
v=1
R
1
u=1
8u3/2 v 3/2
4
−
du dv =
2
1
4u1/2 v 3/2
v=1
4
dv =
u=1
1
1
1
− √
4v 3/2
4 2 v 3/2
dv
√
√
4
2 −2
2− 2
≈ 0.07322330470336311890.
=
=
8
4v 1/2 1
13.9.12: The transformation
u=
2x
,
x2 + y 2
v=
2y
x2 + y 2
yields
u2 + v 2 =
x=
4(x2 + y 2 )
4
= 2
;
(x2 + y 2 )2
x + y2
x2 + y 2 =
1
2u
u · (x2 + y 2 ) = 2
;
2
u + v2
y=
4
;
u2 + v 2
1
2v
v · (x2 + y 2 ) = 2
.
2
u + v2
The circle x2 + y 2 = 2x is thereby transformed into
4
4u
= 2
:
u2 + v 2
u + v2
Similarly, the other three circles are transformed into u =
2(u2 − v 2 )
− 2
(u + v 2 )2
∂(x, y)
= ∂(u, v)
4uv
−
(u2 + v 2 )2
transformation is
1
3,
v = 1, and v =
1
4.
The Jacobian of this
4
= −
.
2 + v 2 )2
(u
2(v 2 − u2 ) − 2
(u + v 2 )2
−
4uv
(u2 + v 2 )2
Note also that
(x2 + y 2 )2 =
u = 1.
16
,
(u2 + v 2 )2
so that
1
(u2 + v 2 )2
.
=
(x2 + y 2 )2
16
Therefore
R
1
dx dy =
2
(x + y 2 )2
1/4
v=1
1/4
1/3
u=1
1/3
=
v=1
u=1
4
(u2 + v 2 )2
· 2
du dv
16
(u + v 2 )2
1
du dv =
4
3
1
2
1
−
· −
· = .
4
3
4
8
13.9.13: The Jacobian of the transformation x = 3r cos θ, y = 2r sin θ is
3 cos θ
∂(x, y)
= ∂(r, θ)
2 sin θ
−3r sin θ = 6r cos2 θ + 6r sin2 θ = 6r.
2r cos θ 1792
The ellipse
y2
x2
+
= 1 is transformed into
9
4
9r2 cos2 θ 4r2 sin2 θ
+
=1:
9
4
the circle r = 1.
The paraboloid has equation
z = x2 + y 2 = 9r2 cos2 θ + 4r2 sin2 θ.
Therefore the volume of the solid is
2π
1
V =
θ=0
=
0
2π
(9r2 cos2 θ + 4r2 sin2 θ) · 6r dr dθ =
2π
0
r=0
27
cos2 θ + 6 sin2 θ
2
dθ
2π
39
15
39
15 1 + cos 2θ
·
θ+
sin 2θ
π ≈ 61.26105674500096815.
=
6+
dθ =
2
2
4
8
2
0
13.9.14: The Jacobian of the transformation x = au, y = bv, z = cw (a, b, and c are positive constants)
is
a
∂(x, y, z)
= 0
∂(u, v, w)
0
The ellipsoid with equation
0
0 = abc.
c
0
b
0
x2
y2
z2
+
+
= 1
a2
b2
c2
becomes the sphere S with equation u2 + v 2 + w2 = 1. Let B denote the ball bounded by that sphere.
Then the volume of the ellipsoid is
V =
1 dx dy dz =
R
abc dV =
B
4
πabc.
3
13.9.15: We are given the transformation u = xy, v = xz, w = yz. Then uvw = x2 y 2 z 2 . Hence
u1/2 v 1/2 w1/2 = xyz = uz :
z=
v 1/2 w1/2
;
u1/2
u1/2 v 1/2 w1/2 = xyz = vy :
y=
u1/2 w1/2
;
v 1/2
u1/2 v 1/2 w1/2 = xyz = wx :
x=
u1/2 v 1/2
.
w1/2
The surface xy = 1 corresponds to the plane u = 1. Similarly, the other surfaces correspond to the planes
u = 4, v = 1, v = 9, w = 4, and w = 9. The Jacobian of the given transformation is
1793
v 1/2
2u1/2 w1/2
∂(x, y, z)
w1/2
= 2u1/2 v 1/2
∂(u, v, w)
v 1/2 w1/2
−
2u3/2
u1/2
2v 1/2 w1/2
u1/2 v 1/2
−
2w3/2
u1/2 w1/2
2v 3/2
u1/2
2v 1/2 w1/2
w1/2
2u1/2 v 1/2
2u1/2 w1/2
−
1
= − 1/2 1/2 1/2 .
2u v w
v 1/2
Therefore the volume bounded by the surfaces is
9
9
4
V =
w=4
9
v=1
9
=
4
1
u=1
1
du dv dw =
2u1/2 v 1/2 w1/2
1
dv dw =
v 1/2 w1/2
4
9
2v 1/2
w1/2
9
4
9
1
9
u1/2
1/2
v w1/2
9
4
dw =
w1/2
4
v=1
4
dv dw
u=1
9
dw = 8w1/2 = 8.
4
13.9.16: We are given the solid bounded by the paraboloids z = x2 + y 2 and z = 4(x2 + y 2 ) and the planes
z = 1 and z = 4. We are also given the transformation
r
cos θ,
t
x=
y=
r
sin θ,
t
z = r2 .
Under this transformation, the plane z = 1 corresponds to r = 1 and the plane z = 4 corresponds to r = 2.
The paraboloid z = x2 + y 2 corresponds to
r2 =
r2
r2
r2
cos2 θ + 2 sin2 θ = 2 ,
2
t
t
t
thus to t = 1. The other paraboloid yields t = 2. Finally, to obtain the entire solid, θ varies from 0 to 2π.
The Jacobian of the given transformation is
∂(x, y, z)
= ∂(r, θ, t)
1
cos θ
t
r
− sin θ
t
1
sin θ
t
r
cos θ
t
2r
0
r
cos
θ
t2
2
2
3
r
= 2r r sin2 θ + r cos2 θ = 2r .
− 2 sin θ 3
3
3
t
t
t
t
0
−
Hence the volume of the solid is
2π
2
2
V =
θ=0
r=1
2
= 2π
1
t=1
2r3
dt dr dθ = 2π
t3
1
2
−
r3
t2
2
dr
t=1
2
1
3 3
3
45
r dr = π r4 =
π ≈ 17.67145867644258696635.
4
2
4
8
1
1794
13.9.17: The substitution x = u + v, y = u − v transforms the rotated ellipse x2 + xy + y 2 = 3 into the
ellipse S in “standard position,” in which its axes lie on the coordinate axes. The resulting equation of S
(in the uv-plane) is 3u2 + v 2 = 3. The Jacobian of this transformation is
1
1 ∂(x, y)
= −2.
= ∂(u, v)
1
−1 Therefore
I=
exp(−x2 − xy − y 2 ) dx dy = 2
R
exp(−3u2 − v 2 ) du dv.
(1)
S
√
The substitution u = r cos θ, v = r 3 sin θ has Jacobian
cos θ
−r sin θ √
√
∂(u, v)
= r 3 (cos2 θ + sin2 θ) = r 3 .
= √
√
∂(r, θ)
3 sin θ
r 3 cos θ This transformation, applied to the bounding ellipse of the region S, yields
3 = 3u2 + v 2 = 3r2 cos2 θ + 3r2 sin2 θ = 3r2 ,
and thereby transforms it into the circle with polar equation r = 1. Then substitution in the second integral
in Eq. (1) yields
I=2
exp(−3u2 − u2 ) du dv = 2
2π
θ=0
S
√
r 3 exp(−3r2 ) dr dθ
1
r=0
1
√
1
2 √ 2
= 4π 3 − exp(−3r ) = π 3 1 − e−3 ≈ 3.44699122256300138528.
6
3
0
13.9.18: Remember that x = x(u, v), y = y(u, v), u = u(x, y), and v = v(x, y). Then
x
∂(x, y) ∂(u, v) u
·
=
∂(u, v) ∂(x, y) yu
xv ux
·
yv vx
xu ux + xv vx
= yu ux + yv vx
uy vy xu uy + xv vy =
yu uy + yv vy ∂x
∂x
∂y
∂x
1
=
∂y 0
∂y ∂x
∂y
0 = 1.
1
13.9.19: Suppose that k is a positive constant. First we need an integration by parts with
u = ρ2
and
dv = ρ exp(−kρ2 ) dρ :
du = 2ρ dρ
and
v=−
1795
1
exp(−kρ2 ).
2k
Thus
1
ρ exp(−kρ ) dρ = − ρ2 exp(−kρ2 ) +
2k
3
2
=−
1
ρ exp(−kρ2 ) dρ
k
1 2
1
ρ exp(−kρ2 ) − 2 exp(−kρ2 ) + C.
2k
2k
Then the improper triple integral given in Problem 19 will be the limit of Ia as a → +∞, where
2π π a
Ia =
ρ3 exp(−kρ2 ) sin φ dρ dφ dθ
θ=0
φ=0
ρ=0
π
= 2π − cos φ
−
φ=0
1
1 2
ρ exp(−kρ2 ) − 2 exp(−kρ2 )
2k
2k
a
ρ=0
1
1
1
= 4π − a2 exp(−ka2 ) − 2 exp(−ka2 ) + 2 .
2k
2k
2k
Because k > 0, it is clear that Ia →
2π
as a → +∞.
k2
13.9.20: Given: The solid ellipsoid R with constant density δ and boundary surface with equation
x2
y2
z2
+
+
= 1
a2
b2
c2
(1)
(where a, b, and c are positive constants). The transformation
x = aρ sin φ cos θ,
y = bρ sin φ sin θ,
z = cρ cos φ
has Jacobian
a sin φ cos θ
∂(x, y, z) = b sin φ sin θ
J=
∂(ρ, φ, θ) c cos φ
−aρ sin φ sin θ bρ sin φ cos θ 0
aρ cos φ cos θ
bρ cos φ sin θ
−cρ sin φ
= abcρ2 cos2 φ sin φ cos2 θ + abcρ2 sin3 φ cos2 θ + abcρ2 cos2 φ sin φ sin2 θ + abcρ2 sin3 φ sin2 θ
= abcρ2 sin φ.
