16.1 SOLUTIONS
1105
CHAPTER SIXTEEN
Solutions for Section 16.1
Exercises
1. Mark the values of the function on the plane, as shown in Figure 16.1, so that you can guess respectively at the smallest
and largest values the function takes on each small rectangle.
X
Lower sum =
f (xi , yi )∆x∆y
= 4∆x∆y + 6∆x∆y + 3∆x∆y + 4∆x∆y
= 17∆x∆y
= 17(0.1)(0.2) = 0.34.
Upper sum =
X
f (xi , yi )∆x∆y
= 7∆x∆y + 10∆x∆y + 6∆x∆y + 8∆x∆y
= 31∆x∆y
= 31(0.1)(0.2) = 0.62.
y
↓ 3
2.4
2.2
5
4
4
6
8
5
7
6
∆y = 0.2
?
2.0
1.0
1.1
∆x = 0.1
-
10
1.2 ← x
Figure 16.1
2. In the subrectangle in the top left in Figure 16.4, it appears that f (x, y) has a maximum value of about 9. In the subrectangle in the top middle, f (x, y) has a maximum value of 10. Continuing in this way, and multiplying by ∆x and ∆y, we
have
Overestimate = (9 + 10 + 12 + 7 + 8 + 10 + 5 + 7 + 8)(10)(5) = 3800.
Similarly, we find
Underestimate = (7 + 7 + 8 + 4 + 5 + 7 + 1 + 3 + 6)(10)(5) = 2400.
Thus, we expect that
2400 ≤
Z
R
f (x, y)dA ≤ 3800.
3. (a) If we take the partition of R consisting of just R itself, we get
Lower bound for integral = minR f · AR = 0 · (4 − 0)(4 − 0) = 0.
Similarly, we get
Upper bound for integral = maxR f · AR = 4 · (4 − 0)(4 − 0) = 64.
1106
Chapter Sixteen /SOLUTIONS
(b) The estimates asked for are just the upper and lower sums. We partition R into subrectangles R (a,b) of width 2 and
height 2, where (a, b) is the lower-left corner of R(a,b) . The subrectangles are then R(0,0) , R(2,0) , R(0,2) , and R(2,2) .
(0, 4)
(4, 4)
(0, 0)
(4, 0)
Figure 16.2
Then we find the lower sum
Lower sum =
X
(a,b)
AR(a,b) · min f =
R(a,b)
X
(a,b)
=4
4 · (Min of f on R(a,b) )
X
(Min of f on R(a,b) )
(a,b)
= 4(f (0, 0) + f (2, 0) + f (0, 2) + f (2, 2))
√
√
√
√
= 4( 0 · 0 + 2 · 0 + 0 · 2 + 2 · 2)
= 8.
Similarly, the upper sum is
Upper sum = 4
X
(Max of f on R(a,b) )
(a,b)
= 4(f (2, 2) + f (4, 2) + f (2, 4) + f (4, 4))
√
√
√
√
= 4( 2 · 2 + 4 · 2 + 2 · 4 + 4 · 4)
√
= 24 + 16 2 ≈ 46.63.
The upper sum is an√overestimate and the lower sum is an underestimate, so we can get a better estimate by averaging
them to get 16 + 8 2 ≈ 27.3.
4. (a) We first find an over- and underestimate of the integral, using four subrectangles. On the first subrectangle (0 ≤ x ≤
3, 0 ≤ y ≤ 4), the function f (x, y) appears to have a maximum of 100 and a minimum of 79. Continuing in this
way, and using the fact that ∆x = 3 and ∆y = 4, we have
Overestimate = (100 + 90 + 85 + 79)(3)(4) = 4248,
and
Underestimate = (79 + 68 + 61 + 55)(3)(4) = 3156.
A better estimate of the integral is the average of the overestimate and the underestimate:
Better estimate =
4248 + 3156
= 3702.
2
(b) The average value of f (x, y) on this region is the value of the integral divided by the area of the region. Since the
area of R is (6)(8) = 48, we approximate
Average value =
1
Area
Z
R
f (x, y)dA ≈
1
· 3702 = 77.125.
48
We see in the table that the values of f (x, y) on this region vary between 55 and 100, so an average value of 77.125
is reasonable.
5. Partition R into subrectangles with the lines x = 0, x = 0.5, x = 1, x = 1.5, and x = 2 and the lines y = 0, y = 1,
y = 2, y = 3, and y = 4. Then we have 16 subrectangles, each of which we denote R (a,b) , where (a, b) is the location of
the lower-left corner of the subrectangle.
16.1 SOLUTIONS
1107
We want to find a lower bound and an upper bound for the volume above each subrectangle. The lower bound for the
volume of R(a,b) is
0.5(Min of f on R(a,b) )
because the area of R(a,b) is 0.5 · 1 = 0.5. The function f (x, y) = 2 + xy increases with both x and y over the whole
region R, as shown in Figure 16.3. Thus,
Min of f on R(a,b) = f (a, b) = 2 + ab,
because the minimum on each subrectangle is at the corner closest to the origin.
y
z
4
R(a, b)
(a, b)
y
x
x
2
Figure 16.4
Figure 16.3
Similarly,
Max of f on R(a,b) = f (a + 0.5, b + 1) = 2 + (a + 0.5)(b + 1).
So we have
Lower sum =
X
0.5(2 + ab) = 0.5
X
(2 + ab)
(a,b)
(a,b)
= 16 + 0.5
X
ab
(a,b)
Since a = 0, 0.5, 1, 1.5 and b = 0, 1, 2, 3, expanding this sum gives
Lower sum = 16 + 0.5 ( 0 · 0 + 0 · 1 + 0 · 2 + 0 · 3
+ 0.5 · 0 + 0.5 · 1 + 0.5 · 2 + 0.5 · 3
+ 1·0+1·1+1·2+1·3
+ 1.5 · 0 + 1.5 · 1 + 1.5 · 2 + 1.5 · 3)
= 25.
Similarly, we can compute the upper sum:
Upper sum =
X
0.5(2 + (a + 0.5)(b + 1)) = 0.5
(a,b)
X
(2 + (a + 0.5)(b + 1))
(a,b)
= 16 + 0.5
X
(a + 0.5)(b + 1)
(a,b)
= 41.
6. Since f (x, y) is measured in micrograms per square meter, and we are integrating over an area measured in square
meters, the units of the integral are micrograms. The integral represents the total quantity of pollution, in micrograms, in
the region R.
7. The exact value of the integral is 40/3.
8. The value of the integral is around −2.4.
Problems
9. The function being integrated is f (x, y) = 1, which is positive everywhere. Thus, its integral over any region is positive.
1108
Chapter Sixteen /SOLUTIONS
10. The function being integrated is f (x, y) = 1, which is positive everywhere. Thus, its integral over any region is positive.
11. The function being integrated is f (x, y) = 5x. Since x > 0 in R, f is positive in R and thus the integral is positive.
12. The function being integrated is f (x, y) = 5x, which is an odd function in x. Since B is symmetric with respect to x, the
contributions to the integral cancel out, as f (x, y) = −f (−x, y). Thus, the integral is zero.
13. The region D is symmetric both with respect to x and y axes. The function being integrated is f (x, y) = 5x, which is an
odd function in x. Since D is symmetric with respect to x, the contributions to the integral cancel out. Thus, the integral
of the function over the region D is zero.
14. The function being integrated, f (x, y) = y 3 + y 5 , is an odd function in y while D is symmetric with respect to y. Then,
by symmetry, the positive and negative contributions of f will cancel out and thus its integral is zero.
15. In a region such as B in which y < 0, the quantity y 3 + y 5 is less than zero. Thus, its integral is negative.
16. The region R is symmetric with respect to y and the integrand is an odd function in y, so the integral over R is zero.
17. The function being integrated, f (x, y) = y − y 3 is always negative in the region B since in that region −1 < y < 0 and
|y 3 | < |y|. Thus, the integral is negative.
18. The function being integrated, f (x, y) = y − y 3 , is an odd function in y while D is symmetric with respect to y. By
symmetry, the integral is zero.
19. The region D is symmetric both with respect to x and y axes. The function being integrated is odd with respect to y in the
region D. Thus, its integral is zero.
20. Since D is a disk of radius 1, in the region D, we have |y| < 1. Thus, −π/2 < y < π/2. Thus, cos y is always positive
in the region D and thus its integral is positive.
21. The function f (x, y) = ex is positive for any value of x. Thus, its integral is always positive for any region, such as D,
with nonzero area.
22. The region D is symmetric both with respect to x and y axes. Looking at the contributions to the integral of the function
f (x, y) = xex , we can see that any contribution made by the point (x, y), where x > 0, is greater than the corresponding
contribution made by (−x, y), since ex > 1 > e−x for x > 0. Thus, the integral of f in the region D is positive.
23. The region D is symmetric both with respect to x and y axes. The function f (x, y) = xy 2 is odd with respect to x, and
thus the contributions to the integral from (x, y) and (−x, y) cancel. Thus the integral is zero in the region D.
24. The function f (x, y) is odd with respect to x, and thus the integral is zero in region B, which is symmetric with respect
to x.
25. We use four subrectangles to find an overestimate and underestimate of the integral:
Overestimate = (15 + 9 + 9 + 5)(4)(3) = 456,
Underestimate = (5 + 2 + 3 + 1)(4)(3) = 132.
A better estimate of the integral is the average of the two:
Z
R
f (x, y)dA ≈
456 + 132
= 294.
2
The units of the integral are milligrams, and the integral represents the total number of mg of mosquito larvae in this 8
meter by 6 meter section of swamp.
26. The question is asking which graph has more volume under it, and from inspection, it appears that it would be the graph
for the mosquitos.
27. Let’s break up the room into 25 sections, each of which is 1 meter by 1 meter and has area ∆A = 1.
We shall begin our sum as an upper estimate starting with the lower left corner of the room and continue across the
bottom and moving upward using the highest temperature, T i , in each case. So the upper Riemann sum becomes
P25
i=1
Ti ∆A = T1 ∆A + T2 ∆A + T3 ∆A + · · · + T25 ∆A
= ∆A(T1 + T2 + T3 + · · · + T25 )
= (1)( 31 +29 + 28 + 27 + 27+
29 +28 + 27 + 27 + 26+
27 +27 + 26 + 26 + 26+
26 +26 + 25 + 25 + 25+
25 +24 + 24 + 24 + 24)
= (1)(659) = 659.
16.1 SOLUTIONS
1109
In the same way, the lower Riemann sum is formed by taking the lowest temperature, t i , in each case:
P25
t ∆A
i=1 i
= t1 ∆A + t2 ∆A + t3 ∆A + · · · + t25 ∆A
= ∆A(t1 + t2 + t3 + · · · + t25 )
= (1)( 27 +27 + 26 + 26 + 25+
26 +26 + 25 + 25 + 25+
25 +24 + 24 + 24 + 24+
24 +23 + 23 + 23 + 23+
23 +21 + 20 + 21 + 22)
= (1)(602) = 602.
So, averaging the upper and lower sums we get: 630.5.
To compute the average temperature, we divide by the area of the room, giving
Average temperature =
630.5
≈ 25.2◦ C.
(5)(5)
Alternatively we can use the temperature at the central point of each section ∆A. Then the sum becomes
P25
i=1
P25
Ti0 ∆A = ∆A i=1 Ti0
= (1)( 29 +28 + 27 + 26.5 + 26+
27 +27 + 26 + 26 + 25.5+
26 +25.5 + 25 + 25 + 25+
25 +24 + 24 + 24 + 24+
24 +23 + 22 + 22.5 + 23)
= (1)(630) = 630.
Then we get
Average temperature =
P25
Ti0 ∆A
630
=
≈ 25.2◦ C.
Area
(5)(5)
i=1
28. The total area of the square R is (1.5)(1.5) = 2.25. See Figure 16.5. On a disk of radius ≈ 0.5 the function has a value of 3
or more, giving a total contribution to the integral of at least (3)·(π ·0.5 2 ) ≈ 2.3. On less than half of the rest of the square
the function has a value between −2 and 0, giving a contribution to the integral of between (1/2·2.25)(−2) = −2.25
R and
0. Since the positive contribution to the integral is therefore greater in magnitude than the negative contribution, R f dA
is positive.
y
2.0
1.5
1.0
0.5
−1
2
0
3
−0.5
−1.0
−1.0 −0.5
4
x
0
1
2
0
0.5
Figure 16.5
1.0
1.5
2.0
1110
Chapter Sixteen /SOLUTIONS
29. The total number of tornados, per year, in a certain region, is the integral of the frequency of tornados in that region. In
order to approximate it, we first subdivide the states into smaller regions of 100 miles 2 , as shown in the figure in the text.
We will find the upper and lower bounds for the frequency of tornados, and then take the average of the two. We do this by
finding the highest frequency of tornados in each subdivision, and then add up all the frequencies, and then do the same
for the low frequencies.
(a) For the high frequency in Texas, going left to right, top to bottom, we get: [(9 + 8) + (9 + 8 + 9 + 9 + 8) + (7 +
7 + 9 + 9 + 8) + (0 + 3 + 4 + 6 + 8 + 7 + 7 + 6) + (0 + 1 + 4 + 6 + 7 + 7) + (2 + 4 + 5) + (2 + 3)]. This equals
182 tornados. For the low frequency, we get: [(7 + 7) + (7 + 7 + 7 + 7 + 5) + (3 + 5 + 6 + 7 + 5) + (0 + 0 + 0 +
1 + 4 + 6 + 7 + 6) + (0 + 0 + 0 + 3 + 5 + 5) + (0 + 1 + 3) + (0 + 0)]. This equals 114 tornados. The average of
the two equals (182 + 114)/2 = 148 tornados.
(b) For the high frequency in Florida, going left to right, top to bottom, we get: [(5 + 5 + 5 + 9) + (9 + 9) + (7 + 7) + (5)].
This equals 61 tornados. For the low frequency, we get: [(5 + 5 + 5 + 5) + (7 + 7) + (7 + 5) + (5)]. This equals 51
tornados. The average of the two is equal to (61 + 51)/2 = 56 tornados.
(c) For the high frequency in Arizona, going left to right, top to bottom, we get: [(1 + 1 + 0) + (1 + 1 + 0) + (0 + 0 + 0) +
(0 + 0)]. This equals 4 tornados. For the low frequency, we get: [(0 + 0 + 0) + (0 + 0 + 0) + (0 + 0 + 0) + (0 + 0)].
This equals 0 tornados. The average of the two is equal to (4 + 0)/2 = 2 tornados.
30. Let R be the region 0 ≤ x ≤ 60,
0 ≤ y ≤ 8. Then
Volume =
Z
w(x, y) dA
R
Lower estimate: 10·2(1+4+8+10+10+8+0+3+4+6+6+4+0+1+2+3+3+2+0+0+1+1+1+1) = 1580.
Upper estimate:
10·2(8+13+16+17+17+16+4+8+10+11+11+10+3+4+6+7+7+6+1+2+3+4+4+3) = 3820.
The average of the two estimates is 2700 cubic feet.
31. (a) The graph of f looks like the graph of g in the xz-plane slid parallel to itself forward and backward in the y direction,
since the value of y does not affect the value of f .
(b) The solid bounded by the graph of f is a cylinder, hence
Z
f dA = Volume under surface =
R
= (d − c)
32. (a)
(b)
(c)
(d)
(e)
Z
b
Length in
y direction
Cross-sectional area
in xz-plane
g(x) dx
a
On the xy-plane, z = 0, so
of the base is x 2 + y 2 = 15, a circle of radius
√the 2equation of the edge
2
The area of the base is π( 15) = 15π meters .
√
2
2
The cross-section at z = 10 has equation
√ 2 x + y = 25, a circle of radius 5.
The area of the cross-section is π( 5) = 5π meters .
At height z, the cross-section is the circle
x2 + y 2 = 15 − z.
√
This is a circle of radius 15 − z, so
√
A(z) = π( 15 − z)2 = (15 − z)π meters2 .
(f) The approximate volume of the slice is A(z)∆z meters3 .
(g) We have
Volume of pile =
33. Let
Z
15
0
(15 − z)π dz =
z2
15z −
2
f (x, y) =
1
1 + a 1 x2 + y 2
g(x, y) =
1
1 + a 2 x2 + y 2
and let
15
225π
π =
meters3 .
2
0
√
15.
16.2 SOLUTIONS
1111
where
0 < aR1 < a2 . For all points (x, y) we have f (x, y) ≥ g(x, y) > 0, so in Riemann sum approximations for
R
f
dA
and R g dA using the same subdivision of R for both integrals, we have
R
It follows that
R
R
R
f dA ≥
X
f (xi , yj )∆x∆y ≥
X
g(xi , yj )∆x∆y.
g dA, and so increasing the value of a decreases the value of the integral
R
34. Take a Riemann sum approximation to
Z
R
X
f dA ≈
R
1
R 1+ax2 +y 2
f (xi , yi )∆A.
i,j
Then, using the fact that |a + b| ≤ |a| + |b| repeatedly, we have:
Z
X
X
f dA ≈ |
f
(x
,
y
)∆A|
≤
|f (xi , yi )∆A|.
i
i
R
i,j
Now |f (xi , yj )∆A| = |f (xi , yj |∆A since ∆A is non-negative, so
Z
X
X
f dA ≤
|f (xi , yj )|∆A.
|f
(x
,
y
)∆A|
=
i
i
R
i,j
i,j
But the last expression on the right is a Riemann sum approximation to the integral
R
R
|f |dA, so we have
Z
Z
X
X
f dA ≈ |f
(x
,
)|∆A
≈
≤
|f |dA.
f
(x
,
y
)∆A
i j
i
j
R
Thus,
R
i,j
i,j
Z
Z
f dA ≤
|f |dA.
R
R
Solutions for Section 16.2
Exercises
1. We evaluate the inside integral first:
Therefore, we have
Z
Z
3
0
Z
4
4
(4x + 3y) dx = (2x2 + 3yx) = 32 + 12y.
0
0
4
(4x + 3y) dxdy =
0
Z
Therefore, we have
Z
2
0
Z
3
(x2 + y 2 ) dy =
0
3
2
2
(x + y ) dydx =
0
3. We evaluate the inside integral first:
Therefore, we have
Z
0
3Z
Z
2
(32 + 12y) dy = (32y + 6y 2 ) = 150.
0
0
2. We evaluate the inside integral first:
Z
Z
0
y3
3
y=3
= 3x2 + 9.
y=0
2
2
(3x2 + 9) dx = (x3 + 9x) = 26.
0
2
(6xy) dy = (3xy 2 ) = 12x.
(6xy) dydx =
0
x2 y +
2
0
3
3
0
Z
3
0
3
(12x) dx = (6x2 ) = 54.
0
dA.
1112
Chapter Sixteen /SOLUTIONS
4. We evaluate the inside integral first:
Z
Therefore, we have
5.
Z
1
4Z
2
f dy dx or
1
Z
1
2Z
Z
1
0
Z
2
(x2 y) dy =
0
2
(x2 y) dydx =
0
Z
x2 y 2
2
1
y=2
= 2x2 .
y=0
(2x2 ) dx =
0
2x3
3
4
f dx dy
1
= 2.
3
0
1
6. This region lies between x = 0 and x = 4 and between the lines y = 3x and y = 12, and so the iterated integral is
Z
4
0
Z
12
f (x, y) dydx.
3x
Alternatively, we could have set up the integral as follows:
Z
12
0
7. The line connecting (−1, 1) and (3, −2) is
Z
y/3
f (x, y) dxdy.
0
3x + 4y = 1
or
y=
So the integral becomes
Z
3
−1
Z
1 − 3x
4
Z
(1−3x)/4
f dy dx or
−2
1
−2
Z
(1−4y)/3
f dx dy
−1
8. The line on the left (through points (0, 0) and (3, 6)) is the line y = 2x; the line on the right (through points (3, 6) and
(5, 0)) is the line y = −3x + 15. See Figure 16.6. One way to set up this iterated integral is:
Z
6
0
Z
(15−y)/3
f (x, y) dxdy.
y/2
The other option for setting up this integral requires two separate integrals, as follows:
Z
3
0
Z
2x
f (x, y) dydx +
0
Z
5
3
Z
−3x+15
f (x, y) dydx.
0
y
y = 2x
y = −3x + 15
R
x
Figure 16.6
9. Two of the sides of the triangle have equations x =
Z
3
1
y−5
y−1
and x =
. So the integral is
2
−2
Z
1 (y−5)
−2
1 (y−1)
2
f dx dy
16.2 SOLUTIONS
10. The line connecting (1, 0) and (4, 1) is
y=
So the integral is
11.
Z
3
1
Z
4
e
x+y
dxdy =
0
Z
3
4
Z
e e dx =
Z
0
y
1
Z
2
f dy dx
(x−1)/3
3
ex (e4 − 1) dx = (e4 − 1)(e2 − 1)e. See Figure 16.7.
x y
1
4
1
(x − 1)
3
1
y
y=x
y=4
x
x
1
3
2
Figure 16.8
Figure 16.7
12.
13.
Z
2
0
Z
x
2
ex dydx =
0
Z
2
0
x
2
ex y dx =
0
Z
5
1
Z
Z
2
2
xex dx =
0
2x
sin x dy dx =
x
=
Z
Z
5
1
5
1
2
1
1 x2 e = (e4 − 1). See Figure 16.8.
2
2
0
2x
sin x · y x dx
sin x · x dx
5
= (sin x − x cos x)1
= (sin 5 − 5 cos 5) − (sin 1 − cos 1) ≈ −2.68.
See Figure 16.9.
y
y
y = 2x
x=
√
y
y=x
y=4
4
y=x
1
x
1
x
5
1
Figure 16.9
14.
Z
Figure 16.10
4
1
2
Z
y
√
x2 y 3 dxdy =
y
Z
4
y3
1
y
x3 dy
3 √y
4
1113
1114
Chapter Sixteen /SOLUTIONS
1
3
=
1
3
=
1
=
3
See Figure 16.10.
Z
4
1
9
(y 6 − y 2 ) dy
y 11/2
y7
−
7
11/2
4
1
47
411/2 × 2
−
7
11
−
1
2
−
7
11
≈ 656.082
15. The region of integration ranges from x = 0 to x = 3 and from y = 0 to y = 2x, as shown in Figure 16.11. To evaluate
the integral, we evaluate the inside integral first:
Z
2x
(x2 + y 2 ) dy =
0
Therefore, we have
Z
3
0
Z
x2 y +
y3
3
2x
y=2x
(2x)3
8x3
14 3
= 2x3 +
=
x .
= x2 (2x) +
3
3
3
y=0
Z
(x2 + y 2 ) dydx =
0
3
0
14 3
x
3
y
6
dx =
14 4 3
x = 94.5.
12
0
x2 + y 2 = 9
y
3
y = 2x
R
−3 −2
x
−3
3
Figure 16.12
Figure 16.11
16. See Figure 16.12.
Z
0
−2
Z
0
−
√
2xy dydx =
9−x2
Z
0
−2
=−
=
=
Z
Z
0
0
−2
−2
0
x y 2 √
−
Z
√
x + y dA =
R
=
=
Z
Z
2
0
2
0
2
3
Z
Z
1
0
x(9 − x2 ) dx
(x3 − 9x) dx
x4
9
− x2
4
2
0
√
x + y dx dy
1
3
2
(x + y) 2 dy
3
0
2
0
dx
9−x2
= −4 + 18 = 14
17.
3
3
3
((1 + y) 2 − y 2 ) dy
−2
x
16.2 SOLUTIONS
2
5 5
2 2
· [(1 + y) 2 − y 2 ]
3 5
0
=
5
5
4
((3 2 − 2 2 ) − (1 − 0))
15
√
4 √
(9 3 − 4 2 − 1) = 2.38176
=
15
=
18. In the other order, the integral is
Z
1
0
Z
2
√
x + y dy dx.
0
First we keep x fixed and calculate the inside integral with respect to y:
Z
2
0
y=2
√
2
x + y dy = (x + y)3/2 3
y=0
=
Then the outside integral becomes
Z
1
0
2
(x + 2)3/2 − x3/2 .
3
i 1
h
2 2
2
2 5/2 3/2
3/2
5/2
(x + 2)
−x
dx =
(x + 2)
− x
3
3 5
5
0
=
2 2 5/2
3
− 1 − 25/2 = 2.38176
·
3 5
Note that the answer is the same as the one we got in Exercise 17.
19.
Z
(5x2 + 1) sin 3y dA =
R
=
=
=
=
Z
Z
1
−1
1
Z
π/3
(5x2 + 1) sin 3y dy dx
0
2
(5x + 1)
−1
2
3
Z
1
1
π/3
− cos 3y 3
0
dx
(5x2 + 1) dx
−1
1
2 5 3
( x + x)
3 3
−1
32
2 10
(
+ 2) =
3 3
9
20. The region of integration, R, is shown in Figure 16.13.
Integrating first over y, as shown in the diagram, we obtain
Z
xy dA =
R
Z
1
0
Z
1−x
xy dy
0
Now integrating with respect to x gives
Z
xy dA =
R
dx =
Z
1
0
1−x
xy 2 2 0
dx =
Z
1
0
1 2 1 3 1 4 1
1
x − x + x =
.
4
3
8
24
0
1
x(1 − x)2 dx
2
1115
1116
Chapter Sixteen /SOLUTIONS
y
1
y
1
y =1−x
y = −x + 1
y =x+1
x
x
−1
1
1
Figure 16.13
Figure 16.14
21. It would be easier to integrate first in the x direction from x = y − 1 to x = −y + 1, because integrating first in the y
direction would involve two separate integrals.
Z
2
(2x + 3y) dA =
R
=
=
=
=
Z
Z
Z
Z
Z
1
0
Z
1
0
1
0
1
0
−y+1
(2x + 3y)2 dx dy
y−1
−y+1
(4x2 + 12xy + 9y 2 ) dx dy
y−1
4 3
x + 6x2 y + 9xy 2
3
−y+1
dy
y−1
8
[ (−y + 1)3 + 9y 2 (−2y + 2)] dy
3
2
9
(−y + 1)4 − y 4 + 6y 3
3
2
−
9
13
2
= − (−1) − + 6 =
3
2
6
1
0
See Figure 16.14.
22. The region is bounded by x = 1, x = 4, y = 2, and y = 2x. Thus
Volume =
Z
4
1
Z
2x
(6x2 y) dydx.
2
To evaluate this integral, we evaluate the inside integral first:
Z
Therefore, we have
Z
4
1
Z
2x
2x
2
(6x2 y) dy = (3x2 y 2 )
2x
(6x2 y) dydx =
2
The volume of this object is 2203.2.
Z
2
= 3x2 (2x)2 − 3x2 (22 ) = 12x4 − 12x2 .
4
1
(12x4 − 12x2 ) dx =
23. To find the average value, we evaluate the integral
Z
3
0
Z
12 5
4
x − 4x3 = 2203.2.
5
1
6
(x2 + 4y) dydx,
0
and then divide by the area of the base region.
To evaluate this integral, we evaluate the inside integral first:
Z
6
0
y=6
(x2 + 4y) dy = (x2 y + 2y 2 )
y=0
= 6x2 + 72.
16.2 SOLUTIONS
Therefore, we have
Z
3
0
Z
Z
6
2
(x + 4y)dydx =
0
1117
3
3
(6x2 + 72)dx = (2x3 + 72x) = 270.
0
0
The value of the integral is 270. The area of the base region is 3 · 6 = 18. To find the average value of the function, we
divide the value of the integral by the area of the base region:
1
Average value =
Area
Z
Z
3
0
6
1
· 270 = 15.
18
(x2 + 4y) dydx =
0
The average value is 15. This is reasonable, since the smallest value of f (x, y) on this region is 0, and the largest value is
32 + 4 · 6 = 33.
24. To find the average value, we first find the value of the integral
Z
We evaluate the inside integral first:
Therefore, we have
Z
4
Z
Z
4
0
Z
3
(xy 2 ) dydx.
0
3
2
(xy ) dy =
0
Z
3
xy 3
3
y=3
= 9x.
y=0
4
9x2 4
(xy ) dydx =
(9x) dx =
= 72.
2
0
0
0
0
The value of the integral is 72. To find the average value, we divide the value of the integral by the area of the region:
2
Average value =
1
Area
Z
4
0
Z
3
72
= 6.
3·4
(xy 2 ) dydx =
0
The average value of f (x, y) on this rectangle is 6. This is reasonable since the smallest value of xy 2 on this region is 0
and the largest value is 4 · 32 = 36.
25. (a) See Figure 16.15.
y
(1/2, 1/2)
y=x
x+y =1
R
x
Figure 16.15
(b) If we integrate with respect to x first, we have
Z
f (x, y) dA =
R
Z
1/2
0
Z
1−y
f (x, y) dx dy.
y
If we integrate with respect to y first, the integral must be split into two parts, so
(c) If f (x, y) = x,
Z
f (x, y) dA =
R
Z
x dA =
R
=
Z
1/2
0
Z
1/2
0
1
2
Z
Z
x
f (x, y) dy dx +
0
Z
1−y
x dx dy =
y
1/2
0
Z
Z
1/2
=
1
.
8
1/2
1/2
0
(1 − y)2 − y 2 dy =
1
= (y − y 2 )
2
0
1
1
2
Z
1−x
f (x, y) dy dx.
0
1−y
x2 2 y
Z
dy
1/2
0
1 − 2y dy
1118
Chapter Sixteen /SOLUTIONS
Alternatively,
Z
x dA =
R
=
=
Z
1/2
0
Z
Z
1/2
0
Z
x
x dy dx +
0
x
xy dx +
0
1/2
x2 dx +
0
1/2
x3 =
3 0
+
Z
Z
Z
1/2
1
1/2
1
1/2
1
Z
1−x
x dy dx
0
1−x
xy dx
0
x(1 − x) dx
x3
x2
−
2
3
1
1/2
1
1
1
1
1
1
+ − − +
= .
=
24
2
3
8
24
8
26. (a)
y
4
x = −(y − 4)/2 or y = −2x + 4
x
2
Figure 16.16
(b)
R 2 R −2x+4
0
0
g(x, y) dy dx.
Problems
27. As given, the region of integration is as shown in Figure 16.17. Reversing the limits gives
Z
1
0
Z
y
x
e
0
x2
dydx =
Z
1
0
x2 1
x Z
dx =
ye 1
x2
0
e−1
e .
=
=
2 0
2
2
xex dx
0
y
1
y=x
x=y
x=1
x
x
1
1
Figure 16.17
Figure 16.18
16.2 SOLUTIONS
1119
28. The function sin (x2 ) has no elementary antiderivative, so we try integrating with respect to y first. The region of integration is shown in Figure 16.18. Changing the order of integration, we get
Z
Z
1
0
1
sin (x2 ) dx dy =
y
=
=
Z
Z
Z
Z
1
0
x
sin (x2 ) dy dx
0
x
sin (x ) · y dx
1
2
0
0
1
sin (x2 ) · x dx
0
1
cos (x2 ) =−
2
0
=−
1
1
cos 1
+ = (1 − cos 1) = 0.23.
2
2
2
29. As given, the region of integration is as shown in Figure 16.19.
Reversing the limits gives
Z
1
0
Z
x2
p
2+
0
x3
dydx =
=
=
Z
Z
1
(y
0
1
x
0
p
2+
x3
x2
) dx
0
p
2
2 + x3 dx
1
√
3
2 √
2
(2 + x3 ) 2 = (3 3 − 2 2).
9
9
0
y
y
1
x=
√
3
y
x = y2
x=1
x=9
x
x
9
1
Figure 16.20
Figure 16.19
30. As given, the region of integration is as shown in Figure 16.20.
Reversing the limits gives
Z
9
0
Z
√
x
2
y sin (x ) dydx =
0
=
Z
0
1
2
=−
=
√x !
9
Z
9
y 2 sin (x2 ) 2
0
x sin (x2 ) dx
0
9
cos (x2 ) 4
0
cos (81)
1
−
= 0.056.
4
4
dx
1120
Chapter Sixteen /SOLUTIONS
31. The region of the integration is shown in Figure 16.21. To make the integration easier, we want to change the order of the
integration and get
Z
1
0
Z
e
ey
x
dx dy =
ln x
=
=
y
Z
Z
Z
e
1
e
1
e
Z
ln x
0
x
dy dx
ln x
ln x
x
· y
ln x 0
x dx =
1
(e, 1)
dx
e
1
x2 = (e2 − 1).
2 1
2
8
y = 2x + 8
R
e
1
32. Order reversed:
8
0
Z
y = −2x + 8
x
x
−4
Figure 16.21
Z
y
4
Figure 16.22
(8−y)/2
f (x, y) dx dy. See Figure 16.22.
(y−8)/2
33. (a) We divide the base region into four subrectangles as shown in Figure 16.23. The height of the object at each point
(x, y) is given by f (x, y) = xy, we label each corner of the subrectangles with the value of the function at that point.
(See Figure 16.23.) Since Volume = Height × Length × Width, and ∆x = 2 and ∆y = 3, we have
Overestimate = (12 + 24 + 6 + 12)(2)(3) = 324,
and
Underestimate = (0 + 6 + 0 + 0)(2)(3) = 36.
We average these to obtain
Volume ≈
324 + 36
= 180.
2
y
6
3
f =0
f = 12
f = 24
f =0
f =6
f = 12
f =0
f =0
f =0
x
2
4
Figure 16.23
(b) We have f (x, y) = xy, so
Volume =
Z
4
0
Z
6
xy dydx =
0
Z
4
0
xy 2
2
y=6
Z
dx
=
y=0
4
0
4
18x dx = 9x2 = 144.
0
The volume of this object is 144. Notice that 144 is between the over- and underestimates, 324 and 36, found in
part (a).
16.2 SOLUTIONS
1121
34. (a) The contour f (x, y) = 1 lies in the xy-plane and has equation
2e−(x−1)
2
−y 2
= 1,
so
−(x − 1)2 − y 2 = ln(1/2)
(x − 1)2 + y 2 = ln 2 = 0.69.
This is the equation of a circle centered at (1, 0) in the xy-plane.
Other contours are of the form
2e−(x−1)
2
−y 2
2
=c
2
−(x − 1) − y = ln(c/2).
Thus, all the contours are circles centered at the point (1, 0).
2
(b) The cross-section
has equation z = f (1, y) = e−y . If x = 1, the base region in the xy-plane extends from
√
√
y = − 3 to y = 3. See Figure 16.24, which shows the circular region below W in the xy-plane. So
Area =
Z
√
3
√
2
e−y dy.
− 3
(c) Slicing parallel to the y-axis, we get
Volume =
y=
√
Z
2
−2
Z √4−x2
−
√
e−(x−1)
2
−2
(1,
√
2
√
y = − 4 − x2
−y 2
dy dx.
4−x2
y
4 − x2
2
−2
3)
x
√
(1, − 3)
Figure 16.24: Region beneath W in the
xy-plane
35. The intersection of the graph of f (x, y) = 25 − x2 − y 2 and xy-plane is a circle x2 + y 2 = 25. The given solid is shown
in Figure 16.25.
Thus the volume of the solid is
V =
=
Z
Z
f (x, y) dA
R
5
−5
Z √25−y2
−
√
25−y 2
(25 − x2 − y 2 ) dx dy.
1122
Chapter Sixteen /SOLUTIONS
f (x, y) = 25 − x2 − y 2
z
z
f (x, y) = 25 − x2 − y 2
z = 16
y
y
x
x
Figure 16.25
Figure 16.26
36. The intersection of the graph of f (x, y) = 25 − x2 − y 2 and the plane z = 16 is a circle, x2 + y 2 = 32 . The given solid
is shown in Figure 16.26.
Thus, the volume of the solid is
Z
V =
Z
=
R
(f (x, y) − 16) dA
Z √9−y2
3
−
−3
√
9−y 2
(9 − x2 − y 2 ) dx dy.
37. The solid is shown in Figure 16.27, and the base of the integral is the triangle as shown in Figure 16.28.
z
y
y=0
4
y−x=4
backside
-
2x + y + z = 4
y−x=4
−4
2x + y = 4
y
4
2
x
x
−4
Figure 16.27
Thus, the volume of the solid is
V =
=
=
2
Figure 16.28
Z
z dA
ZR
Z
(4 − 2x − y) dA
R
4
0
Z
(4−y)/2
y−4
(4 − 2x − y) dx dy.
16.2 SOLUTIONS
38.
Volume =
Z
2
0
Z
Z
2
xy dy dx =
0
=
Z
2
0
2
2x dx
0
2
1 2 xy dx
2
0
2
= x2 =4
1123
0
39. The region of integration is shown in Figure 16.29. Thus
Volume =
Z
1
0
Z
x
(x2 + y 2 ) dy dx =
0
Z
1
0
x2 y +
y3
3
y=x
Z
dx
=
y=0
1
0
1
x4 1
4 3
x dx =
= .
3
3 0
3
y
1
y
x=y
x=1
3
x = y2
x=9
x
x
1
9
Figure 16.30
Figure 16.29
40. The region of integration is shown in Figure 16.30. Thus,
Volume =
=
Z
Z
9
0
9
0
Z
√
x
(x + y) dy dx =
0
x3/2 +
x
2
dx =
Z
9
0
y2
xy +
2
y=√x
dx
y=0
9
2 5/2 x 2349
x
+
= 117.45.
=
5
4 0
20
2
41. The plane 2x + y + z = 4 cuts the xy-plane in the line 2x + y = 4, so the region of integration is the triangle shown in
Figure 16.31. We want to find the volume under the graph of z = 4 − 2x − y. Thus,
Volume =
=
=
Z
Z
Z
2
0
2
0
2
0
Z
−2x+4
0
(4 − 2x − y) dy dx =
Z
2
0
4y − 2xy −
(−2x + 4)2
4(−2x + 4) − 2x(−2x + 4) −
2
(2x2 − 8x + 8) dx =
y2
2
dx
2 3
2 16
x − 4x2 + 8x =
.
3
3
0
−2x+4
dx
0
1124
Chapter Sixteen /SOLUTIONS
y
4
x = −(y − 4)/2 or y = −2x + 4
x
2
Figure 16.31
42. Let R be the triangle with vertices (1, 0), (2, 2) and (0, 1). Note that (3x + 2y + 1) − (x + y) = 2x + y + 1 > 0 for
x, y > 0, so z = 3x + 2y + 1 is above z = x + y on R. We want to find
Z
Volume =
R
((3x + 2y + 1) − (x + y)) dA =
We need to express this in terms of double integrals.
Z
(2x + y + 1) dA.
R
y
(2, 2)
2
y = 1 + 0.5x
R2
1
R1
y = 2x − 2
y =1−x
x
1
O
2
Figure 16.32
To do this, divide R into two regions with the line x = 1 to make regions R 1 for x ≤ 1 and R2 for x ≥ 1. See
Figure 16.32. We want to find
Z
(2x + y + 1) dA =
R
Z
(2x + y + 1) dA +
R1
Z
(2x + y + 1) dA.
R2
Note that the line connecting (0, 1) and (1, 0) is y = 1 − x, and the line connecting (0, 1) and (2, 2) is y = 1 + 0.5x. So
Z
(2x + y + 1) dA =
R1
The line between (1, 0) and (2, 2) is y = 2x − 2, so
Z
(2x + y + 1) dA =
R2
Z
Z
1
0
2
1
Z
Z
1+0.5x
(2x + y + 1) dy dx.
1−x
1+0.5x
(2x + y + 1) dy dx.
2x−2
16.2 SOLUTIONS
1125
We can now compute the double integral for R1 :
Z
1
0
Z
1+0.5x
(2x + y + 1) dy dx =
1−x
=
=
=
Z
Z
1
0
1
0
y2
+y
2xy +
2
21 2
x + 3x
8
1+0.5x
dx
1−x
dx
1
7 3 3 2 x + x dx
8
2
0
19
,
8
and the double integral for R2 :
Z
2
1
Z
1+0.5x
(2x + y + 1) dy dx =
2x−2
=
Z
Z
2
1
1
1+0.5x
(2xy + y 2 /2 + y)
0
−
dx
2x−2
3
39 2
x + 9x +
8
2
dx
2
13 3 9 2 3
= − x + x + x 8
2
2
1
29
=
.
8
19
29
48
+
=
= 6.
8
8
8
43. We want to calculate the volume of the tetrahedron shown in Figure 16.33.
So, Volume =
z
1/c
1/b
1/a
y
=1
ax + by
x
Figure 16.33
We first find the region in the xy-plane where the graph of ax + by + cz = 1 is above the xy-plane. When z = 0
we have ax + by = 1. So the region over which we want to integrate is bounded by x = 0, y = 0 and ax + by = 1.
Integrating with respect to y first, we have
Volume =
Z
1/a
0
Z
=
=
=
(1−ax)/b
z dy dx =
0
Z
Z
1/a
0
1/a
0
1
.
6abc
Z
1/a
0
y
by 2
axy
−
−
c
2c
c
Z
(1−ax)/b
0
1 − by − ax
dy dx
c
y=(1−ax)/b
dx
y=0
1
(1 − 2ax + a2 x2 ) dx
2bc
1126
Chapter Sixteen /SOLUTIONS
44. The region bounded by the x-axis and the graph of y = x − x 2 is shown in Figure 16.34. The area of this region is
A=
Z
1
0
(x − x2 )dx = (
1
1
1
= − = .
2
3
6
1
x3 x2
−
)
2
3 0
y
1
4
x
0
1
1
2
Figure 16.34
So the average distance to the x-axis for points in the region is
R
Average distance =
Z
y dA =
R
=
Therefore the average distance is
1/60
1/6
Z
Z
1
0
1
0
Z
x−x2
y dy
0
!
x4
x2
− x3 +
2
2
R
y dA
area(R)
dx
dx =
1
1
1
1
− +
=
.
6
4
10
60
= 1/10.
45. Assume the length of the two legs of the right triangle are a and b, respectively. See Figure 16.35. The line through (a, 0)
and (0, b) is given by yb + xa = 1. So the area of this triangle is
A=
1
ab.
2
y
b
a
x
Figure 16.35
Thus the average distance from the points in the triangle to the y-axis (one of the legs) is
Average distance =
=
1
A
2
ab
Z
a
0
Z
a
0
Z
b x+b
−a
x dy dx
0
b
− x2 + bx
a
dx
a
b 3 b 2 2
− x + x =
ab
3a
2
0
2 =
2
ab
a b
6
=
a
.