This transformation also transforms the ellipsoidal surface of Eq. (1) into
ρ2 sin2 φ cos2 θ + ρ2 sin2 φ sin2 θ + ρ2 cos2 φ = ρ2 sin2 φ + ρ2 cos2 φ = ρ2 = 1,
and thereby transforms R into the solid ball B of radius 1 and center at the origin. Therefore the mass of
the ellipsoid is
2π M=
θ=0
π
φ=0
1
δabcρ2 sin φ dρ dφ dθ = 2πδabc
ρ=0
π
φ=0
1796
π
1
1
4
sin φ dφ = 2πδabc − cos φ = πδabc.
3
3
3
0
13.9.21: Given: The solid ellipsoid R with constant density δ and boundary surface with equation
x2
y2
z2
+ 2 + 2 = 1
2
a
b
c
(1)
(where a, b, and c are positive constants). The transformation
x = aρ sin φ cos θ,
y = bρ sin φ sin θ,
z = cρ cos φ
has Jacobian
a sin φ cos θ
∂(x, y, z) J=
= b sin φ sin θ
∂(ρ, φ, θ) c cos φ
aρ cos φ cos θ
bρ cos φ sin θ
−cρ sin φ
−aρ sin φ sin θ bρ sin φ cos θ 0
= abcρ2 cos2 φ sin φ cos2 θ + abcρ2 sin3 φ cos2 θ + abcρ2 cos2 φ sin φ sin2 θ + abcρ2 sin3 φ sin2 θ
= abcρ2 sin φ.
This transformation also transforms the ellipsoidal surface of Eq. (1) into
ρ2 sin2 φ cos2 θ + ρ2 sin2 φ sin2 θ + ρ2 cos2 φ = ρ2 sin2 φ + ρ2 cos2 φ = ρ2 = 1,
and thereby transforms R into the solid ball B of radius 1 and center at the origin. Therefore the moment
of inertia of the ellipsoid with respect to the z-axis is
2π
π
1
(ρ2 sin2 φ)(a2 cos2 θ + b2 sin2 θ)δabcρ2 sin φ dρ dφ dθ
Iz =
θ=0
φ=0
2π π
=
0
0
2π
ρ=0
1
(δabcρ5 sin3 φ)(a2 cos2 θ + b2 sin2 θ)
5
1
dφ dθ
ρ=0
π
1
(δabc sin3 φ)(a2 cos2 θ + b2 cos2 θ) dφ dθ
0
0 5
π
2π 1
2
2
2
2
δabc
=
(cos 3φ − 9 cos φ)(a cos θ + b sin θ) dθ
60
0
0
=
4
δabc
=
15
=
0
2π
2π
1
2
2
2
2
δabc 2a θ + 2b θ + a sin 2θ − b sin 2θ
(a cos θ + b sin θ) dθ =
15
0
2
2
2
2
1
δabc(4πa2 + 4πb2 ).
15
Because the mass of the sphere (found in the solution of Problem 20) is M = 43 πδabc, we see that
Iz
1
= (a2 + b2 ),
M
5
and hence that Iz =
1797
1
M (a2 + b2 ).
5
13.9.22: Given u = xy and v =
y
, we have
x
uv = xy ·
y
= y2
x
and thus we choose
x=
u
v
and
x
u
= xy · = x2 ,
v
y
and
y=
√
uv .
(1)
The Jacobian of this transformation is
1
2u1/2 v 1/2
∂(x, y) =
∂(u, v) v 1/2
2u1/2
u1/2
− 3/2
2v
u1/2
2v 1/2
= 1 .
2v
Also, if y = x, then substitution of the equations in (1) yields
(uv)1/2 =
u1/2
;
v 1/2
uv =
u
;
v
v 2 = 1.
So we choose v = 1. (This choice implies that if we have a similar choice with u, we must choose u > 0
because of the equations in (1).) Similarly, y = 2x yields v = 2, xy = 1 yields u = 1, and xy = 2 yields
u = 2. Hence the area of the region of Fig. 13.9.7 is
A=
2
2
1 dx dy =
R
v=1
u=1
1
du dv =
2v
2
1
1
1
dv = ln 2 ≈ 0.3465735902799727.
2v
2
Its moments with respect to the coordinate axes are
2
2
My =
v=1
2
Mx =
v=1
u=1
1 u1/2
·
du dv =
2v v 1/2
2
v=1
u3/2
3v 3/2
2
2
dv =
u=1
1
√ 2
√
√
2−4 2
2 2 −1
5 2 −6
;
dv =
=
3
3v 3/2
3v 1/2
1
2
2
u3/2
1 √
· uv du dv =
dv
1/2
u=1 2v
v=1 3v
u=1
√
√
2
2 √
4 2 − 2 1/2
10 − 6 2 ,
2 2 −1
v
.
dv =
=
=
3
3
3v 1/2
1
1
2
Hence the coordinates of its centroid are
√
√ 10 2 − 12 20 − 12 2
(x, y) =
,
≈ (1.030149480423, 1.456851366485).
3 ln 2
3 ln 2
13.9.23: If u = xy and v = xy 3 , then
uy 2 = xy 3 = v,
so that
1798
y2 =
v
;
u
y=
v 1/2
.
u1/2
Then
x=
u1/2
u
u3/2
= u · 1/2 = 1/2 .
y
v
v
(We do not need the solution in which x and y are negative.) Then
3u1/2
u3/2 −
1/2
2v 3/2 ∂(x, y) 2v
= 3 − 1 = 1 .
=
4v 4v
∂(u, v) 2v
1
v 1/2
− 3/2
2u
2u1/2 v 1/2
We also find by substitution that xy = 2 corresponds to u = 2, xy = 4 corresponds to u = 4, xy 3 = 3
corresponds to v = 3, and xy 3 = 6 corresponds to v = 6. Hence the area of the region shown in Fig. 13.9.8
is
6
4
1 dx dy =
A=
v=3
D
u=2
1
du dv =
2v
6
3
1
dv = ln 2 ≈ 0.6931471805599453.
v
Its moments with respect to the coordinate axes are
√
6 5/2 4
6
u
1 u3/2
32 − 4 2
My =
dv =
dv
· 1/2 du dv =
3/2
5v 3/2
v=3 u=2 2v v
v=3 5v
3
u=2
√
√
6
√
8 2 − 64
72 3 − 40 6
;
=
=
15
5v 1/2
3
√
6 4
6 1/2 4
6
u
1 v 1/2
2− 2
Mx =
· 1/2 du dv =
dv =
dv
1/2
v 1/2
v=3 u=2 2v u
v=3 v
3
u=2
=
6
4
6
√ √
√
4 − 2 2 v 1/2 = 6 6 − 8 3 .
3
Therefore its centroid is located at the point
√
√
√
√ 72 3 − 40 6 6 6 − 8 3
,
≈ (2.570696785449, 1.212631342551).
(x, y) =
15 ln 2
ln 2
13.9.24: If y = ux2 and x = vy 2 , then
y = uv 2 y 4 ;
y3 =
1
;
uv 2
y=
1
u1/3 v 2/3
.
Then it follows that
x = vy 2 =
Next,
v
1
= 2/3 1/3 .
u2/3 v 4/3
u v
2
− 5/3 1/3
3u
v
∂(x, y) =
∂(u, v) 1
−
3u4/3 v 2/3
−
−
1799
1
3u2/3 v 4/3
2
3u1/3 v 5/3
1
= 2 2.
3u v
Next, y = x2 corresponds to u = 1, y = 2x2 corresponds to u = 2, x = y 2 corresponds to v = 1, and
x = 4y 2 corresponds to v = 4. Therefore the area of the region shown in Fig. 13.9.9 is
4
2
1 dx dy =
A=
v=1
R
u=1
1
du dv =
3u2 v 2
4
v=1
−
1
3uv 2
2
dv =
1
u=1
4
4
1
1
1
dv
=
−
= .
2
6v
6v 1
8
Its moments with respect to the coordinate axes are
4
2
My =
v=1
4
=
v=1
u=1
2
Mx =
v=1
=
1
4
4
−
v=1
1
5u5/3 v 7/3
2
dv
u=1
4
3 · 21/2 − 12
4 − 21/3
96 − 36 · 21/3 + 3 · 22/3
;
dv
=
=
7/3
4/3
640
20v
80v
1
4
1
1
·
du dv =
3u2 v 2 u2/3 v 1/3
u=1
1
1
·
du dv =
3u2 v 2 u1/3 v 2/3
4
−
v=1
1
4u4/3 v 8/3
2
dv
u=1
4
96 + 3 · 21/3 − 30 · 22/3
3 · 22/3 − 12
4 − 22/3
.
dv
=
=
8/3
5/3
640
16v
80v
1
Therefore its centroid is located at the point with coordinates
(x, y) =
96 − 36 · 21/3 + 3 · 22/3 96 + 3 · 21/3 − 30 · 22/3
,
80
80
≈ (0.692563066996, 0.651971644883).
13.9.25: Given: The solid ellipsoid R with constant density δ and boundary surface with equation
y2
z2
x2
+
+
= 1
a2
b2
c2
(1)
(where a, b, and c are positive constants). The transformation
x = aρ sin φ cos θ,
y = bρ sin φ sin θ,
z = cρ cos φ
has Jacobian
a sin φ cos θ
∂(x, y, z) J=
= b sin φ sin θ
∂(ρ, φ, θ) c cos φ
aρ cos φ cos θ
bρ cos φ sin θ
−cρ sin φ
−aρ sin φ sin θ bρ sin φ cos θ 0
= abcρ2 cos2 φ sin φ cos2 θ + abcρ2 sin3 φ cos2 θ + abcρ2 cos2 φ sin φ sin2 θ + abcρ2 sin3 φ sin2 θ
= abcρ2 sin φ.
This transformation also transforms the ellipsoidal surface of Eq. (1) into
ρ2 sin2 φ cos2 θ + ρ2 sin2 φ sin2 θ + ρ2 cos2 φ = ρ2 sin2 φ + ρ2 cos2 φ = ρ2 = 1,
1800
and thereby transforms R into the solid ball B of radius 1 and center at the origin. Note also that
x2 + y 2 = a2 ρ2 sin2 φ cos2 θ + b2 r2 sin2 φ sin2 θ = (ρ2 sin2 φ)(a2 cos2 θ + b2 sin2 θ).
Assume that the solid R has constant density δ. Then its moment of inertia with respect to the z-axis is
2π
π
1
(ρ2 sin2 φ)(a2 cos2 θ + b2 sin2 θ)δabcρ2 sin φ dρ dφ dθ
Iz =
θ=0
2π
φ=0
π
1
(δabc sin3 φ)(a2 cos2 θ + b2 sin2 θ) dφ dθ
5
=
0
0
2π
=
0
ρ=0
4
4
1
(δabc)(a2 cos2 θ + b2 sin2 θ) dθ =
πδabc(a2 + b2 ) = M (a2 + b2 )
15
15
5
where M is the mass of the ellipsoid. By symmetry,
Iy =
4
1
πδabc(a2 + c2 ) = M (a2 + c2 )
15
5
and
4
1
πδabc(b2 + c2 ) = M (b2 + c2 ).
15
5
Ix =
13.9.26: Assume that the solid of Problem 16 has constant density δ. By symmetry its centroid lies on
the z-axis. A consequence of the solution of Problem 16 is that the solid has mass M =
45
8 πδ.
Its moment
with respect to the xy-plane is
2π
2
2
Mxy =
2
θ=0
r=1
2
= 2π
1
r 3
t
t=1
· r2 dt dr dθ =
2π
2
2
Iy = Ix = δ
θ=0
2π
r=1
2
=δ
θ=0
2π
r=1
2
=δ
θ=0
=δ
0
2π
r=1
t=1
0
2
−
1
r5
t2
2
dr dθ
t=1
2
1 6
63
3 5
63
r dr = 2π r
=
π.
= 2π ·
4
8
8
4
1
Therefore the centroid of the solid is located at the point 0, 0,
2π
14
5
. Next, by symmetry,
2
r
r 3
4
sin θ + r · 2
dt dr dθ
t
t
r5 (1 + 4r2 t2 − cos 2θ)
−
4t4
2
dr dθ
t=1
3 5
r (5 + 16r2 − 5 cos 2θ) dr dθ = δ
64
765 315 2
+
sin θ
32
64
dθ = δ
2π
θ=0
45
(7 sin 2θ − 150θ)
64
3 8
5 2 2
r +
r sin θ
32
64
2π
=
0
Thus Ix = Iy ≈ (165.6699250916492528)δ. Finally, to compute Iz , note that
x2 + y 2 =
r
t
cos θ
2
+
1801
r
t
sin θ
2
=
r2
.
t2
3375
πδ.
64
2
dθ
r=1
Therefore the moment of inertia of the solid with respect to the z-axis is
2π
2
2
2r5
dt dr dθ = δ
t5
Iz = δ
θ=0
r=1
2π
2
=δ
θ=0
r=1
t=1
2π
θ=0
2
r=1
r5
− 4
2t
2
dr dθ
t=1
2
5 6
15 5
315
r dr = 2πδ
r
πδ ≈ (30.9250526837745272)δ.
=
32
64
32
1
13.9.27: The average distance of points of the ellipsoid from its center at (0, 0, 0) is
d=
1
V
0
4
3 πabc
where V =
2π
π
0
1
(abcρ2 sin φ) (aρ sin φ cos θ)2 + (bρ sin φ sin θ)2 + (cρ cos φ)2 dρ dφ dθ
0
is the volume of the ellipsoid. In particular, if a = 4, b = 3, and c = 2, we find (using
the NIntegrate command in Mathematica) that d ≈ 2.300268522983.
13.9.28: Following the instructions in Problem 28, we have
1
0
1
0
1
dx dy =
1 − xy
1
1
1+
0
=1+
0
∞
n n
x y
∞ dx dy = 1 +
n=1
∞ n=1
0
n=1
1
xn dx ·
1
y n dy
0
1
1
1
1
1
·
= 1 + 2 + 2 + 2 + · · · = ζ(2).
n+1
n+1
2
3
4
13.9.29: Part (a): First note that
1
0
1
0
1
1
−
1 − xy
1 + xy
1
1
2xy
dx dy.
1 − x2 y 2
dx dy =
0
The Jacobian of the substitution u = x2 , v = y 2 is
2x
∂(u, v)
= 0
∂(x, y)
0
0 = 4xy,
2y so
0
1
1
0
2xy
1
dx dy =
1 − x2 y 2
2
0
1
0
1
1
1
· 4xy dx dy =
1 − x2 y 2
2
0
1
0
1
1
1
du dv = ζ(2).
1 − uv
2
Part (b): Addition as indicated in Problem 29, and cancellation of the integrals involving 1/(1 + xy), yields
the equation
1
2
0
0
1
1
1
dx dy = ζ(2) + 2
1 − xy
2
0
1
0
1
1
dx dy,
1 − x2 y 2
which we readily solve for
0
1
0
1
1
dx dy =
1 − x2 y 2
0
1
0
1
1
3
1
dx dy − ζ(2) = ζ(2).
1 − xy
4
4
1802
Part (c): The Jacobian of the transformation T : R2uv → R2xy that we define by x = (sin v)/(cos u),
y = (sin u)/(cos v) is
cos u
cos v
= sin u sin v
−
cos2 u
JT
−
sin u sin v
cos2 v
cos v
cos u
2
2
= 1 − sin u sin v = 1 − tan2 u tan2 v.
cos2 u cos2 v
Reading the limits for the transformed integral from Fig. 14.9.10(a) in the text, we therefore find that
4
3
ζ(2) =
4
3
=
4
=
3
1
0
1
π/2
0
1
dx dy
1 − x2 y 2
0
(π/2)−v
0
π/2
π/2
4 π
1 2
π2
4 π2
− v dv =
v − v
·
=
.
=
2
3 2
2
3 8
6
π
0
−1
4 π/2 (π/2)−v
sin2 u sin2 v
2
2
·
(1
−
tan
u
tan
v)
du
dv
=
1 du dv
1−
cos2 u cos2 v
3 0
0
0
Chapter 13 Miscellaneous Problems
13.M.1: The domain of the given integral is bounded above by the graph of y = x3 , below by the x-axis,
and on the right by the vertical line x = 1. When its order of integration is reversed, the given integral
becomes
1
x=0
x3
y=0
1
√
dy dx =
1 + x2
1
0
=
x3
√
dx
1 + x2
√
1
1 2
2− 2
(x − 2) 1 + x2
≈ 0.1952621458756350.
=
3
3
0
Mathematica can evaluate the given integral without first reversing the order of integration. It obtains
1
y=0
1
x=y 1/3
1
√
dx dy =
1 + x2
1
0
arcsinh(1) − arcsinh(y 1/3 ) dy
y 2/3 − 2 =
1 + y 2/3 + y arcsinh(1) − y arcsinh(y 1/3 )
3
1
0
√
2− 2
.
=
3
13.M.2: The given integral is improper, but the only discontinuity occurs at (0, 0), a corner point of the
domain of the integral. If we define the value of the integrand to be 1 at that point, the integrand will be
continuous on its entire domain and the problem vanishes. The domain of the given integral is bounded
1803
above by the line y = x, below by the x-axis, and on the right by the vertical line x = 1. When its order of
integration is reversed, the given integral becomes
1
x=0
x
y=0
sin x
dy dx =
x
1
y sin x
x
0
1
=
0
x
dx
y=0
1
sin x dx = − cos x
0
= 1 − cos(1) ≈ 0.4596976941318602.
Mathematica can evaluate the given integral without first reversing the order of integration. It obtains
1
y=0
1
sin x
dx dy =
x
x=y
1
1
dy
SinIntegral(x)
0
x=y
1
= y SinIntegral(1) − y SinIntegral(y) − cos y = 1 − cos(1).
0
13.M.3: The domain of the given integral is bounded above by the line y = 1, below and on the right by
the line y = x, and on the left by the y-axis. When its order of integration is reversed, it becomes
1
y
2
1
exp(−y ) dx dy =
y=0
0
x=0
1
y exp(−y ) dy = − exp(−y 2 )
2
2
1
0
1
e−1
1
−
=
≈ 0.3160602794142788.
2 2e
2e
=
Mathematica can evaluate the given integral without first reversing the order of integration:
1
x=0
1
exp(−y 2 ) dy dx =
y=x
1
x=0
1√
π erf(y)
2
1
1
1√ π erf(1) − erf(x) dx
2
dx =
y=x
0
1
1 √
e−1
2
=
.
π x erf(1) − erf(x) − exp(−x ) =
2
2e
0
2
Note: By definition, the error function is erf(x) = √
π
x
exp(−t2 ) dt.
0
13.M.4: The domain of the given integral is bounded above by the line y = 4, below and on the right by
the graph of y = x2/3 , and on the left by the y-axis. When its order of integration is reversed, the integral
takes the form
4
y 3/2
4
4
x cos y dx dy =
y=0
y=0
x=0
=
1 2
x cos y 4
2
1
sin y 4
8
4
=
0
1804
y3/2
dy =
x=0
0
4
1 3
y cos y 4 dy
2
1
sin 256 ≈ −0.1249010042633828.
8
Mathematica can evaluate the given integral without first reversing the order of integration, but all the
antiderivatives that it uses involve the gamma function with complex arguments—we omit the details—and
it expresses the final answer in the form
1 i 1 − exp (512i) · exp (−256i)
16
where i2 = −1. You will need Euler’s formula eiθ = cos θ + i sin θ to show that this answer is equal to the
previous answer.
13.M.5: The domain of the given integral is bounded above by the graph of y = x2 , below by the x-axis,
and on the right by the line x = 2. When its order of integration is reversed, it becomes
2
x=0
x2
y=0
y exp(x2 )
dy dx =
x3
2
y 2 exp(x2 )
2x3
x=0
2
=
0
x 2
dx
y=0
2
1
1
e4 − 1
2
2
x exp(x ) dx =
exp(x ) =
≈ 13.3995375082860598.