3
16.2 SOLUTIONS
1127
Similarly, the average distance from the points in the triangle to the x-axis (the other leg) is
Average distance =
=
=
46. (a) We have
1
Average value of f =
Area of Square
=
1
4
1
=
4
Z
Z
2
0
2
0
Z
2
Z
1
A
2
ab
Z bZ
2
ab
0
−a
y+a
b
Z b
0
ab2
6
a
− y 2 + ay
b
=
b
.
3
dy
f dA
Square
(ax2 + bxy + cy 2 ) dydx =
0
y dx dy
0
8
2ax2 + 2bx + c
3
1 16
16
=
a + 4b +
c
4 3
3
4
4
= a+b+ c
3
3
1
dx =
4
1
4
Z
2
0
ax2 y + bx
y2
y3
+c
2
3
2
8
2 3
ax + bx2 + cx 3
3
0
y=2
dx
y=0
The average value will be 20 if and only if (4/3)a + b + (4/3)c = 20.
(b) Since (4/3)a + b + (4/3)c = 20, we must have b = 20 − (4/3)a − (4/3)c. Any function f (x, y) = ax 2 + (20 −
(4/3)a − (4/3)c)xy + cy 2 where a and c are any real numbers is a correct solution. For example, a = 1, c = 3 leads
to the function f (x, y) = x2 + (44/3)xy + 3y 2 , and a = −3, c = 0 leads to the function f (x, y) = −3x2 + 24xy,
both of which have average value 20 on the given square. See Figures 16.36 and 16.37.
y
y
x
Figure 16.36: f (x, y) = x
2
x
Figure 16.37: f (x, y) = −3x + 24xy
2
+ 44
xy+3y 2
3
47. (a) We have
Average value of f =
=
=
Z
1
f dA
Area of Rectangle Rectangle
1
6
1
6
Z
Z
2
x=0
2
0
Z
3
(ax + by) dydx =
y=0
1
6
Z
2
0
axy + b
2
9
1 3 2 9
3ax + b dx =
ax + bx 2
6 2
2
0
1
= (6a + 9b)
6
3
= a + b.
2
y2
2
y=3
dx
y=0
1128
Chapter Sixteen /SOLUTIONS
The average value will be 20 if and only if a + (3/2)b = 20.
This equation can also be expressed as 2a + 3b = 40, which shows that f (x, y) = ax + by has average value
of 20 on the rectangle 0 ≤ x ≤ 2, 0 ≤ y ≤ 3 if and only if f (2, 3) = 40.
(b) Since 2a + 3b = 40, we must have b = (40/3) − (2/3)a. Any function f (x, y) = ax + ((40/3) − (2/3)a)y where
a is any real number is a correct solution. For example, a = 1 leads to the function f (x, y) = x + (38/3)y, and
a = −3 leads to the function f (x, y) = −3x + (46/3)y, both of which have average value 20 on the given rectangle.
See Figure 16.38 and 16.39.
y
y
x
Figure 16.38: f (x, y) = x +
x
Figure 16.39: f (x, y) = −3x +
38
y
3
46
y
3
48. (a) One solution would be to arrange that the minimum values of f on the square occur at the corners, so that the corner
values give an underestimate of the average. See Figure 16.40.
y
1
y
94
96
1
0.3
0.2
98
0.1
100
0
x
x
1
1
Figure 16.40
Figure 16.41
(b) One solution would be to arrange that the maximum values of f on the square occur at the corners, so that the corner
values give an overestimate of the average. See Figure 16.41.
49. The force, ∆F , acting on ∆A, a small piece of area, is given by
∆F ≈ p∆A,
where p is the pressure at that point. Thus, if R is the rectangle, the total force is given by
F =
Z
p dA.
R
We choose coordinates with the origin at one corner of the plate. See Figure 16.42.
y
(a, b)
b
a
Figure 16.42
x
16.3 SOLUTIONS
Suppose p is proportional to the square of the distance from the corner represented by the origin. Then we have
p = k(x2 + y 2 ),
Thus, we want to compute
F =
Z
Z
for some positive constant k.
k(x2 + y 2 )dA. Rewriting as an iterated integral, we have
R
2
2
k(x + y ) dA =
R
Z bZ
0
=k
a
2
2
k(x + y ) dxdy = k
0
Z b
0
a3
+ ay 2
3
Z b
0
dy = k
b !
a3 y
y 3 +a
3
3
0
k
= (a3 b + ab3 ).
3
Solutions for Section 16.3
Exercises
1.
Z
f dV =
W
=
=
=
=
Z
Z
Z
Z
Z
Z
2
0
Z
2
0
Z
2
0
2
1
−1
Z
3
(x2 + 5y 2 − z) dz dy dx
2
3
1
(x z + 5y z − z 2 ) dy dx
2
−1
2
1
2
2
1
−1
(x2 + 5y 2 −
(x2 y +
0
2
(2x2 +
0
5
) dy dx
2
1
5 3 5 y − y) dx
3
2 −1
10
− 5) dx
3
2
2
5 = ( x3 − x)
3
3 0
=
10
16
−
=2
3
3
2.
Z
f dV =
W
=
=
Z
Z
Z
1
0
1
0
1
Z
Z
1
0
1
Z
2
(ax + by + cz) dz dy dx
0
(2ax + 2by + 2c) dy dx
0
(2ax + b + 2c) dx
0
= a + b + 2c
a x3
+ xy 2 3
0
dy
1129
1130
Chapter Sixteen /SOLUTIONS
3.
Z
f dV =
W
=
=
=
Z
Z
Z
Z
Z bZ
a
0
0
Z bZ
a
0
Z
a
0
Z
a
0
0
c
e−x−y−z dz dy dx
0
c
e−x e−y e−z dz dy dx
0
b
e
−x −y
e
0
c
(−e−z ) dy dx
0
b
e−x e−y (−e−c + 1) dy dx
0
= (1 − e−c )
= (1 − e
−b
Z
a
0
b
e−x (−e−y ) dx
)(1 − e
−c
)
Z
0
a
e
−x
dx
0
= (1 − e−a )(1 − e−b )(1 − e−c )
4.
Z
f dV =
W
=
=
=
Z
Z
Z
π
0
0
Z
πZ
π
Z
π
0
π
0
= −2
= −2
Z
π
sin x cos(y + z) dz dy dx
0
π
sin x sin(y + z) dy dx
0
π
Z0 π Z0 π
0
Z
0
π
Z0 π
0
sin x[sin(y + π) − sin y] dy dx
sin x(−2 sin y) dy dx
π
sin x(− cos y) dx
0
2 sin x dx
π
= −4(− cos x)
0
= (−4)(2) = −8
5. The region is the half cylinder in Figure 16.43.
6. The region is the half cylinder in Figure 16.44.
7. The region is the quarter sphere in Figure 16.45.
z
1
z
z
1
1
1
1
x
y
x
Figure 16.43
1
1
Figure 16.44
y
x
1
1
Figure 16.45
y
1131
16.3 SOLUTIONS
8. The region is the half cylinder in Figure 16.46.
9. The region is the cylinder in Figure 16.47.
10. The region is the hemisphere in Figure 16.48.
z
1
z
1
z
1
1
x 1
x
y
1
y
1
x
1
Figure 16.47
Figure 16.46
1
y
Figure 16.48
11. The region is the hemisphere in Figure 16.49.
12. The region is the quarter sphere in Figure 16.50.
13. The region is the quarter sphere in Figure 16.51.
z
z
1
1
z
1
x
1
x
y
1
1
y
1
1 y
x 1
Figure 16.49
Figure 16.51
Figure 16.50
Problems
14. A slice through W for a fixed value of x is a semi-circle the boundary of which is y 2 = r2 − x2 − z 2 , so the inner integral
is
√
Z
r 2 −x2 −z 2
f (x, y, z) dy.
0
√
√
Lining up these stacks parallel to y-axis gives a slice from z = − r2 − x2 to z = r2 − x2 giving
Z √
Z √
r 2 −x2
−
√
r 2 −x2
r 2 −x2 −z 2
f (x, y, z) dy dz.
0
Finally, there is a slice for each x between −r and r, so the integral we want is
Z Z √
Z √
r 2 −x2
r
−r
−
√
r 2 −x2
r 2 −x2 −z 2
0
f (x, y, z) dy dz dx.
1132
Chapter Sixteen /SOLUTIONS
15. A slice through W for a fixed value of x is a semi-circle the boundary of which is y 2 = r2 − x2 − z 2 , so the inner integral
is
√
Z
r − x2 −z 2
−
f (x, y, z) dy.
√
r − x2 −z 2
Lining up these stacks parallel to y-axis gives a slice from z = 0 to z =
Z √
Z √
r−x2
r−x2 −z 2
−
0
√
√
r2 − x2 giving
f (x, y, z) dy dz.
r−x2 −z 2
Finally, there is a slice for each x between 0 and r, so the integral we want is
Z √
Z Z √
r−x2
r
0
r−x2 −z 2
−
0
√
f (x, y, z) dy dz dx.
r−x2 −z 2
16. A slice through W for a fixed value of x is a semi-circle the boundary of which is y 2 = 4 − z 2 , so the inner integral is
Z √
4−z 2
f (x, y, z) dy.
0
Lining up these stacks parallel to y-axis gives a slice from z = −2 to z = 2 giving
Z Z √
4−z 2
2
f (x, y, z) dy dz.
−2
0
Finally, there is a slice for each x between 0 and 1, so the integral we want is
Z Z Z √
1
4−z 2
2
f (x, y, z) dy dz dx.
−2
0
0
17. A slice through W for a fixed value of y is a semi-circle the boundary of which is z 2 = r2 − x2 , so the inner integral is
Z √
r 2 −x2
f (x, y, z) dz.
0
Lining up these stacks parallel to z-axis gives a slice from x = −r to x = r giving
Z Z √
r 2 −x2
r
−r
f (x, y, z) dz dx.
0
Finally, there is a slice for each y between 0 and 1, so the integral we want is
Z Z Z √
1
0
18. The required volume, V , is given by
V =
=
=
=
r 2 −x2
r
−r
Z
Z
Z
Z
10
0
10
0
10
f (x, y, z) dz dx dy.
0
Z
Z
h
10−x
0
10−x
0
Z
10
dzdydx
x+y
(10 − (x + y)) dydx
10y − xy −
0
10
0
500
=
3
1 2
y
2
1
(10 − x)2 dx
2
iy=10−x
y=0
dx
16.3 SOLUTIONS
1133
19. The pyramid is shown in Figure 16.52. The planes y = 0, and y − x = 4, and 2x + y + z = 4 intersect the plane z = −6
in the lines y = 0, y − x = 4, 2x + y = 10 on the z = −6 plane as shown in Figure 16.53.
z
y
y=0
-
(2, 6, −6)
y − x = 4 backside
z = −6 plane
2x + y + z = 4
−4
y
4
2
y−x=4
x
(−4, 0, −6)
2x + y − 6 = 4
(2, 6, −6)
z = −6 bottom
(5, 0, −6)
x
(5, 0, −6)
(−4, 0, −6)
Figure 16.52
Figure 16.53
These three lines intersect at the points (−4, 0, −6), (5, 0, −6), and (2, 6, −6). Let R be the triangle in the planes
z = −6 with the above three points as vertices. Then, the volume of the solid is
V =
=
=
=
Z
Z
Z
Z
6
0
6
0
6
Z
Z
(10−y)/2
y−4
(10−y)/2
Z
4−2x−y
dz dx dy
−6
(10 − 2x − y) dx dy = 162
y−4
(10−y)/2
dy
(10x − x − xy)
2
0
y−4
6
0
9y 2
(
− 27y + 81) dy
4
= 162
20. Figure 16.54 shows a slice through the region for a fixed x.
3
z
2
z = 3y
1
z=y
y
0
1
Figure 16.54
1134
Chapter Sixteen /SOLUTIONS
The required volume, V , is given by
V =
=
=
=
=
Z
Z
Z
Z
Z
Z
2
1
Z
2
1
Z
2
1
2
1
0
1
Z
3y
dz dy dx
y
z|3y
y dy dx
0
1
2y dy dx
0
1
y 2 0 dx
1
2
dx
1
= 1.
21. Figure 16.55 shows a slice through the region for a fixed value of y.
z
1
z=x
z = x2
x
0
1
Figure 16.55
We break the region into small cubes of volume ∆V = ∆x∆y∆z. A stack of cubes vertically above the point (x, z)
in the xz-plane gives the strip shown in Figure 16.55 and so the inner integral is
Z
x
dz
x2
The plane and the surface meet when x = x2 , giving x(1 − x) = 0, so x = 0 or x = 1. Lining up the stacks parallel to
the z-axis gives a slice from x = 0 to x = 1. Thus, the limits on the middle integral are
Z
1
0
Z
x
dz dx.
x2
Finally, there is a slice for each y between 0 and 3, so the integral we want is
The required volume, V , is given by
Z
V =
3
0
Z
Z
3
0
1
0
Z
Z
1
0
x
dz dx dy.
x2
Z
x
dz dx dy
x2
16.3 SOLUTIONS
Z
=
Z
=
Z
=
3
0
3
Z
0
3
0
Z
1
1135
x − x2 dx dy
0
1
x3 x2
dy
−
2
3 0
1
dy
6
3
1
=
dy
6 0
1
= ·3
6
1
= .
2
22. The required volume, V , is given by
V =
=
=
=
=
Z
Z
Z
Z
Z
5
0
Z
5
0
5
0
5−x
0
dz dy dx
0
(x + y) dy dx
5
x+y
0
1 2 y=5−x
y dx
2
y=0
xy +
0
Z
5−x
x(5 − x) +
1
(5 − x)2
2
125
.
3
dx
23. The required volume, V , is given by
V =
=
=
=
Z
Z
Z
Z
5
0
5
0
5
0
5
Z
Z
3
0
3
Z
x2
dz dy dx
0
x2 dy dx
0
y=3
x2 y dx
y=0
3x2 dx
0
= 125.
24. (a) The vectors ~
u = ~i − ~j and ~v = ~i − ~k lie in the required plane so p
~ =~
u × ~v = ~i + ~j + ~k is perpendicular
~
~
~
to this plane. Let (x, y, z) be a point in the plane, then (x − 1)i + y j + z k is perpendicular to p
~ , so ((x − 1)~i +
~
~
~
~
~
y j + z k ) · (i + j + j ) = 0 and so
(x − 1) + y + z = 0.
Therefore, the equation of the required plane is x + y + z = 1.
(b) The required volume, V , is given by
V =
Z
1
0
Z
0
1−x
Z
1−x−y
dz dy dx
0
1136
Chapter Sixteen /SOLUTIONS
=
=
=
=
Z
Z
Z
Z
=
1
0
Z
1−x
(1 − x − y) dy dx
0
1
y − xy −
0
1
0
1
0
1 2 1−x
y dx
2
0
1 − x − x(1 − x) −
1
(1 − x)2 dx
2
1
(1 − x)2 dx
2
1
.
6
25. (a) The equation of the surface of the whole cylinder along the y-axis is x 2 + z 2 = 1. The part we want is
z=
See Figure 16.56.
p
1 − x2
0 ≤ y ≤ 10.
z
x
1
10
y
Figure 16.56
(b) The integral is
Z
f (x, y, z) dV =
D
Z
10
0
Z
1
−1
Z √1−x2
f (x, y, z) dzdxdy.
0
26. The region of integration is shown in Figure 16.57, and the mass of the given solid is given by
z
6
+ y2 + z6 = 1
or z = −2x − 3y + 6
x
3
2
y
x
3
x
Figure 16.57
Z
δ dV
R
y
2
=1
2
x+2
or y = − 3
3
mass =
+
16.3 SOLUTIONS
=
Z
Z
3
0
Z
2 x+2
−3
0
−2x−3y+6
1137
(x + y) dzdydx
0
−2x−3y+6
Z 3 Z − 2 x+2
3
(x + y)z =
dydx
0
0
0
Z 3 Z − 2 x+2
3
(x + y)(−2x − 3y + 6) dydx
=
=
=
=
=
=
Z
Z
Z
0
Z
3
0
x+2
−2
3
0
3
0
3
0
0
(−2x2 − 3y 2 − 5xy + 6x + 6y) dydx
− 3 x+2
5
−2x y − y − xy 2 + 6xy + 3y 2 dx
2
0
2
3
14 3 8 2
x − x + 2x + 4
27
3
2
dx
3
7 4 8 3
x − x + x2 + 4x 54
9
0
8
21
15
7
· 34 − · 33 + 32 + 12 =
−3=
.
54
9
2
2
27. The pyramid is shown in Figure 16.58. The planes y = 0, and y − x = 4, and 2x + y + z = 4 intersect the plane z = −6
in the lines y = 0, y − x = 4, 2x + y = 10 as shown in Figure 16.59.
z
y
y=0
-
(2, 6, −6)
y − x = 4 backside
z = −6 plane
2x + y + z = 4
−4
y
4
2
y−x=4
x
(−4, 0, −6)
2x + y − 6 = 4
(2, 6, −6)
z = −6 bottom
(5, 0, −6)
x
(5, 0, −6)
(−4, 0, −6)
Figure 16.58
Figure 16.59
These three lines (the edges of the pyramid) intersect the plane z = −6 at the points (−4, 0, −6), (5, 0, −6), and
(2, 6, −6). Let R be the triangle in the plane z = −6 with these three points as vertices. Then, the mass of the solid is
Mass =
=
=
=
=
Z
Z
Z
Z
Z
6
0
6
0
6
0
6
0
6
Z
Z
Z
(10−y)/2
y−4
(10−y)/2
y−4
(10−y)/2
y−4
Z
Z
4−2x−y
δ(x, y, z) dz dx dy
−6
4−2x−y
y dz dx dy
−6
y(10 − 2x − y) dx dy
x=(10−y)/2
y(10x − x2 − xy)
(
0
= 243.
x=y−4
9y 3
− 27y 2 + 81y) dy
4
dy
1138
Chapter Sixteen /SOLUTIONS
28. From the problem, we know that (x, y, z) is in the cube which is bounded by the three coordinate planes, x = 0, y = 0,
z = 0 and the planes x = 2, y = 2, z = 2. We can regard the value x 2 + y 2 + z 2 as the density of the cube. The average
value of x2 + y 2 + z 2 is given by
average value =
=
=
=
=
=
=
=
29. Positive. The function
p
R
V
(x2 + y 2 + z 2 ) dV
volume(V )
R2R2R2
0
0
0
(x2 + y 2 + z 2 ) dxdydz
R 2 R 2 x3
0
3
0
R2R2
0
R2
0
R2
0
0
8
+ 2y 2 + 2z 2 dydz
8
8
y
3
+
16
3
+
32
z
3
64
3
8
3
8
2
+ (y 2 + z 2 )x 0 dydz
2 3
y
3
16
3
2
+ 2z 2 y 0 dz
8
+ 4z 2 dz
8
2
+ 34 z 3 0
+
8
8 32
3
= 4.
x2 + y 2 is positive, so its integral over the solid W is positive.
30. Positive. If (x, y, z) is any point inside the solid W then
its integral over the solid W is positive.
p
x2 + y 2 < z. Thus the integrand z −
p
x2 + y 2 > 0, and so
31. Zero. The value of x is positive above the first and fourth quadrants in the xy-plane, and negative (and of equal absolute
value) above the second and third quadrants. The integral of x over the entire solid cone is zero because the integrals over
the two halves of the cone cancel.
32. Zero. The value of y is positive on the half of the cone above the second and third quadrants and negative (of equal
absolute value) on the half of the cone above the third and fourth quadrants. The integral of y over the entire solid cone is
zero because the integrals over the four quadrants cancel.
33. Positive. Since z is positive on W , its integral is positive.
34. Zero. You can see this in several ways. One way is to observe that xy is positive on part of the cone above the first and
third quadrants (where x and y are of the same sign) and negative (of equal absolute value) on the part of the cone above
the second and fourth quadrants (where x and y have opposite signs). These add up to zero in the integral of xy over all
of W .
Another way to see that the integral is zero is to write the triple integral as an iterated integral, say integrating first
with respect to x. For fixed y and z, the x-integral is over an interval symmetric about 0. The integral of x over such an
interval is zero. If any of the inner integrals in an iterated integral is zero, then the triple integral is zero.
35. Zero. Write the triple integral as an iterated integral, say integrating first with respect to x. For fixed y and z, the x-integral
is over an interval symmetric about 0. The integral of x over such an interval is zero. If any of the inner integrals in an
iterated integral is zero, then the triple integral is zero.
36. Negative. If (x, y, z) is any point inside the cone then z < 2. Hence the function z − 2 is negative on W and so is its
integral.
37. Positive. The function e−xyz is a positive function everywhere so its integral over W is positive.
38. Positive. The function
p
x2 + y 2 is positive, so its integral over the solid W is positive.
39. Positive. If (x, y, z) is any point inside the solid W then
over the solid W is positive.
p
x2 + y 2 < z. Thus z −
40. Positive. The value of x is positive on the half-cone, so its integral is positive.
p
x2 + y 2 > 0, and so its integral
41. Zero. y is positive on the half of the half-cone above the first quadrant in the xy-plane and negative (of equal absolute
value) on the half of the half-cone above the fourth quadrant. The integral of y over W is zero because the integrals over
1139
16.3 SOLUTIONS
each half add up to zero.