2
4
4
0
Mathematica can evaluate the given integral without first reversing the order of integration, but the intermediate antiderivatives involve the exponential integral function
∞ −t
e
Ei(z) =
dt
t
z
and hence we omit the details.
13.M.6: Here we obtain
∞ ∞
∞ y
1 −y
1 −y
e dy dx =
e dx dy
y
y
x=0 y=x
y=0 x=0
y
∞
∞
∞
x −y
e
=
dy =
e−y dy = −e−y
= 1 − lim e−y = 1.
y→∞
y
0
0
x=0
0
13.M.7: The volume is
1 2−y
(x2 + y 2 ) dx dy =
V =
y=0
x=y
1
y=0
=
0
1
1 3
x + xy 2
3
2−y
dy
x=y
4
(2 − 3y + 3y 2 − 2y 3 ) dy =
3
8
4
2
y − 2y 2 + y 3 − y 4
3
3
3
1
=
0
4
.
3
13.M.8: The paraboloids intersect in the circle x2 + y 2 = 16, z = 32. Hence the volume between them is
2π
4
48−r 2
r dz dr dθ = 2π
V =
θ=0
r=0
= 2π 24r2 −
z=2r 2
3 4
r
4
4
0
0
4
(48r − 3r3 ) dr
= 2π(384 − 192) = 384π ≈ 1206.3715789784806036.
1805
13.M.9: By symmetry, the centroid lies on the z-axis. Assume that the solid has unit density. Then its
mass and volume are
2π
π/2
3
ρ2 sin φ dρ dφ dθ
m=V =
θ=0
φ=π/3
ρ=0
π/2 3
1 3
1
= 2π · − cos φ
ρ
·
= 18π · 0 − −
= 9π ≈ 28.2743338823081391.
3
2
π/3
0
The moment of the solid with respect to the xy-plane is
2π
π/2
3
Mxy =
3
π/2
ρ sin φ cos φ dρ dφ dθ = 2π
θ=0
φ=π/3
π/2
= 2π
φ=π/3
ρ=0
φ=π/3
1 4
ρ sin φ cos φ
4
3
dφ
ρ=0
π/2
81
81
81
=
sin φ cos φ dφ = 2π − cos2 φ
π.
4
8
16
φ=π/3
Therefore the centroid of the solid is located at the point with coordinates
9
(x, y, z) = 0, 0,
.
16
13.M.10: The elliptic paraboloids intersect in a curve that lies on the elliptical cylinder with Cartesian
equation x2 + 4y 2 = 4. The intersection of that cylinder with the xy-plane forms the elliptical boundary of
a region R suitable for the domain of the volume integral, which is
V =
(8 − 2x2 − 8y 2 ) dx dy.
R
Let us use the substitution x = 2r cos θ, y = r sin θ. The boundary of R takes the form
4r2 cos2 θ + 4r2 sin2 θ = 4,
and hence is transformed into the circle r = 1. The Jacobian of this transformation is
2 cos θ
−2r sin θ ∂(x, y)
= 2r.
= ∂(r, θ)
sin θ
r cos θ Therefore the volume intgral in Eq. (1) is transformed into
2π
1
(8 − 8r2 cos2 θ − 8r2 sin2 θ) · 2r dr dθ
V =
θ=0
2π
r=0
=
0
0
1
16(r − r3 ) dr dθ = 32π
1 2 1 4
r − r
2
4
1806
1
0
= 8π ≈ 25.1327412287183459.
(1)
13.M.11: First interchange y and z: We are to find the volume bounded by the paraboloid z = x2 + 3y 2
and the cylinder z = 4 − y 2. These surfaces intersect in a curve that lies on the elliptic cylinder x2 + 4y 2 = 4,
bounding the region R in the xy-plane. Hence the volume is
V =
(4 − x2 − 4y 2 ) dx dy.
R
Apply the transformation x = 2r cos θ, y = r sin θ. This transforms R into the region 0 r 1, 0 θ 2π.
Moreover, the Jacobian of this transformation is
2 cos θ
∂(x, y)
= ∂(r, θ)
sin θ
−2r sin θ = 2r.
r cos θ Hence
2π
1
4
1
0
r=0
13.M.12: The volume is
1 2−x2
(4 − z) dz dx =
V =
x=−1
2
(4 − 4r ) · 2r dr dθ = 2π · 4r − 2r
V =
θ=0
2
z=x2
1
−1
4z −
1 2
z
2
= 4π ≈ 12.56637061435917295385.
2−x2
1
dx =
−1
x2
1
(6 − 6x2 ) dx = 6x − 2x3
= 8.
−1
13.M.13: First interchange x and z: We are to find the volume enclosed by the elliptical cylinder 4x2 +y 2 =
4 and between the planes z = 0 and z = y + 2. Let R denote the plane region in which the elliptical cylinder
meets the xy-plane. Then the volume is
(y + 2) dx dy.
V =
R
The transformation x = r cos θ, y = 2r sin θ transforms R into the rectangle 0 r 1, 0 θ 2π. The
Jacobian of this transformation is
−r sin θ = 2r.
2r cos θ cos θ
∂(x, y)
= ∂(r, θ)
2 sin θ
Therefore
2π
1
V =
θ=0
=
0
2π
r=0
(2 + 2r sin θ) · 2r dr dθ =
2π
0
4
2r + r3 sin θ
3
2
1
dθ
r=0
2π
4
4
= 4π ≈ 12.56637061435917295385.
2 + sin θ dθ = 2θ − cos θ
3
3
0
13.M.14: Let R denote the bounded plane region with boundary the ellipse
x2
y2
+
= 1;
a2
b2
1807
(1)
we assume that a > 0, b > 0, and that a < h (this ensures that the plane z = h + x is above R). Then the
volume within the elliptical cylinder and between the planes z = 0 and z = h + x is
(h + x) dx dy.
V =
R
This transforms R into the region 0 r 1,
Apply the transformation x = ar cos θ, y = br sin θ.
0 θ 2π (you can see this by substitution into Eq. (1)). Moreover, the Jacobian of this transformation is
a cos θ
∂(x, y)
= ∂(r, θ)
b sin θ
−ar sin θ = abr(cos2 θ + sin2 θ) = abr.
br cos θ Consequently,
2π
1
V =
0
0
2π
=
0
(h + ar cos θ) · abr dr dθ =
1
1
abh + a2 b cos θ
2
3
dθ =
2π
0
1
1
abhr2 + a2 br3 cos θ
2
3
1
1
abhθ − a2 b sin θ
2
3
1
dθ
0
2π
= πabh.
0
13.M.15: The graph of x4 + x2 y 2 = y 2 in the first quadrant is the graph of
y=√
x2
,
1 − x2
This curve meets the line y = x at the point
1 √
2
2,
0 x < 1.
1
2
√ 2 and, of course, at the point (x, y) = (0, 0).
Conversion of the first equation into polar form yields
r4 cos4 θ + r4 cos2 θ sin2 θ = r2 sin2 θ;
(r2 cos2 θ)(cos2 θ + sin2 θ) = sin2 θ;
r2 cos2 θ = sin2 θ;
thus r = tan θ. Noting that the line y = x has polar equation θ = 14 π, we find that
R
1
(1 + x2 + y 2 )2
π/4
tan θ
dA =
θ=0
=
0
π/4
r=0
r
dr dθ =
(1 + r2 )2
0
π/4
1
−
2(1 + r2 )
tan θ
dθ
r=0
π/4
1
1
π−2
(1 − cos2 θ) dθ =
≈ 0.0713495408493621.
=
2θ − sin 2θ
2
8
16
0
1808
13.M.16: The mass and moments are
1 √x
(x2 + y 2 ) dy dx =
m=
√x
1
1 3/2
1 3
1
x + x5/2 − x4 − x6 dx
dx =
x y+ y
3
3
3
0
x2
0
0
x2
1
2 5/2 2 7/2 1 5
1 7
6
x + x − x −
x
≈ 0.1714285714285714;
=
=
15
7
5
21
35
0
1
2
√x
1 3
My =
x y + xy
x · (x + y ) dy dx =
dx
3
0
x2
0
x2
1
1
1 5/2
2 7/2 2 9/2 1 6
1
1 8
55
x + x7/2 − x5 − x7 dx =
x + x − x −
x
;
=
=
3
3
21
9
6
24
504
0
0
1
√
x
2
2
1
3
√ x
1 2 2 1 4
Mx =
x y + y
y · (x + y ) dy dx =
dx
2
4
0
x2
0
x2
1
1
1 2 1 3 1 6 1 8
1 3 1 4
1 7
1 9
55
=
x + x − x − x dx =
x + x −
x −
x
.
=
4
2
2
4
12
8
14
36
504
0
0
1
√
x
2
2
1
Therefore the centroid of the lamina is located at the point with coordinates
275 275
,
≈ (0.636574074074, 0.636574074074).
(x, y) =
432 432
13.M.17: The mass and and moments are
4+y2
2
2 4+y2
2
2
y dx dy =
dy
xy
m=
2y 2
−2
2
=
−2
Mx =
4+y 2
3
2y 2
4 3 1 5
y − y
3
5
2
y dx dy =
2y 2
−2
My =
(4y 2 − y 4 ) dy =
2
−2
2
−2
4+y 2
2y 2
−2
xy
3
2
=
−2
4+y2
2y 2
128
≈ 8.5333333333333333;
15
2
1
dx =
(4y − y ) dy = y − y 6
6
−2
3
5
4
2
= 0;
−2
2
8 3 4 5
3 6
3 7
4096
2
4
y + y −
y
.
xy dx dy =
=
8y + 4y − y dy =
2
3
5
14
105
−2
−2
2
2
Therefore the centroid of the lamina is located at the point with coordinates
32
, 0 ≈ (4.5714285714285714, 0).
(x, y) =
7
13.M.18: The mass and moments are
2 ln x
2
2 ln x
y
1
ln x
dy dx =
dx
dx =
m=
x
x
x
1
0
1
1
0
2
1
1
2
=
(ln x)
= (ln 2)2 ≈ 0.2402265069591007;
2
2
1
My =
1
Mx =
2
1
ln x
0
2
0
ln x
x
dy dx =
x
y
dy dx =
x
2
1
1
2
2
ln x dx = −x + x ln x
y2
2x
ln x
dx =
1
0
1809
2
1
= −1 + 2 ln 2;
2
1
(ln x)2
1
dx =
(ln x)3 = (ln 2)3 .