42. Positive. Since z is positive on W , its integral is positive.
43. Zero. You can see this in several ways. One way is to observe that xy is positive on part of the cone above the first
quadrant (where x and y are of the same sign) and negative (of equal absolute value) on the part of the cone above the
fourth quadrant (where x and y have opposite signs). These add up to zero in the integral of xy over all of W .
Another way to see that the integral is zero is to write the triple integral as an iterated integral, say integrating first
with respect to y. For fixed x and z, the y-integral is over an interval symmetric about 0. The integral of y over such an
interval is zero. If any of the inner integrals in an iterated integral is zero, then the triple integral is zero.
44. Zero. Write the triple integral as an iterated integral, say integrating first with respect to y. For fixed x and z, the y-integral
is over an interval symmetric about 0. The integral of y over such an interval is zero. If any of the inner integrals in an
iterated integral is zero, then the triple integral is zero.
45. Negative. If (x, y, z) is any point inside the cone then z < 2. Hence the function z − 2 is negative on W and so is its
integral.
46. Positive. The function e−xyz is a positive function everywhere so its integral over W is positive.
47. Orient the region as shown in Figure 16.60 and use Cartesian coordinates with origin at the center of the sphere. The
equation of the sphere is x2 + y 2 + z 2 = 25, and we want the volume between the planes z = 3 and z = 5. The plane
z = 3 cuts the sphere in the circle x2 + y 2 + 32 = 25, or x2 + y 2 = 16.
Z Z √
Z √
16−x2
4
Volume =
−4
−
25−x2 −y 2
√
16−x2
dzdydx.
3
z
Sphere is
z
5
6
x2 + y 2 + z 2 = 25
2
?
6
ICircle is
y
x2 + y 2 = 16
3
?
y
I
x
x2 + y 2 = 16
x
Figure 16.61
Figure 16.60
48. The intersection of two cylinders x2 + z 2 = 1 and y 2 + z 2 = 1 is shown in Figure 16.61. This region is bounded by four
surfaces:
p
p
p
p
z = − 1 − x2 , z = 1 − x2 , y = − 1 − z 2 , and y = 1 − z 2
So the volume of the given solid is
V =
49. Set up axes as in Figure 16.62.
Z
1
−1
Z √1−x2 Z √1−z2
−
√
1−x2
−
√
1−z 2
dy dz dx
1140
Chapter Sixteen /SOLUTIONS
z
2
y
2
x
2
Figure 16.62
The slanting plane has equation z = 2 − x − y and the line where it intersects the xy-plane has equation y = 2 − x.
The mass of the bottom part is δ1 times its volume, VBottom , where
VBottom =
Z 2Z
0
=
=
Z 2Z
Z
0
2
0
2−x
0
2−x
Z
2−x−y
dz dy dx =
0
(2 − x − y) dy dx =
0
Z 2Z
Z
0
2
0
3 2
2−x
0
2−x−y
z
y2
(2 − x)y −
2
−(2 − x) (2 − x)
4 3
dx =
=3m .
2
6
0
2
dy dx
0
2−x
dx
0
Since the cube has volume 23 = 8 m3 , the upper part has volume VTop = 8 − 4/3 = 20/3 m3 . Thus
4
20
δ1 +
δ2 gm.
3
3
Mass = VBottom δ1 + VTop δ2 =
50. The mass m is given by
m=
=
=
=
Z
Z
Z
Z
1 dV =
W
1
0
1
Z
1
Z
1
0
Z
1
0
Z
x+y+1
1 dz dy dx
0
(x + y + 1) dy dx
0
1
xy + y 2 /2 + y)0 dx
0
1
(x + 3/2) dx = 2 gm.
0
Then the x-coordinate of the center of mass is given by
1
x̄ =
2
=
=
1
2
1
2
1
=
2
Z
Z
Z
Z
W
1
0
1
0
1
1
x dV =
2
Z
1
Z
1
0
Z
1
0
Z
x+y+1
x dz dy dx
0
x(x + y + 1) dy dx
0
1
x2 y + xy 2 /2 + xy)0 dx
(x2 + 3/2x) dx = 13/24 cm.
0
An essentially identical calculation (since the region is symmetric in x and y) gives ȳ = 13/24 cm.
16.3 SOLUTIONS
Finally, we compute z̄:
Z
1
2
z̄ =
Z
1
2
=
Z
1
=
2
1
=
12
z dV =
W
1
Z
0
1
1
Z
1
0
Z
1
0
Z
x+y+1
z dz dy dx
0
(x + y + 1)2 /2 dy dx
0
1
(x + y + 1)3 /60 dx
0
Z
1
2
1
((x + 2)3 − (x + 1)3 ) dx = 25/24 cm.
0
So (x̄, ȳ, z̄) = (13/24, 13/24, 25/24).
51. The mass m is given by
m=
=
Z
Z
W
1
0
1
=
3
1
=
3
=
=
Z
1 dV =
1
3
1
3
Z
1
0
(1−x)/2
Z
Z
(1−x)/2
0
(1−x−2y)/3
1 dz dy dx
0
1 − x − 2y
dy dx
3
0
(1−x)/2
dx
(y − xy − y )
0
0
Z 1
Z
Z
1
2
1−x
1−x
−x
−
2
2
0
1
0
1−x
2
(1 − x)2
(1 − x)2
−
2
4
(1 − x)3
12
2 dx
dx
1
= 1/36 gm.
0
Then the coordinates of the center of mass are given by
x̄ = 36
and
ȳ = 36
and
z̄ = 36
Z
x dV = 36
W
Z
Z
y dV = 36
W
z dV = 36
W
Z
Z
Z
1
0
1
0
1
0
Z
Z
Z
(1−x)/2
0
Z
(1−x−2y)/3
x dz dy dx = 1/4 cm.
0
(1−x)/2
0
(1−x)/2
0
Z
Z
(1−x−2y)/3
y dz dy dx = 1/8 cm.
0
(1−x−2y)/3
z dz dy dx = 1/12 cm.
0
52. The volume V of the solid is 1 · 2 · 3 = 6. We need to compute
m
6
Z
m
x + y dV =
6
2
W
2
m
=
6
m
=
2
m
=
2
Z
Z
Z
Z
1
0
1
0
1
0
1
Z
Z
2
0
2
Z
3
x2 + y 2 dz dy dx
0
3(x2 + y 2 ) dy dx
0
2
(x2 y + y 3 /3)0 dx
(2x2 + 8/3) dx = 5m/3
0
1141
1142
Chapter Sixteen /SOLUTIONS
53. The volume of the solid is 8abc, so we need to evaluate
m
8abc
Z
(y 2 + z 2 ) dV =
W
m
8abc
m
=
8abc
m
=
4bc
m
=
2c
Z
Z
Z
Z
c
−c
c
−c
c
−c
c
Z
Z
b
−b
b
Z
a
(y 2 + z 2 ) dx dy dz
−a
2a(y 2 + z 2 ) dy dz
−b
b
(y 3 /3 + yz 2 )−b dz
(b2 /3 + z 2 ) dz
−c
= m(b2 + c2 )/3
54. By the definition, we have that
m
a+b =
V
=
m
V
m
=
V
Z
Z
m
(y + z ) dV +
V
W
2
Solutions for Section 16.4
Exercises
Z 2π Z
√
2.
3.
4.
Z
Z
0
0
ZW
W
π/2
3π/4
π/4
(x2 + y 2 ) dV +
Z
m
V
Z
(2z 2 ) dV
W
(2z 2 ) dV
W
(2z 2 ) dV will be positive, thus a + b > c.
W
f rdr dθ
0
π/2 Z
3π/2
(x2 + z 2 ) dV
W
2
1.
Z
R
Z
(x2 + y 2 + 2z 2 ) dV
m
= c+
V
Since z 2 is always positive, the integral
2
1/2
f rdr dθ
0
Z
Z
2
f rdr dθ
1
2
f rdr dθ
0
5. See Figure 16.63.
y
y
r=4
θ = π/2
r=1
1
R
x
x
−1
θ = −π/2
Figure 16.63
6. See Figure 16.64.
Figure 16.64
16.4 SOLUTIONS
1143
7. See Figure 16.65.
8. See Figure 16.66.
y
θ = 3π/4
y
r=2
y
θ = π/3
r=1
1
2
r=1
x
x
θ = π/6
r=4
θ = 3π/2
r=3
x
Figure 16.66
Figure 16.65
Figure 16.67
9. See Figure 16.67.
10. See Figure 16.68.
y
y
θ = π/4
θ = π/4
r = 2/ sin θ
or r sin θ = 2
or y = 2
x
x
r = 1/ cos θ
or r cos θ = 1
or x = 1
Figure 16.69
Figure 16.68
11. See Figure 16.69.
12. By using polar coordinates, we get
Z
sin(x2 + y 2 )dA =
R
=
Z
Z
2π
0
2π
0
Z
2
sin(r 2 )r dr dθ
0
2
1
− cos(r 2 ) dθ
2
0
Z
2π
1
(cos 4 − cos 0) dθ
2 0
1
= − (cos 4 − 1) · 2π = π(1 − cos 4)
2
=−
13. The region is pictured in Figure 16.70.
y
2
1
1
Figure 16.70
2
x
1144
Chapter Sixteen /SOLUTIONS
Using polar coordinates, we get
Z
2
R
2
(x − y )dA =
Z
Z
π/2
0
Z
2
2
1
2
2
r (cos θ − sin θ)rdr dθ =
15
=
4
15
=
4
=
Z
Z
π/2
0
π/2
2
1 (cos θ − sin θ) · r4 dθ
4 1
2
2
(cos2 θ − sin2 θ) dθ
0
π/2
cos 2θ dθ
0
π/2
15 1
· sin 2θ 4 2
0
= 0.
14. The presence of the term x2 +y 2 suggests that we should convert the integral into polar coordinates. Since
the integral becomes
Z p
x2 + y 2 dxdy =
R
15. (a)
Z
2π
0
Z
3
r2 drdθ =
2
Z
2π
0
y
3
r3 dθ =
3 2
Z
2π
0
p
x2 + y 2 = r,
38π
19
dθ =
.
3
3
y = x/3
1
x
3
Figure 16.71
(b)
Z
1
0
Z
3y
f (x, y) dx dy.
0
(c) For polar coordinates, on the line y = x/3, tan θ = y/x = 1/3, so θ = tan −1 (1/3). On the y-axis, θ = π/2. The
quantity r goes from 0 to the line y = 1, or r sin θ = 1, giving r = 1/ sin θ and f (x, y) = f (r cos θ, r sin θ). Thus
the integral is
Z
π/2
tan−1 (1/3)
Z
1/ sin θ
f (r cos θ, r sin θ)r dr dθ.
0
Problems
√
√
16. By the given limits 0 ≤ x ≤ −1, and − 1 − x2 ≤ y ≤ 1 − x2 , the region of integration is in Figure 16.72.
In polar coordinates, we have
Z
3π/2
π/2
Z
1
r cos θ r dr dθ =
0
Z
3π/2
cos θ
π/2
1
=
3
Z
3π/2
cos θ
1
1 3 r 3
0
dθ
dθ
π/2
3π/2
1
2
1
= (−1 − 1) = −
= sin θ
3
3
3
π/2
1145
16.4 SOLUTIONS
y
√
y
6
y=x
x
√x
6
−1
y = −x
√
− 6
Figure 16.72
Figure 16.73
17. From the given limits, the region of integration is in√
Figure 16.73.
√
In polar coordinates, −π/4 ≤ θ ≤ π/4. Also, 6 = x = r cos θ. Hence, 0 ≤ r ≤ 6/ cos θ. The integral becomes
Z
√
0
6
Z
x
dy dx =
−x
=
Z
Z
π/4
−π/4
Z
π/4
−π/4
√
r dr dθ
0
√6/cos θ !
r2 2 0
π/4
= 3 tan θ 6/cos θ
−π/4
dθ =
Z
π/4
−π/4
6
dθ
2 cos2 θ
= 3 · (1 − (−1)) = 6.
Notice that we can check this answer because the integral gives the area of the shaded triangular region which is
√
(2 6) = 6.
18. From the given limits, the region of integration is in Figure 16.74.
y
2
x=y
√
2
π/4
√
x
2
2
Figure 16.74
So, in polar coordinates, we have,
Z
π/4
0
Z
2
(r2 cos θ sin θ)r dr dθ =
0
Z
=4
π/4
cos θ sin θ
0
Z
π/4
0
sin(2θ)
2
2
1 4 r dθ
4
0
dθ
π/4
= 0 − (−1) = 1.
= − cos(2θ)
0
1
2
·
√
6·
1146
Chapter Sixteen /SOLUTIONS
19. The graph of f (x, y) = 25−x2 −y 2 is an upside down bowl, and the region whose volume we want is contained between
the bowl (above) and the xy-plane (below). We must first find the region in the xy-plane where f (x, y) is positive. To do
that, we set f (x, y) ≥ 0 and get x2 + y 2 ≤ 25. The disk x2 + y 2 ≤ 25 is the region R over which we integrate.
Volume =
Z
2
R
Z
2
(25 − x − y ) dA =
=
Z
2π
0
Z
625
4
625π
=
2
=
0
5
0
(25 − r 2 ) rdr dθ
5
25 2 1 4 r − r dθ
2
4
0
2π
dθ
0
20. First, let’s find where the two surfaces intersect.
p
Z
2π
p
8 − x2 − y 2 =
x2 + y 2
8 − x 2 − y 2 = x2 + y 2
x2 + y 2 = 4
So z = 2 at the intersection. See Figure 16.75.
z
2
y
x2 + y 2 = 4
2
R
x
Figure 16.75
The volume of the ice cream cone has two parts. The first part (which is the volume of the cone)
Z is the volume of the
solid bounded by the plane z = 2 and the cone z =
p
x2 + y 2 . Hence, this volume is given by
R
(2 −
where R is the disk of radius 2 centered at the origin, in the xy-plane. Using polar coordinates, we have:
Z R
2−
p
x2 + y 2
dA =
=
Z
Z
Z
2π
0
0
Z
0
"
2π
4
=
3
2
(2 − r) · r dr dθ
r3
r −
3
2
2π
dθ
0
= 8π/3
2 #
dθ
0
p
x2 + y 2 ) dA,
16.4 SOLUTIONS
1147
TheZsecond part is the volume of the region above the plane z = 2 but inside the sphere x 2 + y 2 + z 2 = 8, which is given
by
(
R
p
8 − x2 − y 2 − 2) dA where R is the same disk as before. Now
Z p
(
R
8 − x2 − y 2 − 2) dA =
=
=
Z
Z
Z
2π
0
2π
0
2π
0
Z
Z
Z
2
(
0
2
p
r
0
8 − r 2 − 2)rdr dθ
p
8 − r 2 dr dθ −
2 !
1
− (8 − r 2 )3/2 3
0
2π
21. (a)
Total Population =
Z
3π/2
π/2
Z
2π
0
dθ −
1
(43/2 − 83/2 ) dθ −
3 0
√
1
= − · 2π(8 − 16 2) − 8π
3
√
2π
=
(16 2 − 8) − 8π
3
√
8π(4 2 − 5)
=
3
√
Thus, the total volume is the sum of the two volumes, which is 32π( 2 − 1)/3.
=−
Z
Z
Z
Z
2
2r dr dθ
0
2π
2
r dθ
2
0
0
2π
4 dθ
0
4
δ(r, θ) rdr dθ.
1
(b) We know that δ(r, θ) decreases as r increases, so that eliminates (iii). We also know that δ(r, θ) decreases as the
x-coordinate decreases, but x = r cos θ. With a fixed r, x is proportional to cos θ. So as the x-coordinate decreases,
cos θ decreases and (i) δ(r, θ) = (4 − r)(2 + cos θ) best describes this situation.
(c)
Z
3π/2
π/2
Z
4
(4 − r)(2 + cos θ) rdr dθ =
1
Z
=9
3π/2
π/2
Z
(2 + cos θ)(2r 2 −
3π/2
(2 + cos θ) dθ
π/2
= 9 2θ + sin θ
4
1 3 r ) dθ
3 1
3π/2
π/2
= 18(π − 1) ≈ 39
Thus, the population is around 39,000.
22. (a) The volume, V , is given by
V =
Converting to polar coordinates gives
V =
Z
2π
0
Z
a
e
−r 2
0
2
Z
e−(x
2
+y 2 )
dA.
x2 +y 2 ≤a2
2π a
2
1 −r2 1
1 −a2
− e
r dr dθ = θ −
e
=
2π
= π(1 − e−a ).
2
2
2
0
0
(b) As a → ∞, the value of e−a → 0, so the volume tends to π.
23. The density function is given by
ρ(r) = 10 − 2r
where r is the distance from the center of the disk. So the mass of the disk in grams is
Z
ρ(r) dA =
R
Z
2π
0
Z
5
0
(10 − 2r)rdr dθ
1148
Chapter Sixteen /SOLUTIONS
=
=
Z
Z
2π
0
2π
0
5r2 −
2 3
r
3
5
dθ
0
125
250π
dθ =
(grams)
3
3
24. A rough graph of the base of the spring is in Figure 16.76, where the coil is roughly of width 0.01 inches. The volume is
equal to the product of the base area and the height. To calculate the area we use polar coordinates, taking the following
integral:
Area =
=
=
Z
Z
4π
0
1
2
1
2
Z
Z
0.26+0.04θ
rdrdθ
0.25+0.04θ
4π
(0.26 − 0.04θ)2 − (0.25 − 0.04θ)2 dθ
0
4π
0.01 · (0.51 + 0.08θ)dθ
0
= 0.0051 · 2π +
Therefore, the volume= 0.0636 · 0.2 = 0.0127 in3 .
4π
1
(0.0008θ 2 ) = 0.0636
4
0
Figure 16.76
25. The charge density is δ = k/r, where k is a constant.
Total charge =
Z
δ dA =
Disk
Z
R
0
Z
2π
0
k
r dθ dr = k
r
Z
R
0
Z
2π
dθ dr = k
0
Thus the total charge is proportional to R with constant of proportionality 2kπ.
Z
R
2π dr = 2kπR.
0
26. (a) We must first decide where to put the origin. We locate the origin at the center of one disk and locate the center of the
second disk at the point (1, 0). See Figure 16.77. (Other choices of origin are possible.)
y
√
x2 + y 2 = 1
(x − 1)2 + y 2 = 1
y
r=1
r = 2 cos θ
3/2
1
2
1
x
π/3
x
√
− 3/2
Figure 16.77
Figure 16.78
By symmetry, the points of intersection of the circles are half-way between the centers, at x = 1/2. The y-values
at these points are given by
r
√
2
p
1
3
y = ± 1 − x2 = ± 1 −
=±
.
2
2
16.5 SOLUTIONS
1149
We integrate in the x-direction first, so that it is not necessary to set up two integrals. The right-side of the circle
x2 + y 2 = 1 is given by
p
x = 1 − y2 .
The left side of the circle (x − 1)2 + y 2 = 1 is given by
x=1−
Thus the area of overlap is given by
Area =
Z
√
3/2
√
− 3/2
p
1 − y2 .
Z √1−y2
1−
dxdy.
√
1−y 2
(b) In polar coordinates, the circle centered at the origin has equation r = 1. See Figure 16.78. The other circle, (x −
1)2 + y 2 = 1, can be written as
x2 − 2x + 1 + y 2 = 1
x2 + y 2 = 2x,
so its equation in polar coordinates is
r2 = 2r cos θ,
and, since r 6= 0,
r = 2 cos θ.
√
√
At the top point of intersection of the two circles, x = 1/2, y = 3/2, so tan θ = 3, giving θ = π/3.
Figure 16.78 shows that if we integrate with respect to r first, we have to write the integral as the sum of two
integrals. Thus, we integrate with respect to θ first. To do this, we rewrite
r
r = 2 cos θ
as
θ = arccos
.
2
This gives the top half of the circle; the bottom half is given by
r
θ = − arccos
.
2
Thus the area is given by
Area =
Z
1
0
Z
arccos(r/2)
r dθdr.
− arccos(r/2)
Solutions for Section 16.5
Exercises
1. (a)
(b)
(c)
(d)
(e)
(f)
2.
A vertical plane perpendicular to the x-axis: x = 2.
A cylinder: r =√3.
A sphere: ρ = 3.
A cone: φ = π/4.
A horizontal plane: z = −5.
A vertical half-plane: θ = π/4.
Z
f dV =
W
=
=
Z
Z
Z
1
−1
1
−1
1
−1
Z
Z
3π/4
π/4
3π/4
Z
4
(r2 + z 2 ) rdr dθ dz
0
(64 + 8z 2 ) dθ dz
π/4
π
(64 + 8z 2 ) dz
2
= 64π +
8
200
π=
π
3
3
1150
3.
Chapter Sixteen /SOLUTIONS
Z
f dV =
W
=
Z
Z
3
−1
3
−1
1
=−
2
= −π
4.