2x
6
6
1
Therefore the centroid of the lamina is located at the point
(x, y) =
−2 + 4 ln 2 ln 2
,
(ln 2)2
3
≈ (1.608042201545, 0.231049060187).
13.M.19: The mass and moments are
π/2
2 cos θ
π/2
kr dr dθ =
m=
θ=−π/2
π/2
r=0
θ=−π/2
2k cos2 θ dθ =
=
θ=−π/2
π/2
2 cos θ
1
k(2θ + sin 2θ)
2
2
My =
1 2
kr
2
2 cos θ
dθ
r=0
π/2
= kπ;
θ=−π/2
π/2
1 3
kr cos θ
3
kr cos θ dr dθ =
θ=−π/2
π/2
=
θ=−π/2
r=0
2 cos θ
kr2 sin θ dr dθ =
Mx =
θ=−π/2
π/2
=
θ=−π/2
dθ
r=0
π/2
1
8
k cos4 θ dθ =
k(12θ + 8 sin 2θ + sin 4θ)
= kπ;
3
12
θ=−π/2
π/2
θ=−π/2
2 cos θ
r=0
π/2
1 3
kr sin θ
3
θ=−π/2
2 cos θ
dθ
r=0
π/2
8
2
3
4
k cos θ sin θ dθ = − k cos θ
= 0.
3
3
θ=−π/2
Therefore the centroid of the lamina is located at the point (1, 0).
13.M.20: The mass and moments are
π/2
2 cos θ
m=
π/2
r dr dθ =
−π/2
π/2
=
−π/2
My =
2
π/2
−π/2
0
−π/2
8
cos3 θ dθ =
3
2 cos θ
1 3
r
3
2 cos θ
dθ
0
2
(9 sin θ + sin 3θ)
9
r3 cos θ dr dθ =
0
π/2
−π/2
π/2
32
≈ 3.5555555555555555;
9
=
−π/2
1 4
r cos θ
4
2 cos θ
dθ
0
π/2
1
64
=
;
4 cos θ dθ =
=
150 sin θ + 25 sin 3θ + 3 sin 5θ
60
15
−π/2
−π/2
Mx =
π/2
π/2
2 cos θ
3
π/2
r sin θ dr dθ =
−π/2
0
−π/2
1 4
r sin θ
4
2 cos θ
dθ
0
π/2
4
4 cos4 θ sin θ dθ = − cos5 θ
= 0.
5
−π/2
−π/2
=
5
π/2
Therefore the centroid of the lamina is located at the point
1810
6
5,
0 .
13.M.21: By the first theorem of Pappus, the y-coordinate y of the centroid must satisfy the equation
2πy ·
4
1
πab = πab2 ,
2
3
4b
.
3π
and it follows immediately that y =
13.M.22: Part (a): By symmetry, x = y, and by the first theorem of Pappus, y must satisfy the equation
2πy
and therefore
1 2 1 2
πb − πa
4
4
4 · 3a2
2a
4(a2 + ab + b2 )
=
=
.
3π(a + b)
3π · 2a
π
b→a
=
2
π(b3 − a3 ),
3
4(a2 + ab + b2 )
4(b3 − a3 )
=
.
2
2
3π(b − a )
3π(a + b)
y=
Part (b): lim
Therefore the centroid of the region is located at
2a 2a
,
.
π π
13.M.23: Assume that the lamina has constant density δ. By symmetry, x = 0. The mass of the lamina
and its moment with respect to the x-axis are
2
4−x2
m=
x=−2
y=0
2
2
x=−2
y=0
=
2
δy dy dx =
x=−2
1
δ dy dx =
δ(4 − x ) dx = δ 4x − x3
3
x=−2
4−x2
Mx =
2
8
1
1
δ 16x − x3 + x5
2
3
5
2
=
−2
1 2
δy
2
4−x2
dx =
y=0
1
δ
2
2
=
−2
2
32
δ
3
and
(16 − 8x2 + x4 ) dx
x=−2
256
δ.
15
Therefore the centroid of the lamina is at the point 0, 85 .
13.M.24: The volume of the solid is
1
1−y
V =
2
x dx dy =
0
0
0
1
1 3
x
3
1−y
dy =
0
0
1
1
1
1
1
3
4
(1 − y) dy = − (1 − y)
.
=
3
12
12
0
13.M.25: The volume of the ice-cream cone is
2π
1
√
5−r 2
V =
1
r dz dr dθ = 2π
θ=0
r=0
r=0
z=2r
2
= − π · (5 − r2 )3/2 + 2r3
3
r 5 − r2 − 2r2 dr
1
0
10 √
π
=
5 − 2 ≈ 2.472098079537133054103626.
3
1811
13.M.26: Assume constant density δ = 1. Clearly x = y = 0. The mass and moment with respect to the
xy-plane are
2π
π/3
a
m=
π/3
1 3
1
1 3
a sin φ dφ = 2π − a cos φ
= πa3
3
3
3
0
π/3
ρ sin φ dρ dφ dθ = 2π
0
Mxy =
2
0
2π
0
0
π/3
0
π/3
= 2π
0
0
a
and
ρ3 sin φ cos φ dρ dφ dθ
0
π/3
1 4 2
1 4
3
a sin φ cos φ dφ = 2π a sin φ
πa4 .
=
4
8
16
0
Therefore the centroid is located at the point
(x, y, z) =
Because the density is 1, the volume is V =
9
a .
0, 0,
16
1 3
πa (numerically the same as the mass).
3
13.M.27: Let δ be the [constant] density of the cone and let h denote its height. Place the cone with its
vertex at the origin and with its axis lying on the nonnegative z-axis. Then its mass is M = 13 πδa2 h. The
side of the cone has cylindrical equation z = hr/a, so the moment of inertia of the cone with respect to the
z-axis is
2π
a
h
Iz =
θ=0
= 2πδ
r=0
δr3 dz dr dθ = 2πδ
z=hr/a
a
r3 h −
r=0
1 4
1 5
r h−
r h
4
5a
a
=
r=0
r4 h
a
dr
1
3
πδa4 h =
M a2 .
10
10
Note that the answer is plausible and dimensionally correct. One of our physics teachers, Prof. J. J. Kyame
of Tulane University (retired), always insisted that we express moment of inertia in terms of mass, as here,
so that the answer can be inspected for plausibility and dimensional accuracy.
13.M.28: The mass is
π/2 π/2 a
m=
(ρ3 sin2 φ cos φ sin θ cos θ) · (ρ2 sin φ) dρ dφ dθ
0
0
π/2 =
0
0
π/2
=
0
0
π/2
1 6 3
a sin φ cos φ sin θ cos θ dφ dθ =
6
π/2
0
1 6 4
a sin φ sin θ cos θ
24
π/2
dθ
0
π/2
1 6 2
1 6
1 6
a sin θ cos θ dθ =
a sin θ
a .
=
24
48
48
0
13.M.29: We are given the solid ellipsoid E with constant density δ = 1 and boundary the surface with
Cartesian equation
x2
y2
z2
+
+
= 1.
a2
b2
c2
1812
Its moment of inertia with respect to the x-axis is then
(y 2 + z 2 ) dV.
Ix =
E
We use the transformation
x = aρ sin φ cos θ,
y = bρ sin φ sin θ,
z = cρ cos φ.
(1)
Under this transformation, E is replaced with the solid B determined by
0 θ 2π,
0 φ π,
0 ρ 1.
The Jacobian of the transformation in (1) is
a sin φ cos θ
∂(x, y, z)
= b sin φ sin θ
∂(ρ, φ, θ)
c cos φ
aρ cos φ cos θ
bρ cos φ sin θ
−cρ sin φ
−aρ sin φ sin θ bρ sin φ cos θ = abcρ2 sin φ.
0
Therefore
Ix =
2π
0
π
0
2π
1
0
π
1
=
0
=
=
1
abc
5
1
abc
5
1
= abc
5
1
= abc
5
0
0
2π
2π
0
π
0
2π 0
π
(b2 sin3 φ sin2 θ + c2 sin φ cos2 φ) dφ dρ
0
0
(b2 sin2 φ sin2 θ + c2 cos2 φ) · abcρ4 sin φ dρ dφ dθ
0
(bρ sin φ sin θ)2 + (cρ cos φ)2 · abcρ2 sin φ dρ dφ dθ
2π
b2 (1 − cos2 φ) sin φ sin2 θ + c2 sin φ cos2 φ dφ dθ
1 2
1
b cos3 φ sin2 θ − b2 cos φ sin2 θ − c2 cos3 φ
3
3
π
dθ
0
2 2
1 2
2
− b (1 − cos 2θ) + b (1 − cos 2θ) + c dθ
3
3
2π
1
1
2
1
1
abc − b2 θ + b2 sin 2θ + b2 θ − b2 sin 2θ + c2 θ
5
3
6
2
3
0
2
1
4
1
4
πabc(b2 + c2 ) = M (b2 + c2 )
= abc − πb2 + 2πb2 + πc2 =
5
3
3
15
5
==
where M = 43 πabc is the mass of E.
1813
13.M.30: The region of Problem 30 is in the first octant, within the cylinder with cylindrical equation
r = a, outside the sphere with spherical equation ρ = a, and below the plane with Cartesian equation z = a
(where a > 0). Thus its volume is
π/2
a
V =
0
0
π/2
=
0
a
√
a2 −r 2
π/2
a
r dz dr dθ =
1 2 1 2
ar + (a − r2 )3/2
2
3
ar − r
a2 − r 2
0
0
dθ =
π 1 3
1
· a =
πa3 .
2 6
12
a
0
dr dθ
To check this answer, begin with a solid sphere of radius a. Circumscribe a right circular cylinder of the
same radius. Subtract the volume of the sphere from that of the cylinder, then divide by 8 to get the volume
of the part of the region in the first octant:
1
V =
8
4 3
1
1 2
3
πa3 .
2πa − πa = · πa3 =
3
8 3
12
13.M.31: The cylinder r = 2 cos θ meets the xy-plane in the circle with equation r = 2 cos θ, − 21 π θ 1
2 π.
With density δ = 1, the moment of inertia of the solid region with respect to the z-axis is
Iz = 2
π/2
−π/2
π/2
0
=
π/2
0
√
0
=4
4
=
15
2 cos θ
4−r 2
r3 dz dr dθ = 4
π/2
0
0
2 cos θ
0
r3
4 − r2 dr dθ
2 cos θ
1
4
2
2
(3r − 4r − 32) 4 − r
dθ
15
0
(64 − 64 sin θ − 32 sin θ cos2 θ + 96 sin θ cos4 θ) dθ
π/2
128
8
(15π − 26) ≈ 12.0171461995217912.
=
480θ + 450 cos θ − 25 cos 3θ − 9 cos 5θ
225
225
0
The student who obtains the incorrect answer
128
15 π
may well have overlooked the fact that
4 − 4 cos2 θ = |2 sin θ|.
13.M.32:
The area element dA = r dr dθ moves around a circle of radius x = r cos θ, and hence of
circumference 2πr cos θ. So the volume generated is
π/2
2a cos θ
V =
2
2πr cos θ dr dθ = 2
−π/2
0
=2
0
π/2
0
16 3
πa cos4 θ
3
dθ =
π/2
2 3
πr cos θ
3
2a cos θ
dθ
0
1 3
πa (12θ + 8 sin 2θ + sin 4θ)
3
1814
π/2
0
= 2π 2 a3 .
13.M.33: The area element r dr dθ moves around a circle of radius y = r sin θ, and therefore of circumference 2πr sin θ. Hence the volume swept out is
1+cos θ
2 3
πr sin θ
dθ
θ=0 r=0
θ=0 3
r=0
π
π
2
1
8
π(1 + cos θ)3 sin θ dθ = − π(1 + cos θ)4 = π ≈ 8.37758040957278196923.
=
6
3
0 3
0
π
1+cos θ
V =
13.M.34:
2πr2 sin θ dr dθ =
Assume that b 0.
π
Then the area element r dr dθ moves around a circle having radius
b + x = b + r cos θ, and hence the volume swept out is
2π
a
V =
2π
2π(b + r cos θ)r dr dθ =
0
0
0
2π
=
0
2
πbr + πr3 cos θ
3
2
a
dθ
0
2π
2 3
2 3
2
2
= 2π 2 a2 b.
πa b + πa cos θ dθ = πa bθ + πa sin θ
3
3
0
13.M.35: The moment of inertia of the torus of Problem 34 with respect to the line x = −b (where b 0),
its natural axis of symmetry, is
a
3
2
πδ b3 r2 + 2b2 r3 cos θ + br4 cos2 θ + r5 cos3 θ
dθ
2
5
θ=0 r=0
θ=0
r=0
2π 3
2
= πδ
a2 b3 + 2a3 b2 cos θ + a4 b cos2 θ + a5 cos3 θ dθ
2
5
0
I=
2π
a
2πδ(b + r cos θ)3 · r dr dθ =
2π
=
2π
1
πδa2 90a2 bθ + 120b3 θ + 36a3 sin θ + 240ab2 sin θ + 45a2 b sin 2θ + 4a3 sin 3θ
120
0
=
1
1
1
πδa2 (180πa2 b + 240πb3 ) = π 2 δa2 b(3a2 + 4b2 ) = M (3a2 + 4b2 )
120
2
4
where M = 2π 2 δa2 b is the mass of the torus.
13.M.36: The average distance of points of a circular disk of radius a from its center is
d=
1
πa2
0
2π
a
r2 dr dθ =
0
a
1 3
2
2a3
2
r
·
= 2 = a.
2
a
3
3a
3
0
13.M.37: Use the disk bounded by the circle with polar equation r = 2a sin θ, 0 θ π. Then the origin
is a point on the boundary of the disk, and the average distance of points of this disk from the origin is
1
d=
πa2
π
θ=0
2a sin θ
r=0
1
r dr dθ =
πa2
2
=
2a3
9πa2
π
0
8 3 3
a sin θ dθ
3
π
cos 3θ − 9 cos θ
1815
=
0
32
a ≈ (1.131768484209)a.
9π
13.M.38: Use the circular disks bounded by r = 2 cos θ and r = 4 cos θ, − 12 π θ 1
2 π.
The two circles
are tangent at the origin, and the average distance of points outside the small circle and inside the large
circle from the origin is
1
3π
d=
π/2
−π/2
4 cos θ
r2 dr dθ =
2 cos θ
1
3π
π/2
−π/2
1 3
r
3
4 cos θ
dθ =
2 cos θ
=
14
9 sin θ + sin 3θ
27π
1
9π
π/2
π/2
56 cos3 θ dθ
−π/2
224
≈ 2.6407931298210782.
27π
=
−π/2
13.M.39: We use the ball bounded by the surface with spherical-coordinates equation ρ = a. Then the
average distance of points of this ball from its center is
d=
=
3
4πa3
2π
θ=0
3
4πa3
π
φ=0
2π
θ=0
π
φ=0
a
ρ3 sin φ dρ dφ dθ
ρ=0
π
1 4
3
1 4
3
1
3
a sin φ dφ dθ =
a
·
2π
·
−
cos
φ
=
· 2π · a4 = a.
3
4
4πa3
4
4πa
2
4
0
13.M.40: We use the ball of radius a centered at the point (0, 0, a) on the z-axis. The average distance
of points of this ball from the origin (its “south pole”) is then
3
d=
4πa3
2π
0
π/2
0
2a cos φ
0
3
ρ sin φ dρ dφ dθ =
· 2π ·
4πa3
3
π/2
4a4 sin φ cos4 φ dφ
0
π/2
3 4
6
4 4
3
5
= 3 · a4 = a.
= 3 − a cos φ
2a
5
2a 5
5
0
13.M.41: We will use the spheres with spherical-coordinates equations ρ = 2 cos φ and ρ = 4 cos φ, which
have a mutual point of tangency at the origin. Then the average distance of points outside the smaller and
inside the larger sphere from the origin is
3
d=
28π
3
=
14
13.M.42:
2π
θ=0
π/2
0
π/2
φ=0
4 cos φ
3
· 2π ·
ρ sin φ dρ dφ dθ =
28π
ρ=2 cos φ
3
π/2
φ=0
1 4
ρ sin φ
4
4 cos φ
dφ
ρ=2 cos φ
π/2
3
18
5
≈ 2.5714285714285714.
60 sin φ cos φ dφ =
=
−12 cos φ
14
7
0
4
Place the cone with its vertex at the origin and its axis on the nonnegative z-axis. Then a
spherical-coordinates equation of its side is
φ = arctan
1816
R
H
and an equation of its base (at the top) is ρ = H sec φ. Hence the average distance of points of the cone
from its vertex is
3
d=
πR2 H
2π
0
arctan(R/H)
0
=
H sec φ
3
· 2π ·
ρ sin φ dρ dφ dθ =
πR2 H
3
0
1 4 3
6
H sec φ
R2 H 12
arctan(R/H)
arctan(R/H)
0
1 4 3
H sec φ tan φ dφ
4
(H 2 + R2 )3/2 − H 3
1 6
2
2 3/2
4
·
·
H(H
=
+
R
)
−
H
.
R2 H 12
2R2
=
0
Thus the average distance of points of the cone from its vertex is
L3 − H 3
2R2
d =
where L =
√
R2 + H 2 is the slant height of the cone.
13.M.43: The part of the paraboloid that lies between the two given planes also is the part between the
cylinders r = 2 and r = 3. Let R denote the part of the xy-plane between those two cylinders. Then the
surface area in question is
2
2
2
A=
r + (rzr ) + (zθ ) dr dθ =
= 2π
θ=0
R
2π
1
(1 + 4r2 )3/2
12
3
=
2
2
4
r + 4r dr dθ = 2π
3
3
r(1 + 4r2 )1/2 dr
2
r=2
√ 1 √
π 37 37 − 17 17 ≈ 81.1417975124065455.
6
13.M.44: Let D denote the circular disk x2 + y 2 4. Then the surface area is
2
2
1 + (zx ) + (zy ) dx dy =
1 + 4x2 + 4y 2 dx dy
A=
D
D
2π =
0
2
r(1 + 4r2 )1/2 dr dθ = 2π
0
1
(1 + 4r2 )1/2
12
2
=
0
1 √
π 17 17 − 1 ≈ 36.1769031974114084.
6
13.M.45: Let R be the region in the φθ-plane determined by the inequalities φ2 φ φ1 and 0 θ 2π,
where
cos φ2 =
z2
a
and
cos φ1 =
z1
.
a
Thus the part of the sphere ρ = a for which the spherical coordinates φ and θ satisfy these inequalities is
the part of the sphere between the planes z = z1 and z = z2 . Thus the formula in Problem 18 of Section
14.8 yields the area of this surface to be
a2 sin φ dφ dθ =
A=
= 2πa
2π
θ=0
R
2
φ2
= 2πa
+ cos φ
φ1
2
φ1
a2 sin φ dφ dθ
φ=φ2
z2 − z1
a
1817
= 2πa(z2 − z1 ) = 2πah
because h = z2 − z1 .
13.M.46:
The graph of z(r, θ) =
√
4 − r2 is the top half of the sphere, so we will need to double the
area integral. Let D be the plane region bounded by the circle with Cartesian equation x2 + y 2 = 2x; the
polar-coordinates equation of this circle is r = 2 cos θ, − 21 π θ 1
2 π.
The area of the part of the sphere
above this circle plus the area of the part below it is then
2
2
2
A=2
r + (rzr ) + (zθ ) dr dθ = 2
−π/2
D
π/2
2 1/2
−2(4 − r )
=4
0
2 cos θ
π/2
dθ = 4
2 cos θ
0
= 4·4·
0
2r
dr dθ
(4 − r2 )1/2
4 − 2(4 − 4 cos2 θ)1/2 dθ
π/2
(4 − 4 sin θ) dθ = 4 4θ + 4 cos θ
0
0
0
π/2
=4
π/2
π
− 16 = 8(π − 2) ≈ 9.13274123.