Z
Z
Z
Z
Z
2π
0
2π
0
3
−1
3
−1
Z
1
(sin (r 2 )) rdr dθ dz
0
1
1
(− cos r 2 ) dθ dz
2
0
Z
2π
(cos 1 − cos 0) dθ dz
0
(cos 1 − 1) dz = −4π(cos 1 − 1) = 4π(1 − cos 1)
Z
f dV =
W
Z
=
Z
=
0
Z
f dV =
W
=
=
=
Z
Z
0
0
0
Z
Z
Z
Z
7
6
Z
2π
0
Z
2π
0
2π
π
π/2
1 2
· ρ sin φ dφ dθ dρ
ρ
π
ρ sin φ dφ dθ dρ
π/2
ρ dθ dρ
0
5
ρ dρ = 25π
0
Z
Z
2π
7
=
6
=
Z
0
7
3
Z
5
2π
7
3
Z
5
= 2π
5.
Z
5
Z
π/4
0
π/4
0
2π
Z
0
Z
2π
0
2π
0
2π
π
4
Z
2
(sin φ)ρ2 sin φ dρ dφ dθ
1
2
ρ2 sin2 φ dρ dφ dθ
1
sin2 φ dφ dθ
0
π/4
0
1 − cos 2φ
dφ dθ
2
π/4
1
(φ − sin 2φ)
2
0
(
0
dθ
1
π
− ) dθ
4
2
7π(π − 2)
π
1
7
= · 2π( − ) =
6
4
2
12
6. Using Cartesian coordinates, we get:
Z
7. Using cylindrical coordinates, we get:
8. Using cylindrical coordinates, we get:
Z
Z
3
0
1
0
4
0
Z
Z
Z
1
0
2π
0
π/2
0
Z
Z
5
f dz dy dx
0
4
f · rdr dθ dz
0
Z
2
0
f · rdr dθ dz
16.5 SOLUTIONS
1151
9. Using spherical coordinates, we get:
Z
π
0
Z
Z
π
0
3
f · ρ2 sin φ dρ dφ dθ
2
10. Using spherical coordinates, we get:
Z
2π
0
Z
π/6
0
Z
3
f · ρ2 sin φ dρ dφ dθ
0
11. We use Cartesian coordinates, oriented as shown in Figure 16.79. The slanted top has equation z = mx, where m is the
slope in the x-direction, so m = 1/5. Then if f is an arbitrary funtion, the triple integral is
Z
Other answers are possible.
5
0
Z
2
0
Z
x/5
f dzdydx.
0
z
(5, 0, 1)
1
y
2
?
(5, 2, 1)
(5, 2, 0)
5
x
Figure 16.79
12. We choose cylindrical coordinates oriented as in Figure 16.80. The cone has equation z = r. Since we have a half cone
scooped out of a half cylinder, θ varies between 0 and π. Thus, if f is an arbitrary function, the integral is
Other answers are possible.
Z
π
0
Z
2
0
Z
r
f r dzdrdθ.
0
z
z=r
2
π/4
x
2
Figure 16.80
13.
y
z
x
Figure 16.81
R is one eighth of a sphere of radius 1, below the xy-plane and under the first quadrant.
1152
Chapter Sixteen /SOLUTIONS
14. (a) The region of integration is the region between the cone z = r, the xy-plane and the cylinder r = 3. In spherical
coordinates, r = 3 becomes ρ sin φ = 3, so ρ = 3/ sin φ. The cone is φ = π/4 and the xy-plane is φ = π/2. See
Figure 16.82. Thus, the integral becomes
Z
2π
0
Z
Z
π/2
π/4
3/ sin φ
ρ2 sin φ dρdφdθ.
0
z
z=r
r=3
π
4
x
3
Figure 16.82: Region of integration is
between the cone and the xy-plane
(b) The original integral is easier to evaluate, so
Z
2π
0
Z
3
0
Z
r
r dzdrdθ =
0
Z
2π
0
Z
3
Z
z=r
zr
drdθ =
z=0
0
Problems
2π
0
Z
3
0
3
r3 = 18π.
r drdθ = 2π ·
3 0
2
√
15. In spherical coordinates, the spherical cap is part of the surface ρ = 2. If α is the angle at the vertex of the cone, we
have tan(α/2) = 2/2 = 1, so α/2 = π/4. Since the cone is below the xy-plane, the angle φ ranges from 3π/4 to π.
Thus, the integral is given by
√
Z
2π
0
Z
π
3π/4
Z
2
f (ρ, φ, θ)ρ2 sin φ dρ dφ dθ.
0
√
16. In cylindrical coordinates, the spherical cap has equation z = − 2 − r 2 . If α is the angle at the vertex of the cone, we
have tan(α/2) = 2/2 = 1, so α/2 = π/4. The cone has equation z = −r. Thus, the integral is
Z Z Z √
0
0
2−r 2
−
1
2π
g(r, θ, z)r dz dr dθ.
r
p
17. In rectangular coordinates, the spherical cap has equation z = − 2 − x2 − y 2 . If α is the angle at the vertex of the
p
cone, we have tan(α/2) = 2/2 = 1, so α/2 = π/4. The cone has equation z = − x2 + y 2 . Thus, the integral is
Z √ Z √
Z √
2−x2
2
√
− 2
−
√
2−x2
x2 +y 2
−
−
√
h(x, y, z) dz dy dx.
2−x2 −y 2
18. Orient the cone as shown in Figure 16.83 and use cylindrical coordinates with the origin at the vertex of the cone. Since√
the
angle at the vertex of the cone is a right angle, the angles AOB and COB are both π/4. Thus, OB = 5 cos π/4 = 5/ 2.
The curved surface of the cone has equation z = r, so
Volume =
=
Z
2π
0
Z
2π
0
Z
5/
√
2
0
Z
√
5/ 2
0
Z
√
5/ 2
r dz dr dθ
r
z=5/√2
Z 2π Z 5/√2 5
rz dr dθ =
r √ − r dr dθ
2
0
0
z=r
16.5 SOLUTIONS
1153
√
2π 5/ 2
5 52
r3 53
5 r2
√
√
= θ
= 2π √ · 2 −
−
3 2 2
2 2
2·3· 2
0
0
3 3
5
= 2π · √
2 2
1
1
−
2
3
5 π
= √ = 46.28 cm3 .
6 2
z
C
B
A
π/4 π/4
5
^ √ r
5/ 2
O
Figure 16.83
19. (a) In Cartesian coordinates, the bottom half of the sphere x2 + y 2 + z 2 = 1 is given by z = −
Z
Z
Z Z √
1−x2
1
0
dV =
W
−
0
0
dz dy dx.
√
1−x2 −y 2
p
1 − x2 − y 2 . Thus
√
(b) In cylindrical coordinates, the sphere is r 2 + z 2 = 1 and the bottom half is given by z = − 1 − r 2 . Thus
Z
dV =
W
Z
Z
π/2
0
(c) In spherical coordinates, the sphere is ρ = 1. Thus,
Z
dV =
W
Z
π/2
0
Z
1
0
π
π/2
Z
0
−
Z
√
r dz dr dθ.
1−r 2
1
ρ2 sin φ dρ dφ dθ.
0
20. (a) Since the cone has a right angle at its vertex, it has equation
p
z=
x2 + y 2 .
The sphere has equation x2 + y 2 + z 2 = 1, so the top half is given by
z=
The cone and the sphere intersect in the circle
p
1 − x2 − y 2 .
x2 + y 2 =
See Figure 16.84. Thus
Z
W
dV =
Z
√
1/ 2
−1/
√
2
1
,
2
1
z= √ .
2
Z √(1/2)−x2 Z √1−x2 −y2
−
√
(1/2)−x2
√
dz dy dx.
x2 +y 2
z
1
6
ICircle is
√1
2
π
4
x2 + y 2 =
?
1
2
y
I
x
x2 + y 2 =
Figure 16.84
1
2
1154
Chapter Sixteen /SOLUTIONS
(b) In cylindrical coordinates, the cone has equation z = r and the sphere has equation z =
Z
Z Z √ Z √
1−r 2
1/ 2
2π
dV =
W
√
1 − r 2 . Thus
r dz dr dθ.
0
0
r
(c) In spherical coordinates, the cone has equation φ = π/4 and the sphere is ρ = 1. Thus
Z
dV =
W
Z
2π
0
Z
π/4
0
Z
1
ρ2 sin φ dρ dφ dθ.
0
21. (a) Since the cone has a right angle at its vertex, it has equation
p
z = x2 + y 2 .
√
Figure 16.85 shows the plane with equation z = 1/ 2. The plane and the cone intersect in the circle x2 + y 2 = 1/2.
Thus,
√
√
Z
dV =
W
Z
1/ 2
√
−1/ 2
Z
(1/2)−x2
−
√
(1/2)−x2
Z
1
dz dy dx.
√
x2 +y 2
z
√1
2
√1
2
π
4
1
y
x
Figure 16.85
(b) In cylindrical coordinates the cone has equation z = r, so
Z
dV =
W
Z
2π
0
Z
1/
0
√
2
Z
1
r dz dr dθ.
r
√
√
(c) In spherical coordinates, the cone has equation φ = π/4 and the plane z = 1/ 2 has equation ρ cos φ = 1/ 2.
Thus
Z
Z 2π Z π/4 Z 1/(√2 cos φ)
dV =
ρ2 sin φ dρ dφ dθ.
W
0
0
0
22. The region is a solid cylinder of height 1, radius 1 with base on the xy-plane and axis on the z-axis. Use cylindrical
coordinates:
Z 1 Z 2π Z 1
Z 1 Z 1 Z √1−x2
1
1
dy
dx
dz
=
r dr dθ dz
√
2 + y 2 )1/2
r
(x
2
0
0
0
−1 − 1−x
0
=
=
Z
Z
1
0
1
0
Z
Z
2π
0
1
r dθ dz
0
2π
dθ dz = 2π.
0
16.5 SOLUTIONS
1155
23. The region of integration is half of a ball centered at the origin, radius 1, on the x ≥ 0 side. Since the integral is symmetric,
we can integrate over the quarter unit ball (x ≥ 0, y ≥ 0) and multiply the result by 2. Use spherical coordinates:
Z 1 Z √1−x2 Z √1−x2 −z2
1
dy dz dx
√
√
(x2 + y 2 + z 2 )1/2
0
− 1−x2 − 1−x2 −z 2
=2
=2
=
Z
Z
Z
π/2
0
π/2
0
π/2
0
Z
Z
π
0
π
0
Z
1
0
1 2
ρ sin φ dρ dφ dθ
ρ
1
ρ2 sin φ dφ dθ
2 0
π
(− cos φ) dθ
0
= (−(−1) − 0) · π = π.
24. Use cylindrical coordinates: when r 2 = x2 + y 2 = 1, then x2 + y 2 + z 2 = 1 + z 2 = 2 so z = ±1. The region W is
shown in Figure 16.86.
Z
Z Z Z √
2−z 2
2π
1
2
2
(x + y ) dV =
W
=
=
Z
−1
1
−1
2π
4
π
=
2
0
Z
Z
1
2π
0
1
r2 · r drdθdz
√2−z2
Z 1 Z 2π
r4 1
(2 − z 2 )2 − 1 dθdz
dθdz
=
4 1
4 −1 0
3 − 4z 2 + z 4 dz
−1
z5
4
3z − z 3 +
3
5
z
1
= 28π .
15
−1
Cylinder
x2 + y 2 = 1
Sphere
1
x2 + y 2 + z 2 = 2
y
x
−1
Figure 16.86
25. (a) The angle φ takes on values in the range 0 ≤ φ ≤ π. Thus, sin φ is nonnegative everywhere in W 1 , and so its integral
is positive.
(b) The function φ is symmetric across the xy plane, such that for any point (x, y, z) in W 1 , with z 6= 0, the point
(x, y, −z) has a cos φ value with the same magnitude but opposite sign of the cos φ value for (x, y, z). Furthermore,
if z = 0, then (x, y, z) has a cos φ value of 0. Thus, with cos φ positive on the top half of the sphere and negative on
the bottom half, the integral will cancel out and be equal to zero.
26. (a) The integral is negative. In W2 , we have 0 < z < 1. Thus, z 2 − z is negative throughout W2 and thus its integral is
negative.
(b) On the top half of the sphere, z is nonnegative, but x can be both positive and negative. Thus, since W 2 is symmetric
with respect to the yz plane, the contribution of a point (x, y, z) will be canceled out by its reflection (−x, y, z).
Thus, the integral is zero.
1156
Chapter Sixteen /SOLUTIONS
27. The region whose volume we want is shown in Figure 16.87:
z
6
5
2
?
π
6
θ=
θ=
π
3
y
x
Figure 16.87
Using cylindrical coordinates, the volume is given by the integral:
V =
=
Z
Z
2
0
2
0
Z
Z
25
2
=
Z
Z
π/3
π/6
π/3
π/6
2
0
2
Z
Z
5
r dr dθ dz
0
5
r2 dθ dz
2 0
π/3
dθ dz
π/6
25
π
π
=
−
dz
2 0
3
6
25 π
25π
=
· ·2=
.
2 6
6
2
2
2
2
28. (a) In cylindrical coordinates,
the cone
√
√ is z = r and the sphere is r + z = 4. The surfaces intersect where z + z =
2
2z = 4. So z = 2 and r = 2.
Z Z √ Z √
2π
4−r 2
2
Volume =
r dzdrdθ.
0
0
r
(b) In spherical coordinates, the cone is φ = π/4 and the sphere is ρ = 2.
Volume =
29. (a)
(b)
Z
Z
2π
0
2π
0
Z
Z
π
0
2
0
Z
Z
Z
2π
0
Z
π/4
0
Z
2
ρ2 sin φ dρdφdθ.
0
2
ρ2 sin φ dρdφdθ.
1√
−
4−r 2
√
4−r 2
r dzdrdθ −
Z
2π
0
Z
1
0
Z √1−r2
−
√
r dzdrdθ.
1−r 2
30. Orient the region as shown in Figure 16.88 and use cylindrical coordinates with the origin at the center of the sphere. The
equation of the sphere is x2 + y 2 + z 2 = 25, or r 2 + z 2 = 25. If z = 3, then r 2 + 32 = 25, so r 2 = 16 and r = 4.
√
√
Volume =
Z
2π
0
Z
4
0
Z
25−r 2
r dzdrdθ =
3
Z
2π
0
Z
4
0
z=
rz z=3
25−r 2
drdθ
16.5 SOLUTIONS
=
Z
2π
0
= 2π
Z
4
(r
0
p
1157
2π 4
2 3/2
2
(25
−
r
)
3r
−
−
25 − r 2 − 3r) drdθ = θ 3
2
0
0
125
27
−
3
3
− 24 =
52π
= 54.45 cm3 .
3
z
Sphere
5
2
r 2 + z 2 = 25
6
?
6
5
3
?
4
r
-
Figure 16.88
31. Orient the region as shown in Figure 16.89 and use spherical coordinates with the origin at the center of the sphere. The
equation of the sphere is x2 + y 2 + z 2 = 25, or ρ = 5. The plane z = 3 is the plane ρ cos φ = 3, so ρ = 3/ cos φ. In
Figure 16.89, angle AOB is given by
cos φ =
3
,
5
so
φ = arccos(3/5).
The volume is given by
2π Z
p=5
arccos(3/5)
3
ρ
sin φ ρ2 sin φ dρdφdθ = θ dφ
V =
3
0
3/ cos φ
0
0
0
p=3/ cos φ
Z arccos(3/5) Z
2π
Z
arccos(3/5)
0
Z
5
9
125
−
3
cos3 φ
= 2π
= 2π
Z
arccos(3/5)
0
125
sin φ dφ −
3
sin φ dφ
Z
arccos(3/5)
0
9
sin φ dφ
cos3 φ
arccos(3/5)
arccos(3/5)
125
1
cos φ φ
= 2π
−
−9
3
2 cos2
0
0
1
125 3
9
−1
−1 −
3
5
2 (3/5)2
2
9 16
125
−
−
= 2π −
3
5
2 9
52π
50
−8 =
≈ 54.45 cm3 .
= 2π
3
3
−
z
2
3
Sphere
ρ=5
6
?A
6
φ
= 2π
B
5
?
O
Figure 16.89
x
!
1158
Chapter Sixteen /SOLUTIONS
32. The density function can be rewritten as δ(ρ, φ, θ) = ρ. So the mass is
Z
δ(P ) dV =
W
=
Z
Z
2π
0
2π
Z
Z
π/4
0
π/4
3
0
ρ · ρ2 sin φ dρ dφ dθ
81
sin φ dφ dθ
4
0
Z 2π √
2
+ 1) dθ
(−
2
0
√
√
2
81
· 2π · (−
+ 1) =
π(− 2 + 2)
2
4
0
81
=
4
81
4
=
Z
33. Using spherical coordinates:
M =
=
=
Z
Z
π
0
π
0
27
4
Z
Z
Z
2π
0
2π
0
π
0
Z
Z
3
(3 − ρ)ρ2 sin φ dρ dθ dφ
0
ρ4
ρ −
4
3
2π
3
sin φ dθ dφ
0
sin φ dθ dφ
0
π
27
27
=
· 2π · (− cos φ) =
π · [−(−1) + 1] = 27π.
4
2
0
34. (a) We use spherical coordinates. Since δ = 9 where ρ = 6 and δ = 11 where ρ = 7, the density increases at 2 gm/cm 3
for each cm increase in radius. Thus, since density is a linear function of radius, the slope of the linear function is 2.
Its equation is
δ − 11 = 2(ρ − 7) so δ = 2ρ − 3.
(b) Thus,
Mass =
(c) Evaluating the integral, we have
π Mass = 2π − cos φ
0
Z
2π
0
Z
π
0
Z
7
6
(2ρ − 3)ρ2 sin φ dρ dφ dθ.
7 !
2ρ4
3ρ3 −
4
3 6
35. The distance from a point (x, y, z) to the origin is given by
R p
R
= 2π · 2(425.5) = 1702π gm = 5346.991 gm.
p
x2 + y 2 + z 2 . Thus we want to evaluate
x2 + y 2 + z 2 dV
where R is the region bounded by the hemisphere z =
We will use spherical coordinates.
Vol(R)
p
8 − x2 − y 2 and the cone z =
p
x2 + y 2 . See Figure 16.90.
16.5 SOLUTIONS
1159
z
2
y
x2 + y 2 = 4
x
Figure 16.90
√
In spherical coordinates, the quantity ρ goes from 0 to 8, and θ goes from 0 to 2π, and φ goes from 0 to π/4
(because the angle of the cone is π/4). Thus we have
Z p
x2 + y 2 + z 2 dV =
R
=
=
Z
Z
Z
2π
0
2π
0
2π
0
Z
Z
Z
π/4
0
π/4
0
π/4
Z
√
8
ρ(ρ2 sin φ) dρdφdθ
0
√8
ρ4 sin φ · dφdθ
4 0
16 sin φ dφdθ
0
π/4
16(− cos φ)
dθ
=
0
0
Z 2π √ Z
2π
=
0
16 1 −
2
2
dθ
√ 2
π
2
√
From Problem 20 of Section 16.4 we know that Vol(R) = 32π( 2 − 1)/3, therefore
= 32 1 −
Average distance =
R p
R
x2 + y 2 + z 2 dV
Vol(R)
√
32 1 − 22 π
3
√
=
= √ .
[32( 2 − 1)π/3]
2
36. (a) First we must choose a coordinate system, since none is given. We pick the xy-plane to be the fixed plane and the
z-axis to be the line perpendicular to the plane. Then the distance from a point to the plane is |z|, so the density at a
point is given by
Density = ρ = k|z|.
Using cylindrical coordinates for the integral, we find
Z Z Z √
a2 −r 2
a
2π
Mass =
0
0
−
√
a2 −r 2
k|z|r dzdrdθ.
1160
Chapter Sixteen /SOLUTIONS
(b) By symmetry, we can evaluate this integral over the top half of the sphere, where |z| = z. Then
√
Z 2π Z a Z √a2 −r2
Z 2π Z a 2 z= a2 −r2
z Mass = 2
r
kzr dzdrdθ = 2k
drdθ
2 z=0
0
0
0
0
0
=k
Z
2π
0
= 2πk
Z
a
2
2
r(a − r ) drdθ = k2π
0
a4
a4
−
2
4
πka4
.
=
2
r2 2 r4
a −
2
4
a
0
37. (a) We use the axes shown in Figure 16.91. Then the sphere is given by r 2 + z 2 = 25, so
Z Z Z √
2π
25−r 2
5
Volume =
0
−
1
r dzdrdθ.
√
25−r 2
(b) Evaluating gives
Volume = 2π
Z
5
1
z=√25−r2
Z 5 p
rz 2r 25 − r 2 dr
dr
=
2π
√
1
z=− 25−r 2
5
2
= 2π −
(25 − r 2 )3/2 3
1
√
4π
3/2
(24)
= 64 6π = 492.5 mm3 .
=
3
r 2 + z 2 = 25
x2 + y 2 + z 2 = 25
y
z
5
1
5
x
a
b
y
x
Figure 16.92
Figure 16.91
38. We must first decide on coordinates. We pick spherical coordinates with the common center of the two spheres as the
origin. We imagine the half-melon with the flat side horizontal and the positive z-axis going through the curved surface.
See Figure 16.92. The volume is given by the integral
Volume =
Z
2π
0
Z
π/2
0
Z
b
ρ2 sin φ dρdφdθ.
a
Evaluating gives
Volume =
Z
2π
0
Z
π/2
0
p=b
ρ3 π/2
dφdθ = 2π(− cos φ)
sin φ
3 p=a
0
b3
a3
−
3
3
=
2π 3
(b − a3 ).
3
To check our answer, notice that the volume is the difference between the volumes of two half spheres of radius a and b.
These half spheres have volumes 2πb3 /3 and 2πa3 /3, respectively.
16.5 SOLUTIONS
1161
39. The total volume of the cone is 31 πr2 h = 13 π · 12 · 1 = 31 π, so the total mass is 13 π (since the density is always 1). The
center of mass z-coordinate is given by
Z
3
z̄ =
z dV
π C
Using cylindrical coordinates to evaluate this integral gives
z̄ =
=
Z
3
π
Z
3
π
Z
3
=
π
40. (a) The mass m of the cone is given by
R
C
2π
0
2π
0
2π
Z
Z
1
0
1
Z
z
zr dr dz dθ
0
z3
dz dθ
2
0
3
1
dθ =
8
4
0
δ dV . In cylindrical coordinates this is
Z
m=
Z
=
Z
=
(b) The center of mass z-coordinate is given by
2π
0
2π
0
2π
Z
Z
1
0
1
Z
0
z
z 2 r dr dz dθ
0
z4
dz dθ
2
1
π
dθ =
10
5
0
5
z̄ =
π
Z
C
z · z 2 dV
Using cylindrical coordinates to evaluate this integral gives
5
π
z̄ =
5
=
π
5
=
π
Z
Z
Z
Z
2π
0
Z
2π
0
2π
1
0
1
0
Z
z
z 3 r dr dz dθ
0
z5
dz dθ
2
1
5
dθ =
12
6
0
Comparing this answer with the center of mass in Problem 39, where the density was constant, it makes sense
that the center of mass would be higher in this problem, since more mass is concentrated near the top of the cone.
41. We first need to find the mass of the solid, using cylindrical coordinates:
m=
=
=
Z
Z
Z
2π
0
2π
0
2π
Z
Z
0
1
0
Z √z/a
r dr dz dθ
0
1
0
z
dz dθ
2a
π
1
dθ =
4a
2a
It makes sense that the mass would vary inversely with a, since increasing a makes the paraboloid skinnier. Now for
the z-coordinate of the center of mass, again using cylindrical coordinates:
2a
z̄ =
π
=
2a
π
2a
=
π
Z
Z
Z
2π
0
2π
0
2π
0
Z
Z
1
0
zr dr dz dθ
0
1
0
Z √z/a
z2
dz dθ
2a
1
2
dθ =
6a
3
1162
Chapter Sixteen /SOLUTIONS
42. The volume of the hemisphere is 23 πa3 so its mass is 23 πa3 b. To find the location of the center of mass, we place the
base of the hemisphere on the xy-plane with the origin at its center, so we can describe it in spherical coordinates by
0 ≤ ρ ≤ a, 0 ≤ φ ≤ π2 and 0 ≤ θ ≤ 2π. Then the x-coordinate of the center of mass is, integrating using spherical
coordinates:
x̄ =
since the first integral
R 2π
0
3
2πa3 b
Z
Z
a
0
Z
π
2
0
2π
ρ sin(φ) cos(θ) · ρ2 sin(φ) dθ dφ dρ = 0
0
cos(θ) dθ is zero. A similar computation shows that ȳ = 0. Now for the z-coordinate:
Z
3
z̄ =
2πa3 b
a
0
3
=
· 2π
2πa3 b
=
=
Z
3
a3 b
a
π
2
0
Z
a
0
Z
Z
2π
ρ cos(φ) · ρ2 sin(φ) dθ dφ dρ
0
π
2
0
ρ3 cos(φ) sin(φ) dφ dρ
π
sin2 (φ) 2
dρ
2 0
ρ3
Z0
3
2a3 b
Z
a
ρ3 dρ =
0
3a
8b
So the x and y-coordinates are located at the center of the base, while the z-coordinate is located
the base.
3a
8b
above the center of
43. The sum of the three moments of inertia I for the ball B will be
Z
Z
Z
3
3
3
2
2
2
2
3I =
(y + z ) dV +
(x + z ) dV +
(x2 + y 2 ) dV
4πa3 B
4πa3 B
4πa3 B
=
3
4πa3
Z
(2x2 + 2y 2 + 2z 2 ) dV,
B
which, in spherical coordinates is
3
2πa3
Z
(x2 + y 2 + z 2 ) dV =
B
3
2πa3
3
= 3
a
=
6
a3
Thus 3I = 65 a2 , so I = 25 a2 .
Z
Z
a
Z0 a
a
0
Z
Z
π
0
π
Z
2π
0
ρ2 · ρ2 sin(φ) dθ dφ dρ
4
ρ sin(φ) dφ dρ
0
ρ4 dρ =
0
6 2
a .
5
44. First we need to find the volume of the cone. In spherical coordinates we find:
V =
Z
a
0
Z
π
3
0
Z
2π
ρ2 sin(φ) dθ dφ dρ =
0
πa3
3
Now, to find the moment of inertia about the z-axis we need to compute the integral
this in spherical coordinates as
3
πa3
Z
x2 + y 2 dV =
W
3
πa3
3
=
πa3
6
= 3
a
Z
Z
Z
a
0
a
0
a
0
6 5
= 3
a 24
Z
Z
Z
Z
π
3
0
π
3
0
π
3
Z
Z
R
W
x2 + y 2 dV . We can do
2π
(ρ2 sin2 (φ) cos2 (θ) + ρ2 sin2 (φ) sin2 (θ)) · ρ2 sin(φ) dθ dφ dρ
0
2π
ρ4 sin3 (φ) dθ dφ dρ
0
4
ρ sin3 (φ) dφ dρ
0
a
ρ4 dρ =
0
3
πa3
a2
.
4
16.5 SOLUTIONS
1163
45. Assume the base of the cylinder sits on the xy-plane with center at the origin. Because the cylinder is symmetric about the
z-axis, the force in the horizontal x or y direction is 0. Thus we need only compute the vertical z component of the force.
We are going to use cylindrical coordinates; since the force is G · mass/(distance) 2 , a piece of the cylinder of volume
dV located at (r, θ, z) exerts on the unit mass a force with magnitude G(δ dV )/(r 2 + z 2 ). See Figure 16.93.
Vertical component
G(δ dV )
Gδ dV
z
Gδz dV
· cos φ = 2
·√
.
= 2
= 2
r + z2
r + z2
(r + z 2 )3/2
of force
r2 + z 2
Adding up all the contributions of all the dV ’s, we obtain
Vertical force =
=
=
=
Z
Z
Z
Z
H
0
H
0
H
0
H
Z
Z
Z
Z
2π
0
2π
0
R
Gδzr
drdθdz
(r2 + z 2 )3/2
0
1
(Gδz) − √
2
r + z2
2π
0
(Gδz) ·
2πGδ
0
= 2πGδ(z −
√
−√
1
1
+
z
R2 + z 2
z
1− √
R2 + z 2
p
= 2πGδ(H −
R2
+
z2)
R
dθdz
0
H
dθdz
dz
0
p
R2
+
H2
+ R) = 2πGδ(H + R −
p
R2 + H 2 )
r2 + z 2
z
φ
Figure 16.93
46. In the system used in this book the volume element is dV = ρ 2 sin φ dρ dφ dθ. In the system shown in the problem, φ
and θ have been interchanged and ρ changed to r. So the volume element is dV = r 2 sin θ dr dθ dφ.
47. The charge density is δ = kz, where k is a constant. In cylindrical coordinates,
Total charge =
Z
δ dV =
Cylinder
= kπ
Z
Z hZ
0
h
R
0
Z
R2 z dz = k(πR2 )
0
2π
kzr dθ dr dz = k
0
Z hZ
0
kπ 2 2
h2
=
R h .
2
2
R
2πzr dr dz
0
Thus, the total charge is proportional to R2 h2 with constant of proportionality kπ/2.
48. The charge density is δ = k/ρ. Integrating in spherical coordinates,
Total charge =
Z
2π
0
Z πZ
0
R
0
k 2
ρ sin φ dρ dφ dθ = k
ρ
R2
= 2πkR2 .
= 4πk
2
Z
2π
0
Z
Thus, the total charge is proportional to R2 with constant of proportionality 2πk.
π
0
R2
sin φ dφ dθ
2
1164
Chapter Sixteen /SOLUTIONS
49. Using spherical coordinates,
Stored energy =
1
2
Z bZ π Z
a
q2
=
8π
Z
0
b
2π
0
1
q2
dρ =
2
ρ
8π
a
Z bZ π Z
q2
32π 2 E 2 ρ2 sin φ dθ dφ dρ =
a
1
1
−
.
a
b
0
2π
1
sin φ dθ dφ dρ
ρ2
0
50. Use cylindrical coordinates, with the z-axis being the axis of the cable. Consider a piece of cable of length 1. Then
1
Stored energy =
2
Z bZ 1Z
0
b
a
q2
=
4π
Z
a
2π
q2
E r dθ dz dr =
8π 2 2
0
Z bZ 1Z
a
0
2π
0
1
dθ dz dr
r
1
q2
q2
b
dr =
(ln b − ln a) =
ln .
r
4π
4π a
So the stored energy is proportional to ln(b/a) with constant of proportionality q 2 /4π.
Solutions for Section 16.6
Exercises
1. We have p(x, y) = 0 for all points (x, y) satisfying x ≥ 3, since all such points lie outside the region R. Therefore the
fraction of the population satisfying x ≥ 3 is 0.
2. The fraction is 0, since
R1
1
xy dx = 0, so
R∞ R1
−∞
p(x, y) dx dy =
1
R1R1
0
1
xy dx dy=0.
3. Since x + y ≤ 3 for all points (x, y) in the region R, the fraction of the population satisfying x + y ≤ 3 is 1.
4. Since p(x, y) = 0 for any (x, y) with x < 0 and also p(x, y) = 0 for any (x, y) with y > 1 or y < 0, the fraction of the
population is given by the double integral:
Z
1
0
Z
1
xy dx dy =
0
Z
1
0
1
x2 y dy =
2 0
Z
1
0
1
y
1
y 2 = .
dy =
2
4 0
4
5. Since p(x, y) = 0 for all (x, y) outside the rectangle R, the population is given by the volume under the graph of p over
the region inside the rectangle R and to the right of the line x = y. Therefore the fraction of the population is given by
the double integral:
Z
1
0
Z
2
xy dx dy =
y
Z
1
0
2
x2 y dy =
2 y
Z
1
0
2y −
y3
2
dy =
y2 −
y4
8
1
= 7.
8
0
6. Since p(x, y) = 0 for all (x, y) outside the rectangle R, the population is given by the volume under the graph of p over
the region inside the rectangle R and below the line x + y = 1. This is the same as the region bounded by the x-axis, the
y-axis, and the line x + y = 1. Therefore the fraction of the population is given by the double integral:
Z
1
0
Z
1−y
xy dx dy =
0
Z
1
0
1−y
x2 y 2 0
dy =
Z
1
0
(1 − y)2 y
dy =
2
y2
y3
y4
−
+
4
3
8
7. The fraction of the population is given by the double integral:
Z
1/2
0
Z
1
xy dx dy =
0
Z
1/2
0
1
x2 y dy =
2 0
Z
1/2
0
1/2
y
y 2 dy =
2
4 0
=
1
= 1.
24
0
1
.
16
16.6 SOLUTIONS
1165
8. We are looking for points inside the circle x2 + y 2 = 1 and inside the rectangle R. In the first quadrant, all of the circle
and its interior lies inside the rectangle R. Thus the fraction of the population we want is given by the volume under the
graph of p over the region inside the circle x2 + y 2 = 1 in the first quadrant. We evaluate this double integral using polar
coordinates:
Z
π/2
0
Z
1
(r cos θ)(r sin θ) r dr dθ =
0
Z
π/2
0
Making the substitution w = sin θ, we get:
Z
1
r4
1
cos θ sin θ dθ =
4
4
0
π/2
cos θ sin θ dθ =
0
Z
1
w dw =
0
Z
π/2
cos θ sin θ dθ.
0
1
.
2
Thus the fraction is (1/4)(1/2) = 1/8.
9. (a)
(b)
(c)
(d)
The entries in this table are non-negative and their sum is 1.
Add up the values along the row x = 2: 0.2 + 0.1 + 0 = 0.3.
Add up the columns with y = 1 and y = 2: 0.3 + 0.2 + 0.1 + 0 + 0.2 + 0.1 + 0 + 0 = 0.9.
Add up the values in the grid corresponding x = 1, 2, 3 and y = 1, 2: 0.3 + 0.2 + 0.2 + 0.1 + 0.1 + 0 + 0 + 0 = 0.9.
10. No, p is not a joint density function. Since p(x, y) = 0 outside the region R, the volume under the graph of p is the same
as the volume under the graph of p over the region R, which is 2 not 1.
11. Yes, p is a joint density function. The values of p(x, y) are nonnegative, since p(x, y) = 1/2 for all points inside R and
p(x, y) = 0 for all other points. The volume under the graph of p over the region R is (1/2)(5 − 4)(0 − (−2)) = 1.
12. No, p is a not joint density function, because p(x, y) < 0 for some points (x, y) in the region R. For example,
p(−0.7, 0.1) = −0.6.
13. Yes, p is a joint density function. Since x ≤ y everywhere in the region R, we have p(x, y) = 6(y − x) ≥ 0 for all x
and y in R, and p(x, y) = 0 for all other (x, y). To check that p is a joint density function, we check that the total volume
under the graph of p over the region R is 1:
Z
p(x, y) dA =
R
Z
1
0
Z
y
6(y − x) dx dy =
0
Z
1
0
6 yx −
x2
2
y
Z
dy =
0
1
0
1
3y 2 dx = y 3 = 1.
0
14. Yes, p is a joint density function. In the region R we have 1 ≥ x 2 + y 2 , so p(x, y) = (2/π)(1 − x2 − y 2 ) ≥ 0 for all x
and y in R, and p(x, y) = 0 for all other (x, y). To check that p is a joint density function, we check that the total volume
under the graph of p over the region R is 1. Using polar coordinates, we get:
Z
2
p(x, y)dA =
π
R
Z
2π
0
Z
1
2
(1 − r )r dr dθ =
π
2
0
Z
2π
0
r2
r4
−
2
4
1
Z
dθ = 2
π
0
2π
0
1
dθ = 1.
4
15. Yes, p is a joint density function. Since e−x−y is always positive, p(x, y) = xye−x−y ≥ 0 for all x and y in R, and
hence for all x and y. To check that p is a joint density function, we check that the total volume under the graph of p over
the region R is 1. Since e−x−y = e−x e−y , we have
Z
xye−x−y dA =
R
Z
∞
0
Z
∞
xye−x−y dx dy =
0
Z
∞
ye−y
0
Z
∞
xe−x dx
0
Using integration by parts:
Z
Thus
Z
∞
0
xye
R
b
xe−x dx = lim (−xe−x − e−x ) = (0 − 0) − (0 − 1) = 1.
b→∞
−x−y
dA =
Z
∞
0
ye
0
−y
Z
∞
0
xe
−x
dx
dy =
Z
∞
0
ye−y dy = 1.
dy.
1166
Chapter Sixteen /SOLUTIONS
16. (a)
Z
1
0
Z
1
1/3
2
(x + 2y) dx dy =
3
=
Z
Z
1
2 1 2
( x + 2xy)1/3 dy
3 2
1
0
1
h
0
Z
1
4
4
2
( + y) dy
3 0 9
3
2 4
2 2 1
=
y+ y
0
3 9
3
20
2 10
=
.
=
3 9
27
=
i
1
2
2 1
( + 2y) − (
+ y) dy
3 2
18
3
(b) It is easier to calculate the probability that x < (1/3) + y does not happen, that is, the probability that x ≥ (1/3) + y,
and subtract it from 1. The probability that x ≥ (1/3) + y is
Z
1
1/3
Z
x−(1/3)
0
2
(x + 2y) dy dx =
3
Z
=
2
3
1
1/3
2
=
3
Z
Z
x−(1/3)
2
(xy + y 2 )0
dx
3
1
1/3
(x(x −
1
1
) + (x − )2 ) dx
3
3
1
1/3
(2x2 − x +
1
) dx
9
2 2
1
1 1
= ( x3 − x2 + x)1/3
3 3
2
9
i
h
1
1
2
1
1
2 2
−
+
)
( − + )−(
=
3 3
2
9
81
18
27
= 44/243.
Thus, the probability that x < (1/3) + y is 1 − (44/243) = 199/243.
x<
1
3
+y
1
3
Figure 16.94
Problems
17. (a) We know that
R∞ R∞
−∞
−∞
f (x, y)dydx = 1 for a joint density function. So,
1=
Z
∞
−∞
Z
∞
f (x, y)dydx =
−∞
=
hence k = 8.
Z
1
0
1
k
8
Z
1
kxydydx
x
16.6 SOLUTIONS
(b) The region where x < y <
√
1167
x is sketched in Figure 16.95
y
y=x
√
y=
1
x
x
0
1
Figure 16.95
So the probability that (x, y) satisfies x < y <
Z
1
0
Z
√
√
x is given by:
x
8xy dy dx =
x
=
Z
Z
1
0
1
√x
4x(y 2 )x dx
4x(x − x2 )dx
0
1
1 3 1 4 x − x =4
3
4
0
=4
=
1
3
1
1
−
3
4
This tells us that in choosing points from the√region defined by 0 ≤ x ≤ y ≤ 1, that 1/3 of the time we would
pick a point from the region defined by x < y < x. These regions are shown in Figure 16.95.
18. (a) For a density function,
1=
Z
∞
−∞
Z
∞
f (x, y) dy dx =
−∞
=
Z
Z
So k = 3/8.
(c)
19. Since
Z
1
0
Z
Z
2−y
0
1/2
0
3 2
x dx dy =
8
Z
1
0
3 2
x dx dy =
8
XX
x
Z
y
1
0
Z
2
1
kx2 dy dx
0
kx2 dx
0
1 15
1
−1
(2 − y)3 dy =
(2 − y)4 0 =
8
32
32
1/2
0
0
Z
kx3 2
8k
=
.
3 0
3
=
(b)
2
1 3 1
x 0 dy =
8
f (x, y) ∆x ∆y ≈
Z
Z
1/2
0
1
1
dy =
.
8
16
f (x, y) dx dy
R
and since x never exceeds 1, and we can assume that no one lives to be over 100, so y does not exceed 100, we have
Fraction of
=
policies
Z
f (x, y) dx dy =
R
where R is the rectangle: 0.8 ≤ x ≤ 1, 65 ≤ y ≤ 100.
Z
100
65
Z
1
f (x, y) dx dy,
0.8
1168
Chapter Sixteen /SOLUTIONS
20. (a) The area of S is (2)(4) = 8. Because the density function p(x, y) is constant on S and the total volume under a
density function above the xy-plane is 1, p(x, y) = 1/8 for (x, y) in S, and p(x, y) = 0 for (x, y) outside S.
(b) The probability that (x, y) is in T is
Z
T
1
f (x, y) dy dx =
8
Z
dy dx =
T
α
area(T )
= .
8
8
21. (a) Since the exponential function is always positive and λ is positive, p(t) ≥ 0 for all t, and
Z
∞
p(t)dt = lim −e
−λt
b→∞
0
b
= lim −e−bt + 1 = 1.
b→∞
0
(b) The density function for the probability that the first substance decays at time t and the second decays at time s is
p(t, s) = λe−λt µe−µs = λµe−λt−µs ,
for s ≥ 0 and t ≥ 0, and is zero otherwise.
(c) We want the probability that the decay time t of the first substance is less than or equal to the decay time s of the
second, so we want to integrate the density function over the region 0 ≤ t ≤ s. Thus, we compute
Z
∞
0
Z
∞
λµe
−λt −µs
e
ds dt =
t
=
=
Z
∞
Z0 ∞
Z0 ∞
∞
λe−λt (−e−µs )t dt
λe−λt e−µt dt
λe(−λ+µ)t dt
0
=
−λ −(λ+µ)t ∞
λ
e
=
.
0
λ+µ
λ+µ
So for example, if λ = 1 and µ = 4, then the probability that the first substance decays first is 1/5 .
22. (a)
Z
(b)
Z
π+ π
6
12
θ= π
6
Z
π
6
θ=0
Z
4
p(r, θ)r dr dθ
1
r= cos
θ
4
p(r, θ)r dr dθ +
1
r= cos
θ
Z
2π
6
π
θ= π
+ 12
6
Z
4
p(r, θ)r dr dθ
1
r= sin
θ
23. (a) If t ≤ 0, then F (t) = 0 because the average of two positive numbers can not be negative. If 1 < t then F (t) = 1
because Rthe average of two numbers each at most 1 is certain to be less than or equal to 1. For any t, we have
F (t) = R p(x, y)dA where R is the region of the plane defined by (x + y)/2 ≤ t. Since p(x, y) = 0 outside the
unit square, we need integrate only over the part of R that lies inside the square, and since p(x, y) = 1 inside the
square, the integral equals the area of that part of the square. Thus, we can calculate the area using area formulas.