2
The student who obtains the incorrect answer 2π 2 likely forgot that
√
4 − 4 cos2 θ = 2 sin θ if −π < θ < 0.
13.M.47: Position the cone with its vertex at the origin and its axis on the nonnegative z-axis. The side
of the cone has Cartesian equation z = x2 + y 2 , and hence
dS = 1 + (zx )2 + (zy )2 dx dy =
1+
x2
√
x2
y2
+ 2
dx dy = 2 dx dy.
2
2
+y
x +y
Let S be the square with vertices at (±1, ±1). Because the area of S is 4, we see with no additional
computations that the area of the part of the cone that lies directly above S is
√
√
2 dx dy = 4 2 ≈ 5.6568542494923802.
S
13.M.48: Let z(x, y) = 12 x2 . Let D be the disk in the xy-plane bounded by the circle x2 + y 2 = 1. Then
the area of the part of the parabolic cylinder that lies over D is
2
2
1 + (zx ) + (zy ) dA = 4
A=
D
The antiderivative of
1
x=0
√
1−x2
2
1 + x dy dx = 4
y=0
0
1
1 − x4 dx.
√
1 − x4 is known to be nonelementary. (See Sherman K. Stein and Anthony Barcellos:
Calculus and Analytic Geometry (McGraw-Hill: New York, 1992), page 460, Problems 164–174.) Therefore
we next used Mathematica for a numerical integration. The command
NIntegrate[ 4∗Sqrt[1 - x∧4], {x,0,1}, AccuracyGoal → 24, WorkingPrecision → 30 ]
yielded the approximation A ≈ 3.49607673905615974729.
1818
13.M.49: Given: x = x(t), y = y(t), z = z, a t b, 0 z h(t): Let
r(t, z) = x(t), y(t), z ,
Then
rt × rz
rt = x (t), y (t), 0 and rz = 0, 0, 1 .
so that
i
= x (t)
0
k
0 = y (t), −x (t), 0 ,
1
j
y (t)
0
and hence
|rt × rz | =
2 2
x (t) + y (t) .
Therefore the area of the “fence” is
b
A=
t=a
h(t) 2 2 1/2
x (t) + y (t)
dz dt.
z=0
13.M.50:
The “fence” stands above the plane curve r = a sin θ, 0 θ π. Thus we take the height
√
function h(θ) = a2 − r2 as the upper limit of integration in the formula in Problem 49, where
z=
a2 − r2 = a2 − a2 sin2 θ = a cos θ
provided that 0 θ 12 π. Moreover, we have
y(θ) = a sin2 θ,
x(θ) = a sin θ cos θ,
and z = z
where 0 θ π and 0 z |a cos θ|. Thus
x (θ)
2
2
+ y (θ) = a2 (cos2 θ − sin2 θ)2 + 4a2 sin2 θ cos2 θ
= a2 (cos4 θ + 2 sin2 θ cos2 θ + sin4 θ) = a2 (cos2 θ + sin2 θ)2 = a2 .
We double the integral to allow for the fact that an equal height of the “fence” stands below the xy-plane,
and double it again in order to restrict the range of θ to the interval 0 θ 12 π. Thus the area in question
is
π/2
h(θ)
π/2
θ=0
z=0
a cos θ
a dz dθ = 4a
A=4
π/2
1 dz dθ = 4a
0
0
0
π/2
a cos θ dθ = 4a2 sin θ
= 4a2 .
0
13.M.51: We are given the region R bounded by the curves x2 − y 2 = 1, x2 − y 2 = 4, xy = 1, and xy = 3,
of constant density δ. Its polar moment of inertia is
(x2 + y 2 ) δ dx dy.
I0 =
R
1819
The hyperbolas bounding R are u-curves and v-curves if we let u = xy and v = x2 − y 2 . If we make this
substitution, then
4u2 + v 2 = 4x2 y 2 − (x2 − y 2 )2 = (x2 + y 2 )2 ,
√
4u2 + v 2 for x2 + y 2 in the integral for I0 . Moreover, it is not necessary
and therefore we will substitute
to solve for x and y in terms of u and v because of a result in Section 13.9 (see the proof in Problem 18
there). Thus
x = −2(x2 + y 2 ),
−2y y
∂(u, v)
= ∂(x, y)
2x
and therefore
∂(x, y)
1
1
=−
=− √
.
2
∂(u, v)
2(x2 + y 2 )
2 4u + v 2
Therefore
4
3
I0 =
v=1
u=1
√
4 3
4u2 + v 2
1
√
δ du dv = 3δ.
δ du dv =
2 4u2 + v 2
1
1 2
13.M.52: The equations u = x − y, v = x + y are easy to solve for
x=
u+v
,
2
y=
−u + v
.
2
(1)
Then the curves that bound the region R are transformed as follows:
x+y =1
becomes
v = 1;
y=0
becomes
u = v;
x=0
becomes
u = −v.
The Jacobian of the transformation in (1) is
1
∂(x, y) 2
=
∂(u, v) 1
−
2
1
2
1
2
1 1
1
= + = .
4 4
2
Therefore
1 v
u
u v
x−y
1 1
1
exp
du dv =
exp
dv
v exp
dx dy =
x+y
v
2 v=0
v
v=0 u=−v 2
u=−v
R
=
1
2
0
1
1
1
e2 − 1
1
1 1 2
v e−
≈ 0.5876005968219007.
v· e−
=
dv =
e
2 2
e
4e
0
13.M.53: We use the transformation
x = aρ sin φ cos θ,
y = bρ sin φ sin θ,
1820
z = cρ cos φ.
We saw in the solution of Problem 29 that the Jacobian of this transformation is
∂(x, y, z)
= abcρ2 sin φ.
∂(ρ, φ, θ)
Moreover, it follows from work shown in the solution of Problem 29 that the density function takes the form
δ(ρ, φ, θ) = 1 − ρ2 . Finally, the ellipsoidal surface
x2
y2
z2
+ 2 + 2 =1
2
a
b
c
is transformed into the surface ρ = 1, and therefore the mass of the solid ellipsoid is
2π π 1
π 1
2
2
m=
(1 − ρ )abcρ sin φ dρ dφ dθ = 2πabc
(ρ2 − ρ4 ) sin φ dρ dφ
θ=0
φ=0
= 2πabc
0
π
ρ=0
φ=0
4
2
sin φ dφ =
πabc − cos φ
15
15
π
=
0
ρ=0
8
πabc.
15
13.M.54: If r2 = u1/2 cos 2θ = v 1/2 sin 2θ, then r4 = u cos2 2θ = v sin2 2θ. Hence
u=
Thus
r4
cos2 2θ
and
v=
r4
.
sin2 2θ
r8
uv
sin2 2θ cos2 2θ
r4
=
·
=
= r4
u+v
sin2 2θ cos2 2θ r4 cos2 2θ + r4 sin2 2θ
cos2 2θ + sin2 2θ
and
tan 2θ =
Therefore
1
θ = arctan
2
r2 u1/2
u1/2
sin 2θ
= 1/2 · 2 = 1/2 .
cos 2θ
r
v
v
u1/2
v 1/2
and
r=
uv
u+v
1/4
.
To find the Jacobian of this transformation from the uv-plane to the rθ-plane, note first that
cos 2θ =
r2
u1/2
and
sin 2θ =
r2
.
v 1/2
After the next computation, we will use a result in Section 13.9 (see Problem 18 there).
3
4r sec2 2θ
4r4 sec2 2θ tan 2θ ∂(u, v) = −16r7 sec3 2θ csc3 2θ,
=
∂(r, θ)
4
2
4r3 csc2 2θ
−4r csc 2θ cot 2θ and thus
sin3 2θ cos3 2θ
∂(r, θ)
r12
=−
=
−
∂(u, v)
16r7
16r7 u3/2 v 3/2
=−
r6
16ru3/2 v 3/2
=−
u3/2 v 3/2
1
=−
.
16r(u + v)3/2 u3/2 v 3/2
16r(u + v)3/2
1821
The transformation we are using has the following effect on the lemniscates that bound the region R:
Therefore the area of R is
A=
r dr dθ =
r2 = 4 sin 2θ
becomes
v 1/2 = 4;
r2 = 3 sin 2θ
becomes
v 1/2 = 3;
r2 = 4 cos 2θ
becomes
u1/2 = 4;
r2 = 3 cos 2θ
becomes
u1/2 = 3.
16
v=9
R
16
1
du dv =
16(u + v)3/2
u=9
16
9
1
−
8(u + v)1/2
16
dv
9
16
1
1
1
1/2
1/2
−
dv =
(v + 9) − (v + 16)
4
(v + 9)1/2
(v + 16)1/2
9
9
√
10 − 7 2
≈ 0.025126265847083664597.
=
4
1
=
8
16
√
3 centered at the origin has equation x2 + y 2 + z 2 = 3, and
hence the upper hemisphere has equation z = 3 − x2 − y 2 . Next,
13.M.55: The spherical surface with radius
1 + (zx )2 + (zy )2 = 1 +
x2
y2
3
+
=
.
3 − x2 − y 2
3 − x2 − y 2
3 − x2 − y 2
We integrate over the unit square in the xy-plane, quadruple the result to find the area of the part of the
surface above the 2-by-2 square, then double it to account for the spherical surface below the xy-plane. Thus
the surface area is
1
1
√
3
1
√
3 arctan
y
dy dx = 8
3 − x2 − y 2
3 − x2 − y 2
x=0 y=0
x=0
1√
1 √
1
1
√
√
=8
3 arctan
8 3 arcsin
dx =
dx.
2 − x2
3 − x2
0
x=0
A=8
1
y=0
Now use integration by parts with
√
1
,
u = 8 3 arcsin √
3 − x2
√
8x 3
√
dx,
du =
(3 − x2 ) 2 − x2
dv = dx;
v = x.
Thus we find that
1
√ 1
√
x2
1
√
−8 3
dx
A = 8x 3 arcsin √
2
2
3 − x2 0
0 (3 − x ) 2 − x
√ 1
√
√ 1
√
x2
x2
1
√
√
dx = 2π 3 − 8 3
dx.
= 8 3 arcsin √ − 8 3
2
2
2
2
2
0 (3 − x ) 2 − x
0 (3 − x ) 2 − x
1822
√
√
2 sin θ, dx = 2 cos θ dθ. This yields
√ π/4
√
2 sin2 θ
A = 2π 3 − 8 3
dθ
(3 − 2 sin2 θ)
0
√
√ π/4 1 − cos 2θ
√ π/2 1 − cos φ
√
dφ
dθ
=
2π
3
−
4
3
= 2π 3 − 8 3
3 − 2 sin2 θ
0
φ=0 2 + cos φ
Now make the substitution x =
where φ = 2θ. Now substitute
φ
u = tan ,
2
sin φ =
2u
,
1 + u2
cos φ =
1 − u2
,
1 + u2
2 du
1 + u2
dφ =
(see the discussion immediately following Miscellaneous Problem 134 of Chapter 7). This yields
√
√
√ 1
√ 1
1
3
4u2
A = 2π 3 − 4 3
du
=
2π
−
3
−
8
3
du
2
2
u2 + 3 u2 + 1
0 (u + 1)(u + 3)
0
1
√ π √
√ √
√
√
u
2 3 −3
3 arctan √
= 2π 3 − 8 3
− arctan u = 2π 3 − 8 3 ·
12
3
0
√
3 − 1 ≈ 9.19922175645144125328.
= 4π
13.M.56: We will find the volume of the part of the solid that lies in the first octant, then multiply by 8.
Thus the volume is
a
(a2/3 −x2/3 )3/2
V =8
x=0
(a2/3 − x2/3 − y 2/3 )3/2 dy dx.
y=0
The substitution y = b sin3 θ transforms the integrand into
(a2/3 − x2/3 − b2/3 sin2 θ)3/2 ,
and hence will be useful provided that b2/3 = a2/3 − x2/3 ; for this reason, we let b = (a2/3 − x2/3 )3/2 . Then
the substitution
y = (a2/3 − x2/3 )3/2 sin3 θ,
dy = 3(a2/3 − x2/3 )3/2 sin2 θ cos θ dθ
yields
a
π/2
V =8
0
0
a
= 24
0
0
π/2
(a2/3 − x2/3 )3/2 cos3 θ · 3(a2/3 − x2/3 )3/2 sin2 θ cos θ dθ dx
(a2/3 − x2/3 )3 sin2 θ cos4 θ dθ dx = 24
0
a
(a2/3 − x2/3 )3
0
π/2
(cos4 θ − cos6 θ) dθ dx.
Then Formula 117 of the long table of integrals yields
a
a
1 3 π
1 3 5 π
π 2/3
2/3
2/3 3
· ·
− · · ·
(a − x2/3 )3 dx
(a − x )
V = 24
dx = 24
2
4
2
2
4
6
2
32
0
0
a
3
16 3
9
9
1
3
4
= π a2 x − a4/3 x5/3 + a2/3 x7/3 − x3 = π ·
a =
πa3 .
4
5
7
3
4
105
35
0
1823
13.M.57: We will find the volume of the part of the solid that lies in the first octant, then multiply by 8.
Thus the volume of the entire solid is
a
(a1/3 −x1/3 )3
V =8
x=0
(a1/3 − x1/3 − y 1/3 )3 dy dx.
y=0
The substitution y = b sin6 θ transforms the integrand into
(a1/3 − x1/3 − b1/3 sin2 θ)3 ,
and hence will be useful provided that b1/3 = a1/3 − x1/3 . Thus we choose b = (a1/3 − x1/3 )3 , and the
substitution
y = (a1/3 − x1/3 )3 sin6 θ,
dy = 6(a1/3 − x1/3 )3 sin5 θ cos θ dθ
then yields
a
π/2
V =8
x=0
π/2
= 48
0
0
a
(a
= 48
(a1/3 − x1/3 ) cos2 θ
3
· 6(a1/3 − x1/3 )3 sin5 θ cos θ dθ
θ=0
a
1/3
0
(a1/3 − x1/3 )6 sin5 θ cos7 θ dθ
1/3 6
−x
)
1
(−600 cos 2θ − 75 cos 4θ
122880
π/2
+ 100 cos 6θ + 30 cos 8θ − 12 cos 10θ − 5 cos 12θ)
a
= 48
0
1
2
(a1/3 − x1/3 )6 dx =
120
5
a
0
dx
0
(a1/3 − x1/3 )6 dx
a
2 2
9 5/3 4/3
45 2/3 7/3 9 1/3 8/3 1 3
2 1 3
1 3
4/3 5/3
2
a x − a x + x
a =
a .
=
= ·
a x − a x + 9a x − 10ax +
5
2
7
4
3
5 84
210
0
13.M.58: The average squared distance of points of the ellipsoid from its center (at (0, 0, 0)) is
1
V
0
2π
0
π
0
1
(a sin φ cos θ)2 + (b sin φ sin θ)2 + (c cos φ)2 · ρ2 · abcρ2 sin φ dρ dφ dθ
(1)
where V = 43 πabc is the volume of the ellipsoid. To obtain this integral, we began with the solid ellipsoid E
given in Problem 58. We set up the triple integral
1
V
(x2 + y 2 + z 2 ) dV,
E
then converted it—using ellipsoidal coordinates, as in Problems 20, 21, 25, and 27 of Section 13.9—to the
integral shown in (1). (The expression abcρ2 sin φ that appears there is the Jacobian of the transformation.)
Then we evaluated the integral in (1) using Mathematica, as follows.
1824
Integrate[ ((a∗Sin[phi]∗Cos[theta])∧2 + (b∗Sin[phi]*Sin[theta])∧2
+ (c∗Cos[phi])∧2)*(a*b*c*rho∧4∗Sin[phi]), rho ]
1
(abcρ5 sin φ)(c2 cos2 φ + a2 cos2 θ sin2 φ + b2 sin2 φ sin2 θ)
5
(% /.
rho → 1) - (% /.
rho → 0)
1
(abc sin φ)(c2 cos2 φ + a2 cos2 θ sin2 φ + b2 sin2 φ sin2 θ)
5
Simplify[ Integrate[ %, phi ] ]
1
(abc cos φ)(−10a2 − 10b2 − 4c2 + 2(a2 + b2 − 2c2 ) cos 2φ + (a2 − b2 ) cos 2(φ − θ)
120
− 10a2 cos 2θ + 10b2 cos 2θ + a2 cos 2(φ + θ) − b2 cos 2(φ + θ))
Simplify[ (% /.
phi → Pi) - (% /.
phi → 0) ]
2
abc(a2 + b2 + c2 + (a2 − b2 ) cos 2θ
15
Integrate[ %, theta ]
1
(2a3 bcθ + 2ab3 cθ + 2abc3 θ + a3 bc sin 2θ − ab3 c sin 2θ))
15
Factor[ (% /.
theta → 2∗Pi) - (% /.
theta → 0) ]
4
πabc(a2 + b2 + c2 )
15
%/((4/3)∗Pi∗a∗b∗c)
1 2
(a + b2 + c2 )
5
13.M.59:
Locate the unit cube C as shown in the next figure, with one vertex at the origin and the
opposite vertex at the point (1, 1, 1) in space. Let L be the line through these two points; we will rotate C
around the line L to generate the solid S. We also install a coordinate system on L; it becomes the w-axis,
√
with w = 0 at the origin and w = 3 at the cube vertex with Cartesian coordinates (1, 1, 1). Thus distance
is measured on the w-axis in exactly the same way it is measured on the three Cartesian coordinate axes.
1825
z
P
r
w
Q
y
x
v 1. We first deal with the case 23 v 1. Then the point with
√
Cartesian coordinates P (v, v, v) is the point on the w-axis where w = v 3 , and the plane normal to the
Choose a point v on the x-axis with
1
2
w-axis at this point intersects C in a triangle, also shown in the preceding figure. It is clear that the plane
√
has equation x + y + z = 3v = w 3 and that it meets one edge of the cube at the point Q(1, 1, z) for some z
√
between 0 and 1. In fact, because (1, 1, z) satisfies the equation of the plane, it follows that z = −2 + w 3 .
When the cube is rotated around L, the resulting solid S meets the plane x + y + z = 3v in a circular disk
centered at P (v, v, v), and the radius of this disk is the distance from P (v, v, v) to Q(1, 1, z), which is
r =
√
(v − 1)2 + (v − 1)2 + (w 3 − 2 − v)2 = 2(v − 1)2 + (3v − 2 − v)2
=
Now we turn to the case
1
2
√
√ √
6(v − 1)2 = (1 − v) 6 = 1 − 13 w 3 · 6 .
v 23 . In this case the plane through P (v, v, v) meets the surface of the
cube in a semi-regular hexagon, one in which each interior angle is 2π/3 and whose sides are of only two
different lengths a and b, alternating as one moves around the hexagon. Such a hexagon is shown in the
figure below.
One of the vertices of the hexagon is located at the point R(1, y, 0) on one edge of the cube. The distance
from P (v, v, v) to this point R is the radius r of the circular disk in which the plane normal to the w-axis
at P meets the solid S. It is easy to show that y = 3v − 1, and it follows that the distance in question is
√
2
2
2
2
r = (v − 1) + (2v − 1) + v = 6v − 6v + 2 = 2w2 − 2w 3 + 2 .
1826
z
P
w
r
x
y
R
1
2
v 1, we obtain only half of the solid S, so we now can find
√
the volume V of S as follows. We shift to coordinates on the w-axis and remember that dw = 3 dv. The
By considering only values of v for which
result is that
2/3
V =2
√
√
π(2w2 − 2w 3 + 2) 3 dw + 2
v=1/2
1
v=2/3
The adjusted limits of integration are w =
1
2
√
3 to
2
3
√ 2 √
6π 1 − 13 w 3
3 dw.
√
3 in the first integral and w =
second. Then Mathematica promptly reports that
π
V =√
3
≈ 1.813799364234217850594078257642155732.
1827
2
3
√
√
3 to 3 in the