For 0 ≤ t ≤ 1, we draw the line (x + y)/2 = t, which has x- and y-intercepts of 2t. Figure 16.96 shows that for
0 < t ≤ 1/2,
1
F (t) = Area of triangle = · 2t · 2t = 2t2 .
2
In Figure 16.97, when x = 1, we have y = 2t − 1. Thus, the vertical side of the unshaded triangle is 1 − (2t − 1) =
2 − 2t. The horizontal side is the same length, so for 1/2 < t ≤ 1,
F (t) = Area of Square − Area of triangle = 12 −
The final result is:
0

2t2
F (t) =
 1 − 2(1 − t)2
1
1
(2 − 2t)2 − 1 − 2(1 − t)2 .
2
if t ≤ 0
if 0 < t ≤ 1/2
.
if 1/2 < t ≤ 1
if 1 < t
16.7 SOLUTIONS
1169
y
2t
y
1
x+y
2
1
=t
2t
2t
2 − 2t -
6
2t − 1
2 − 2t
2t − 16
?
x
1
1
Figure 16.96: For 0 < t ≤
Figure 16.97: For
1
2
?
x+y
2
=t
x
2t
1
<t≤1
2
(b) The probability density function p(t) of z is the derivative of its cumulative distribution function. We have
0

if t ≤ 0
4t
if 0 < t ≤ 1/2
.
p(t) =
 4 − 4t if 1/2 < t ≤ 1
0
if 1 < t
See Figure 16.98.
(c) The values of x and y and equally likely to be near 0, 1/2, and 1. Notice from the graph of the density function in
Figure 16.98 that even though x and y separately are equally likely to be anywhere between 0 and 1, their average
z = (x + y)/2 is more likely to be near 1/2 than to be near 0 or 1.
p(t)
2
x
0
0.5
1
Figure 16.98
Solutions for Section 16.7
Exercises
1. We have
Therefore,
2. We have
Therefore,
xs xt 5 2 ∂(x, y)
= = −1.
= ∂(s, t)
ys yt 3 1 ∂(x, y) ∂(s, t) = 1.
xs xt 2s −2t ∂(x, y)
=
= 4s2 + 4t2 .
= ∂(s, t)
ys yt 2t 2s ∂(x, y) 2
2
∂(s, t) = 4s + 4t .
Notice we can drop the absolute value signs because in this case the Jacobian is nonnegative for all s and t.
1170
Chapter Sixteen /SOLUTIONS
3. We have
Therefore,
xs xt es cos t −es sin t ∂(x, y)
= (es cos t)2 + (es sin t)2 .
=
= ∂(s, t)
ys yt es sin t es cos t ∂(x, y) 2s
2
2 2s
∂(s, t) = e (cos t + sin t) = e .
Notice we can drop the absolute value signs because the Jacobian in this case is positive for all s and t.
4. We have
xs xt 3s2 − 3t2 −6st ∂(x, y)
= 9(s2 − t2 )2 + 36s2 t2 .
=
= ∂(s, t)
6st
3s2 − 3t2 ys yt Therefore, multiplying out and simplifying
∂(x, y) 4
2 2
4
2 2
2
2 2
∂(s, t) = 9 s − 2s t + t + 4s t = 9(s + t ) .
Notice we can drop the absolute value sign since the Jacobian in this case is nonnegative for all s and t.
5. We have
xs xt xu 3 1 2 ∂(x, y, z)
= ys yt yu = 1 5 −1 .
∂(s, t, u)
zs zt zu 2 −1 1 This 3 × 3 determinant is computed the same way as for the cross product, with the entries 3, 1, 2 in the first row playing
the same role as ~i , ~j , ~k . We get
∂(x, y, z)
= ((5)(1) − (−1)(−1))3 + ((−1)(2) − (1)(1))1 + ((1)(−1) − (2)(5))2 = −13.
∂(s, t, u)
6. We have
xr xθ xz cos θ −r sin θ
∂(x, y, z)
= yr yθ yz = sin θ r cos θ
∂(r, θ, z)
zr zθ zz 0
0
0 0 .
1
This 3 × 3 determinant is computed the way it is for the cross product, with the entries cos θ, −r sin θ, 0 in the first row
playing the same role as ~i , ~j , ~k . We get
∂(x, y, z)
= (r cos θ − 0) cos θ + (0 − sin θ)(−r sin θ) + 0
∂(s, t, u)
= r cos2 θ + r sin2 θ = r.
7. The square T is defined by the inequalities
0 ≤ s = ax ≤ 1
0 ≤ t = by ≤ 1
that correspond to the inequalities
0 ≤ x ≤ 1/a = 10
that define R. Thus a = 1/10 and b = 1.
0 ≤ y ≤ 1/b = 1
8. The square T is defined by the inequalities
0 ≤ s = ax ≤ 1
0 ≤ t = by ≤ 1
that correspond to the inequalities
0 ≤ x ≤ 1/a = 1
that define R. Thus a = 1 and b = 4.
0 ≤ y ≤ 1/b = 1/4
16.7 SOLUTIONS
1171
9. The square T is defined by the inequalities
0 ≤ s = ax ≤ 1
0 ≤ t = by ≤ 1
that correspond to the inequalities
0 ≤ x ≤ 1/a = 50
0 ≤ y ≤ 1/b = 10
that define R. Thus a = 1/50 and b = 1/10.
10. The disc S is defined by the inequality
2
2
s 2 + t 2 = a 2 x2 + b 2 y 2 ≤ 1
that corresponds to the inequality x + y ≤ 152 or equivalently
1
1 2
x + 2 y2 ≤ 1
152
15
that defines R. Thus a2 = 1/152 and b2 = 1/152 . We have a = 1/15, b = 1/15.
11. The disc S is defined by the inequality
s 2 + t 2 = a 2 x2 + b 2 y 2 ≤ 1
that corresponds to the inequality x2 /4 + y 2 /9 ≤ 1 that defines R. Thus a2 = 1/4 and b2 = 1/9. We have a = 1/2,
b = 1/3.
12. Inverting the change of variables gives x = s − at, y = t.
The four edges of R are
y = 0, y = 3, y =
1
1
x, y = (x − 10).
4
4
The change of variables transforms the edges to
t = 0, t = 3, t =
1
1
1
10
1
s − at, t = s − at −
.
4
4
4
4
4
These are equations for the edges of a rectangle in the st-plane if the last two are of the form: s = (Constant). This
happens when the t terms drop out, or a = −4. With a = −4 the change of variables gives
Z Z ∂(x, y) ∂(s, t) ds dt
T
over the rectangle
T : 0 ≤ t ≤ 3, 0 ≤ s ≤ 10.
13. Inverting the change of variables gives x = s − at, y = t.
The four edges of R are
1
1
y = 0, y = 5, y = − x, y = − (x − 10).
3
3
The change of variables transforms the edges to
1
1
1
10
1
.
t = 0, t = 5, t = − s + at, t = − s + at +
3
3
3
3
3
These are equations for the edges of a rectangle in the st-plane if the last two are of the form: s = (Constant). This
happens when the t terms drop out, or a = 3. With a = 3 the change of variables gives
Z Z ∂(x, y) ∂(s, t) ds dt
T
over the rectangle
T : 0 ≤ t ≤ 5, 0 ≤ s ≤ 10.
1172
Chapter Sixteen /SOLUTIONS
14. Inverting the change of variables gives x = s − at, y = t.
The four edges of R are
y = 15, y = 35, y = 2(x − 10) + 15, y = 2(x − 30) + 15.
The change of variables transforms the edges to
t = 15, t = 35, t = 2s − 2at − 5, t = 2s − 2at − 45.
These are equations for the edges of a rectangle in the st-plane if the last two are of the form: s = (Constant). This
happens when the t terms drop out, or a = −1/2. With a = −1/2 the change of variables gives
Z Z ∂(x, y) ∂(s, t) ds dt
T
over the rectangle
5
45
≤s≤
.
2
2
T : 15 ≤ t ≤ 35,
Problems
15. Given T = {(s, t) | 0 ≤ s ≤ 3, 0 ≤ t ≤ 2} and
x = 2s − 3t
y = s − 2t
The shaded area in Figure 16.99 is the corresponding region R in the xy-plane.
y
(6, 3)
1
x
2
y=
y=
(0, 0)
2
x
3
y=
−1
x
R2
R1
y=
2
x
3
(0, −1)
−1
1
x
2
(−6, −4)
Figure 16.99
Since
2 −3 =
1 −2 = −1,
∂(x, y) ∂(s, t) = 1.
∂x
∂(x, y)
∂s
= ∂y
∂(s, t)
∂s
Thus we get
∂x
∂t
∂y
∂t
Z Z
∂(x, y) ds dt =
∂(s, t) T
Since
Z
dx dy =
R
=
thus
Z
Z
dx dy +
R1
0
−6
Z
1
x+1
6
Z
R
dx dy =
R2
dx +
Z
6
0
Z
3
ds
0
0
dx
−6
Z
2
dt = 6.
0
Z
1
− x+1
6
2x
3
dy +
1 x−1
2
6
dx
0
dx = 3 + 3 = 6,
Z ∂(x, y) dx dy =
∂(s, t) ds dt.
T
Z
Z
1x
2
dy
2 x−1
3
16.7 SOLUTIONS
16.
y
2
R
x
4
Figure 16.100
Given T = {(s, t) | 0 ≤ s ≤ 2, s ≤ t ≤ 2} and
x = s2
y = t,
√
x ≤ y ≤ 2}.
R = {(x, y) | 0 ≤ x ≤ 4,
∂x
∂(x, y)
∂s
= ∂y
∂(s, t)
∂s
∂(x, y) ∂(s, t) = 2s
Z 2
Z 2
Z ∂(x, y) ∂(s, t) ds dt = 2
T
So,
Z
s ds
0
∂x
∂t
∂y
∂t
2s 0 =
0 1 = 2s,
since
dt = 2
s
dx dy =
R
Z
h
4
dx
0
Z
Z
17. Given
R
2
0
2
√
dy =
x
2
= 2x − x3/2
3
Thus
Z
0 ≤ s ≤ 2.
i4
0
s(2 − s) ds = 2 s2 −
Z
4
0
=8−
(2 −
s3
3
2
0
=
8
.
3
√
x) dx
16
8
= .
3
3
Z ∂(x, y) dx dy =
∂(s, t) ds dt.
T

 x = ρ sin φ cos θ

y = ρ sin φ sin θ
z = ρ cos φ,
sin φ cos θ ρ cos φ cos θ −ρ sin φ sin θ = sin φ sin θ ρ cos φ sin θ ρ sin φ cos θ cos φ
−ρ sin φ
0
ρ cos φ cos θ −ρ sin φ sin θ + ρ sin φ sin φ cos θ −ρ sin φ sin θ = cos φ ρ cos φ sin θ ρ sin φ cos θ
sin φ sin θ ρ sin φ cos θ ∂x
∂ρ
∂(x, y, z)
= ∂y
∂ρ
∂(ρ, φ, θ)
∂z
∂x
∂φ
∂y
∂φ
∂z
∂ρ ∂φ
∂x
∂θ
∂y
∂θ
∂z
∂θ
= cos φ(ρ2 cos2 θ cos φ sin φ + ρ2 sin2 θ cos φ sin φ)
+ρ sin φ(ρ sin2 φ cos2 θ + ρ sin2 φ sin2 θ)
= ρ2 cos2 φ sin φ + ρ2 sin3 φ
= ρ2 sin φ.
1173
1174
Chapter Sixteen /SOLUTIONS
18. Given
we have
Since
∂s
∂(s, t)
= ∂x
∂t
∂(x, y)
∂x
So
s=
t=
∂x
∂(x, y)
∂s
= ∂y
∂(s, t)
∂s
∂s
∂y
∂t
∂y
1
= 13
− 5
2
13
3
26 26
1
(x + 2y)
13
1
(3y − 5x).
26
1
5
2
1
= 3
+
=
.
26
13
26
13
26
x = 2s + t
y = s − t,
∂x
∂(x, y)
∂s
= ∂y
∂(s, t)
∂s
hence
We get
Z
(x + y) dA =
R
3 −4 =
5 2 = 26,
∂x
∂t
∂y
∂t
∂(x, y) ∂(s, t)
1
·
= 26 ·
= 1.
∂(s, t) ∂(x, y)
26
19. Given
we have
x = 3s − 4t
y = 5s + 2t,
Z
T
∂x
∂t
∂y
∂t
2 1 =
1 −1 = −3,
∂(x, y) ∂(s, t) = 3.
Z
Z
∂(x, y) dsdt = (3s)(3) ds dt = 9 s ds dt,
∂(s, t) 3s T
T
where T is the region in the st-plane corresponding to R.
t
y
(0, 3)
(1, 3)
(2, 1)
T
x
(0, 0)
R
(5, −2)
(0, 0)
(3, −3)
(1, 0)
Figure 16.101
Figure 16.102
Now, we need to find T .
As
x = 2s + t
y = s−t
or
s = 13 (x + y)
t = 13 (x − 2y),
so from the above transformation and Figure 16.101, T is the shaded area in Figure 16.102. Therefore
Z
(x + y) dA = 9
R
Z
1
s ds
0
Z
3
0
1
dt = (27)( ) = 13.5.
2
s
16.7 SOLUTIONS
20. Given
we have
x = 2s
y = 3t ,
∂x
∂(x, y)
∂s
= ∂y
∂(s, t)
∂s
hence
So
Z
2
1 0 1
=2 =
0 1 6
3
∂x
∂t
∂y
∂t
∂(x, y) 1
∂(s, t) = 6 .
2
(x + y ) dA =
R
1175
Z T
t2
s2
+
4
9
1
6
ds dt,
where R is the region bounded by the curve 4x2 + 9y 2 = 36 and T is the corresponding region in the st-plane. Since the
curve 4x2 + 9y 2 = 36 corresponds to the curve s2 + t2 = 36 under the change of coordinates, T is the region bounded
by the curve s2 + t2 = 36. So
Z
Z
1
24
(x2 + y 2 ) dA =
R
s2 ds dt +
T
1
54
Now polar coordinates (r, θ) are used to evaluate the above integral. As
So
1
24
Z
s = r cos θ
t = r sin θ
1
54
Thus,
Z
t2 ds dt =
T
1
54
s2 ds dt =
T
Z
27
2
Z
sin2 θ dθ
0
Z
Z
1
24
2π
RR
and
t2 ds dt.
T
s2 = r2 cos2 θ
t2 = r2 sin2 θ,
∂(s, t) cos θ −r sin θ =
∂(r, θ) sin θ r cos θ = r.
=
and
Z
2π
cos2 θ dθ
0
Z
1
2
2π
6
r3 dr =
0
27
2
Z
(1 + cos 2θ) dθ =
0
6
r3 dr = 6
0
Z
Z
2π
sin2 θ dθ = 3
0
2π
cos2 θ dθ
0
Z
27
(2π) = 13.5π
4
2π
0
(1 − cos 2θ) dθ = (3)(2π) = 6π.
(x2 + y 2 ) dA = 13.5π + 6π = 19.5π.
R
21. The area of the ellipse is
dx dy where R is the region x2 + 2xy + 2y 2 ≤ 1. We must change variables in both the
R
area element dA = dx dy and the region R.
Inverting the variable change gives x = s − t, y = t. Thus
∂x
∂(x, y)
∂s
= ∂y
∂(s, t)
∂s
Therefore
∂x
∂t
∂y
∂t
1 −1 =
0 1 = 1.
∂(x, y) ds dt = ds dt.
dx dy = ∂(s, t) The region of integration is
x2 + 2xy + 2y 2 = (s − t)2 + 2(s − t)t + 2t2 = s2 + t2 ≤ 1.
Let T be the unit disc s2 + t2 ≤ 1. We have
Z Z
dx dy =
R
Z Z
ds dt = Area of T = π.
T
1176
Chapter Sixteen /SOLUTIONS
22. We must change variables in the area element dA = dx dy, the integrand sin(x + y) and the region R.
Inverting the variable change gives x = (s + t)/2, y = (t − s)/2. Thus
∂x
∂(x, y)
∂s
= ∂y
∂(s, t)
∂s
Therefore
The integrand is sin(x + y) = sin t.
The region of integration is
1/2 1/2 1
=
−1/2 1/2 = 2 .
∂(x, y) ds dt = 1 ds dt.
dx dy = ∂(s, t) 2
t−s
2
sin(x + y) dx dy =
Z Z
x2 + y 2 =
Let T be the disc s2 + t2 ≤ 2 of radius
Z Z
∂x
∂t
∂y
∂t
√
s+t
2
2
+
2. We have
R
2
T
=
s2 + t 2
≤ 1.
2
1
(sin t) ds dt = 0.
2
The final integral is zero by symmetry, the integral over the part of the disc where t < 0 cancelling the integral over the
part where t > 0.
23. We must change variables in the area element √
dA = dx dy, the integrand x and the region R.
Inverting the variable change gives x = s − t, y = s where we use the positive square root because the region R
is in the first quadrant where x ≥ 0. Thus
∂x ∂x 1/(2√s − t) −1/(2√s − t) ∂(x, y)
∂s ∂t = = √1 .
= ∂y
∂y 2 s−t
∂(s, t)
1
0
∂s ∂t
Therefore
∂(x, y) ds dt = √ 1
dx dy = ds dt.
∂(s, t) 2 s−t
√
The integrand is x = s − t.
The region of integration can be transformed by examination of its boundaries. The left and right boundaries of R
are given by y − x2 = t = 0 and y − x2 = t = −9. The bottom and top boundaries of R are given by y = s = 0 and
y = s = 16.
Let T be the rectangle 0 ≤ s ≤ 16, −9 ≤ t ≤ 0 of area (9)(16) = 144. We have
Z Z
x dx dy =
R
Z Z
√
T
1
1
s−t √
ds dt = (Area of T ) = 72.
2
2 s−t
24. We must change variables in the differential dr dθ, the integrand 1/r 2 and the region R.
Inverting the variable change gives r = 1/s, θ = t. Thus
Therefore
∂r
∂(r, θ)
∂s
= ∂θ
∂(s, t)
∂s
∂r
∂t
∂θ
∂t
−1/s2
=
0
0 1
= − 2.
s
1
∂(r, θ) ds dt = 1 ds dt.
dr dθ = ∂(s, t) s2
The integrand is 1/r 2 = s2 .
The region R : 1 ≤ r < ∞, 0 ≤ θ ≤ 2π transforms to the rectangle T : 0 ≤ s ≤ 1, 0 ≤ t ≤ 2π of area 2π in the
st-plane.
We have
Z Z
Z Z
Z 2π Z 1
1
2 1
r
dr
dθ
=
s
ds
dt
=
ds dt = 2π.
r3
s2
R
T
0
0
1177
SOLUTIONS to Review Problems for Chapter Sixteen
25. Given
we have
∂s
∂(s, t)
= ∂x
∂t
∂(x, y)
∂x
Since
s = xy
t = xy 2 ,
∂s
∂y
∂t
∂y
y x 2
= 2
y 2xy = xy = t.
∂(s, t) ∂(x, y)
·
= 1,
∂(x, y) ∂(s, t)
∂(x, y)
=t
∂(s, t)
So
Z
xy 2 dA =
R
Z
T
∂(x, y) 1
∂(s, t) = t
so
Z
Z
∂(x, y) 1
ds dt =
t(
)
ds
dt
=
ds dt,
∂(s, t) t
T
T
t where T is the region bounded by s = 1, s = 4, t = 1, t = 4.
Then
Z
Z
4
xy 2 dA =
R
26. Let
we get
Hence
1
s = x−y
t = x + y,
Z
ds
4
dt = 9.
1
that is
1
∂(x, y)
= 12
∂(s, t)
−2
Z
Z
1
2
1
2
x = 21 (s + t)
y = 12 (t − s),
1
= .
2
Z
x−y
s ∂(x, y) 1
s
I=
cos
dx dy =
cos
cos
ds dt,
∂(s, t) ds dt = 2
x
+
y
t
t
R
T
T
where R is the triangle bounded by x + y = 1, x = 0, y = 0 and T is its image which is the triangle bounded by t = 1,
s = −t, s = t.
Then
I=
=
1
2
1
2
Z
Z
1
0
1
0
Z
t
cos
−t
s
t
ds dt =
t · 2 sin 1 dt = sin 1
Z
1
2
1
Z
1
0
t[sin(1) − sin(−1)] dt
t dt =
0
sin 1
= 0.42.
2
Solutions for Chapter 16 Review
Exercises
1. See Figure 16.103.
y
x2
y
1
−1
+
y2
=1
x2 + y 2 = 4 -
2
1
x
x
−2
−1
Figure 16.103
2. See Figure 16.104.
Figure 16.104
1178
Chapter Sixteen /SOLUTIONS
3. See Figure 16.105.
y
√
x=− y
x=
√
y
y=4
y
y = sin x or
x = sin−1 y
1
y=1
x
π
2
x
Figure 16.105
Figure 16.106
4. See Figure 16.106.
5. The region is the half cylinder in Figure 16.107.
6. We use Cartesian coordinates, oriented so that the cube is in the first quadrant. See Figure 16.108. Then, if f is an arbitrary
function, the integral is
Z Z Z
2
3
5
f dxdydz.
0
0
0
Other answers are possible. In particular, the order of integration can be changed.
7. If we imagine the disk lying horizontally, as in Figure 16.109, we can use cylindrical coordinates with the origin at the
center of the flat base. Then, if f is an arbitrary function, the triple integral is
Z
0
Z
2
0
Z
3
f r drdzdθ.
0
z
z
2
(5, 3, 2)
6
y
1
x
1
y
3 cm
2 cm
(5, 3, 0)
y
?
5
x
Figure 16.107
x
Figure 16.108
Figure 16.109
8. We use spherical coordinates as in Figure 16.110. Then if f is an arbitrary function, the triple integral is
Z
2π
0
Z
π
π/2
Z
5
f ρ2 sin φ dρdφdθ.
2
Other answers are possible.
9. We use spherical coordinates, as in Figure 16.111. Then if f is an arbitrary function, the triple integral is
Other answers are possible.
Z
π/2
0
Z
0
π
Z
5
f ρ2 sin φ dρdφdθ.
0
z
1
2π
SOLUTIONS to Review Problems for Chapter Sixteen
10. Integrating with respect to x first we get
Z
Integrating with respect to y first we get
Z
See Figure 16.112
0
−2
Z
4
0
Z
−y+4
f (x, y) dx dy
y
−2
2
2x+4
f (x, y) dy dx +
0
Z
4
0
Z
−x+4
f (x, y) dy dx.
0
z
y
5
4
z
y = 2x + 4 or
x = y2 − 2
y
y
0
2
5
x
R
x
5
x
−5
−2
4
Figure 16.112
Figure 16.111
Figure 16.110
y = −x + 4 or
x = −y + 4
11. Compute in polar coordinates:
Z p
x2 + y 2 dA =
R
=
=
12.
Z
10
0
Z
Z
Z
Z
π
0
π
0
π
0
0.1
xe
xy
dy dx =
0
=
=
Z
2
Z
Z
r · r dr dθ
1
r3
3
2
dθ
1
8
1
−
3
3
dθ =
7π
.
3
10
exy |00.1 dx
0
10
0
(e0.1x − e0 ) dx
e0.1x
0.1
10
− x 0
= (10e1 − 10 − 10e0 )
= 10e − 20 = 10(e − 2).
13.
Z
1
0
Z
4
3
(sin(2 − y)) cos (3x − 7) dx dy =
Z
1
0
4
sin (3x − 7) (sin (2 − y))
dy
3
3
Z
1
1
sin (2 − y) dy
= (sin 5 − sin 2)
3
0
1
= (sin 5 − sin 2) [cos (2 − y)]10
3
1
= (sin 5 − sin 2)(cos 1 − cos 2).
3
1179
1180
Chapter Sixteen /SOLUTIONS
14.
Z
1
Z
Z
y
3
(sin x)(cos x)(cos y) dx dy =
0
0
1
0
1
4
=
Z
y
sin4 x (cos y)
dy
4
0
1
(sin4 y)(cos y) dy
0
1
sin5 y =
20 0
sin5 1
.
20
=
sin (xy)
,
x
15. First use integration by parts, with y as the variable, u = x2 y, u0 = x2 , v =
Z
4
3
Z
1
2
x y cos (xy) dy dx =
0
=
=
Z
Z
Z
4
3
4
[xy
sin (xy)]10
−
Z
v 0 = cos (xy). Then,
1
x sin (xy) dy
0
dx
x sin x + [cos (xy)]10 dx
3
4
3
(x sin x + cos x − 1) dx.
Now use integration by parts again, with u = x, u0 = 1, v = − cos x, v 0 = sin x. Then,
Z
Z
4
3
(x sin x + cos x − 1) dx =
[−x cos x]43
+
4
cos x dx +
3
= (−x cos x + 2 sin x − x)|43
Z
4
3
(cos x − 1) dx
= −4 cos 4 + 2 sin 4 + 3 cos 3 − 2 sin 3 − 1.
Thus,
Z
4
3
Z
1
0
x2 y cos (xy) dy dx = −4 cos 4 + 2 sin 4 + 3 cos 3 − 2 sin 3 − 1.
16.
Z
1
0
Z
z
0
Z
2
(y + z)7 dx dy dz =
0
Z
1
0
=
Z
1
0
=
Z
1
0
=
=
255
4
255
4
Z
z
2(y + z)7 dy dz
0
z
2(y + z)8 dz
8
0
(2z)8 − z 8
dz
4
Z
1
z 8 dz
0
z9
9
255 1
·
=
4
9
85
.
=
12
1
0
SOLUTIONS to Review Problems for Chapter Sixteen
1181
17. The region is shown in Figure 16.113.
y
1
√
0
1 − x2
1
x
√
− 1 − x2
−1
Figure 16.113
The integral has the same values in the upper and lower quarter circles, so we integrate over just the upper circle and
multiply by 2. We convert the integral to polar coordinates.
Z
Z
Z
Z Z √
1−x2
1
0
−
√
e−(x
2
+y 2 )
1
π/2
π/2
2
e−r r dr dθ =
dy dx = 2
1−x2
0
0
=
=
Z
0
1
0
π/2
0
2
(−e−r ) dθ
1 − e−1 dθ
π
(1 − e−1 ).
2
18. We use cylindrical coordinates. The cone has radius r = 2 when z = 4, so its equation is z = 2r. Thus, the integral is
Z
2π
0
Z
2
0
Z
4
f (r, θ, z)r dz dr dθ.
2r
19. We use spherical coordinates. The cone has radius 2 when z = 4, so ρ sin φ = 2 when ρ cos φ = 4. Thus tan φ = 1/2,
so φ = arctan(1/2). The top of the cone, z = 4, is given by ρ cos φ = 4. Thus, the integral is
Z
2π
0
Z
arctan(1/2)
0
Z
4/ cos φ
g(ρ, φ, θ)ρ2 sin θ dρ dφ dθ.
0
20. In rectangular coordinates, a cone has equation z = k
√
22 = 2, we have k = 2. Thus, the integral is
Z 2 Z √4−x2 Z
−2
−
√
4−x2
p
2
x2 + y 2 for some constant k. Since z = 4 when
4
√
h(x, y, z) dz dy dx.
x2 +y 2
21. From Figure 16.114, we have the following iterated integrals:
x2 + y 2 + z 2 = 1
z
y
x
Figure 16.114
p
x2 + y 2 =
1182
Chapter Sixteen /SOLUTIONS
(a)
Z
f dV =
Z
1
Z √1−x2 Z √1−x2 −y2
f (x, y, z) dzdydx
√
1−x2 0
√
√
Z
Z 1 Z 1−y2 Z 1−x2 −y2
f (x, y, z) dzdxdy
(b)
f dV =
√
R
−1 − 1−y 2 0
√
√
Z 1 Z 1−y2 Z 1−y2 −z2
Z
f dV =
(c)
f (x, y, z) dxdzdy
√
R
−1 0
− 1−y 2 −z 2
Z √
Z Z √
Z
R
−1
−
1−x2
1
(d)
1−x2 −z 2
f (x, y, z) dydzdx
√
− 1−x2 −z 2
√
√
Z
Z 1 Z 1−z2 Z 1−x2 −z2
(e)
f dV =
f (x, y, z) dydxdz
√
√
R
− 1−z 2 − 1−x2 −z 2
0
√
√
Z
Z 1 Z 1−z2 Z 1−y2 −z2
(f)
f (x, y, z) dxdydz
f dV =
√
√
f dV =
R
−1
0
R
0
−
1−z 2
1−y 2 −z 2
−
22. The integral is over the region 0 ≤ x2 + y 2 ≤ 3, 1 ≤ z ≤ 4 − x2 − y 2 . Using cylindrical coordinates, we get
Z
2π
0
Z
√
0
3
Z
4−r 2
1
rdz dr dθ =
z2
1
=
=
Z
Z
Z
Z
2π
0
2π
0
2π
0
2π
Z
Z
√
0
√
3
3
4−r2
r (− )
z 1
(−
0
dr dθ
r
r
+ )dr dθ
4 − r2
1
1
1
ln(4 − r 2 ) + r2
2
2
√ 3
dθ
0
3
1
1
( ln 1 + − ln 4 − 0)dθ
2
2
2
0
3
= 2π( − ln 2) = π(3 − 2 ln 2)
2
=
23. The region is a hemisphere 0 ≤ x2 +y 2 +z 2 ≤ 32 , z ≥ 0, so spherical coordinates are appropriate. Recall the conversion
formula x = ρ sin φ cos θ. Then the integral in spherical coordinates becomes
Z
2π
0
=
Z
2π
0
=
Z
2π
0
=
243
5
243
=
5
=
243
5
Z
Z
Z
Z
Z
Z
π/2
0
π/2
0
π/2
0
2π
0
2π
Z
Z
Z
3
(ρ sin φ cos θ)2 ρ2 sin φ dρ dφ dθ
0
3
ρ4 sin3 φ cos2 θ dρ dφ dθ
0
243
sin3 φ cos2 θ dφ dθ
5
π/2
cos2 θ · sin φ(1 − cos2 φ) dφ dθ
0
0
1
cos θ − cos φ + cos3 φ
3
2
2π
cos2 θ[−(−1) +
0
243 2
=
·
5
3
Z
2π
0
π2
dθ
0
1
(−1)] dθ
3
1 + cos 2θ
dθ
2
2π
81
1
162π
81
(θ + sin 2θ) =
(2π + 0) =
=
5
2
5
5
0
SOLUTIONS to Review Problems for Chapter Sixteen
24. The integral is over the region x, y ≥ 0, x2 + y 2 ≤ 1, 0 ≤ z ≤
Z
π/2
0
Z
1
0
Z
r
(z + r) rdz dr dθ =
0
=
=
Z
Z
Z
p
1183
x2 + y 2 . Using cylindrical coordinates, we get
Z
π/2
0
Z
π/2
0
1
0
1
0
Z
r
(rz + r 2 ) dz dr dθ
0
1
( r3 + r3 ) dr dθ
2
1
3 4 r dθ
8 0
π/2
0
3 π
3π
= · =
8 2
16
25. W is a cylindrical shell, so cylindrical coordinates should be used. See Figure 16.115.
z
1-16
4
?
y
x
Figure 16.115
Z
W
z
dV =
(x2 + y 2 )3/2
=
=
=
=
Z
Z
Z
Z
Z
4
0
4
0
4
0
4
0
4
0
Z
Z
Z
Z
2π
2
z
rdr dθ dz
r3
2
z
dr dθ dz
r2
1
0
2π
0
Z
1
2π
0
2π
0
Z
2
z (− ) dθ dz
r 1
z
dθ dz
2
4
z
1
· 2π dz = π · z 2 = 8π
2
2
0
Problems
26. Positive, since e−x is always positive.
27. Negative, since y 3 is negative on B, where y < 0.
28. Positive, since (x + y 2 ) is positive on R, where x > 0.
29. Can’t tell, since y 3 is both positive and negative for x < 0.
30. Can’t tell, since x < 0 and y 2 ≥ 0 on L, where x < 0.
31. Zero. The solid sphere is symmetric and z is positive on the top half and negative (of equal absolute value) on the bottom
half. The integral of z over the entire solid is zero because the integrals over each half add up to zero.
32. Zero. x is positive on the hemisphere x2 +y 2 +z 2 ≤ 1, x > 0 and negative (of equal absolute value) on x2 +y 2 +z 2 ≤ 1,
x < 0. The integral of x over the entire solid is zero because the integrals over each half add up to zero.
1184
Chapter Sixteen /SOLUTIONS
33. Zero. You can see this in several ways. One way is to observe that xy is positive on the part of the sphere above and below
the first and third quadrants (where x and y are of the same sign) and negative (of equal absolute value) on the part of the
sphere above and below the second and fourth quadrants (where x and y have opposite signs). These add up to zero in the
integral of xy over all of W .
Another way to see that the integral is zero is to write the triple integral as an iterated integral, say integrating first
with respect to x. For fixed y and z, the x-integral is over an interval symmetric about 0. The integral of x over such an
interval is zero. If any of the inner integrals in an iterated integral is zero, then the triple integral is zero.
34. Zero. Write the triple integral as an iterated integral, say integrating first with respect to x. For fixed y and z, the x-integral
is over an interval symmetric about 0. The integral of x over such an interval is zero. If any of the inner integrals in an
iterated integral is zero, then the triple integral is zero.
35. Zero. Write the triple integral as an iterated integral, say integrating first with respect to x. Then sin( π2 xy) is integrated
over an integral symmetric about the origin, and this integral is zero because sin( π2 xy) is an odd function. Since the
innermost integral is zero so is the triple integral.
36. Positive. Since x2 + y 2 + z 2 ≤ 1 on the solid, we know x2 + y 2 ≤ 1. This means that |xy| ≤ 1, since if |x| ≤ |y| then
|xy| ≤ |y 2 | = y 2 , and if |y| ≤ |x|, then |xy| ≤ |x2 | = x2 . So | π2 xy| ≤ π2 on the solid, and hence cos( π2 xy) is positive
on the solid W and so is its integral.
37. Negative. Since z 2 − 1 ≤ 0 in the sphere, its integral is negative.
38. Positive. The function e−xyz is a positive function everywhere so its integral over W is positive.
39. Positive.
p
x2 + y 2 + z 2 is positive on W , so its integral is positive.
40. (a) The equation of the curved surface of this half cylinder along the x-axis is (y − 1) 2 + z 2 = 1. The part we want is
p
z=
(b) The integral
Z
1 − (y − 1)2
f (x, y, z) dV =
D
Z
10
0
Z
0≤y≤2
2
0
0 ≤ x ≤ 10.
Z √1−(y−1)2
f (x, y, z) dz dy dx.
0
41. The region of integration is shown in Figure 16.116, and the mass of the given solid is given by
z
12
z = 12 − 4x − 3y
4
y
4x + 3y = 12
1
(12 − 4x)
or y = 3
3
x
Figure 16.116
Mass =
=
Z
Z
δ dV
R
3
0
Z
1 (12−4x)
3
0
Z
12−4x−3y
x2 dzdydx
0
SOLUTIONS to Review Problems for Chapter Sixteen
=
=
=
Z
Z
Z
3
0
0
Z
3Z
3
1 (12−4x)
3
0
z=12−4x−3y
x z
dydx
2
z=0
1 (12−4x)
3
0
x2 (12 − 4x − 3y) dydx
y= 13 (12−4x)
3
x (12y − 4xy − y 2 )
2
2
0
1185
3
8 5 x = 8x − 4x +
15
3
4
108
.
=
5
dx
0
0
42. We use spherical coordinates because we are integrating over a sphere and the density has spherical symmetry. D = 2ρ.
M=
Z
2π
0
Z
π
0
Z
3
(2ρ)ρ2 sin φ dρ dφ dθ.
0
43. Let the lower left part of the forest be at (0, 0). Then the other corners have coordinates as shown. The population density
function is then given by
ρ(x, y) = 10 − 2y
The equations of the two diagonal lines are x = −2y/5 and x = 6 + 2y/5. So the total rabbit population in the forest is
Z
5
0
Z
y
6+ 2
5
−2
y
5
(10 − 2y) dx dy =
=
Z
Z
5
(10 − 2y)(6 +
0
5
0
4
y) dy
5
8 2
y ) dy
5
(60 − 4y −
5
8 3 = (60y − 2y −
y )
15
0
2
8
· 125
15
2750
550
=
=
≈ 183
15
3
= 300 − 50 −
44. We must first decide on coordinates. We pick cylindrical coordinates with the z-axis along the axis of the cylinders. The
insulation stretches from z = 0 to z = l. See Figure 16.117. The volume is given by the integral
Volume =
Z
2π
0
Z lZ
0
a+h
r drdzdθ.
a
Evaluating the integral gives
Volume =
Z
2π
0
Z
l
0
r2 a+h
l
dzdθ = 2πz 2 a
0
(a + h)2
a2
−
2
2
= πl((a + h)2 − a2 ).
To check our answer, notice that the volume is the difference between the volume of two cylinders of radius a and a + h.
These cylinders have volumes πl(a + h)2 and πla2 , respectively.
1186
Chapter Sixteen /SOLUTIONS
z
6
l
? h
-
a
-
y
a+h
a
x
Figure 16.117
45. The plane (x/p) + (y/q) + (z/r) = 1 cuts the axes at the points (p, 0, 0); (0, q, 0); (0, 0, r). Since p, q, r are positive,
the region between this plane and the coordinate planes is a pyramid in the first octant. Solving for z gives
z=r
1−
x
y
−
p
q
rx
ry
−
.
p
q
=r−
The volume, V , is given by the double integral
V =
Z R
rx
ry
r−
−
p
q
dA,
where R is the region shown in Figure 16.118. Thus
V =
=
=
=
=
Z
Z
Z
Z
p
0
p
Z
0
q−qx/p
0
p
r
0
p
0
rx
ry
r−
−
p
q
ry 2
rxy
−
ry −
p
2q
qx
q−
p
dydx
y=q−qx/p !
dx
y=0
r
qx
− x q−
p
p
rq 2
2rqx
rqx2
+ 2 −
rq −
p
p
2q
r
−
2q
qx
q−
p
2x
x2
1−
+ 2
p
p
dx
p
rqx3
rqx
rqx2
rqx3 rqx2
+ 2 −
+
− 2
rqx −
p
p 3
2
p2
2p 3 0
= pqr − pqr +
pqr
pqr
pqr
pqr
pqr
−
+
−
=
.
3
2
2
6
6
y
q
(x
y
+ =1
p
q
qx
y=q−
p
R
p
Figure 16.118
x
2 !
dx
1187
SOLUTIONS to Review Problems for Chapter Sixteen
46. We must first decide on coordinates. We imagine the vertex of the cone downward, at the origin, with the flat base in the
plane z = h, as in Figure 16.119. Then, using cylindrical coordinates as in Figure 16.120, we see that the curved surface
of the cone has equation z = hr/a. Thus the volume is given by
Volume =
Evaluating gives
Z
Volume =
2π
0
= 2π
Z
a
0
Z
z=h
rz 2π
0
Z
Z
a
0
h
r dzdrdθ.
hr/a
drdθ =
z=hr/a
hr3
hr2
−
2
3a
Z
2π
0
Z
a
0
hr2
hr −
a
drdθ
2
a
a
πha2
a2
=
−
.
= 2πh
2
3
3
0
z
z
z = hr/a
h
-
a
6
h
?
y
a
x
Figure 16.119
r
Figure 16.120
47. We use spherical coordinates. Since the density, δ, is equal to the distance from the point to the origin, we have
δ = ρ gm/cm3 .
Therefore the mass of the hemisphere is given by
Mass =
Z
=4
2π
0
Z
Z
2π
0
π/2
0
Z
Z
2
0
ρ · ρ2 sin φ dρdφdθ =
π/2
sin φ dφdθ = 4
0
Z
2π
0
Z
0
2π
Z
π/2
0
2
ρ4 sin φ dφdθ
4 0
π/2
dθ = 4 · 2π · 1 = 8π gm.
(− cos φ)
0
48. Since the hole resembles a cylinder, we will use cylindrical coordinates. Let the center of the sphere be at the origin, and
let the center of the hole be the z-axis (see Figure 16.121).
√
a2 − R 2
R
z
R
a
y
x
Figure 16.121
1188
Chapter Sixteen /SOLUTIONS
√
√
Then we will integrate from z = − a2 − R2 to z = a2 − R2 , and each cross-section will be an annulus. So the
volume is
Z √a2 −R2 Z 2π Z √a2 −z2
Z √a2 −R2 Z 2π
1 2
(a − z 2 − R2 ) dθ dz
r dr dθ dz =
√
√
2
− a2 −R2 0
R
− a2 −R2 0
Z √
a2 −R2
=π
−
√
a2 −R2
(a2 − z 2 − R2 ) dz
= π (a2 − R2 )(2
=
3
4π 2
(a − R2 ) 2
3
p
a2 − R 2 ) −
3
1
(2(a2 − R2 ) 2 )
3
49. We must first decide on coordinates. We pick Cartesian coordinates with the smaller sphere centered at the origin, the
larger one centered at (0, 0, −1). A vertical cross-section of the region in the xz-plane is shown in Figure 16.122. The
smaller sphere has equation x2 + y 2 + z 2 = 1. The larger sphere has equation x2 + y 2 + (z + 1)2 = 2.
x2 + y 2 + z 2 = 1
z
x2 + y 2 + (z + 1)2 = 2
x
−1
Figure 16.122
Let R represent the region in the xy-plane which lies directly underneath (or above) the region whose volume we
want. The curve bounding this region is a circle, and we find its equation by solving the system:
x2 + y 2 + z 2 = 1
x2 + y 2 + (z + 1)2 = 2
Subtracting the equations gives
(z + 1)2 − z 2 = 1
2z + 1 = 1
z = 0.
2
2
2
2
Since z = 0, the two surfaces intersect in the xy-plane in the
pcircle x + y = 1. Thus R is x + y ≤ 1.
The top half of the small sphere is represented by z = 1 − x2 − y 2 ; the top half of the large sphere is represented
p
by z = −1 + 2 − x2 − y 2 . Thus the volume is given by
Z Z √
Z √
1−x2
1
Volume =
−1
−
√
1−x2
1−x2 −y 2
−1+
√
dzdydx.
2−x2 −y 2
Starting to evaluate the integral, we get
Volume =
Z
1
−1
Z √1−x2 p
−
√
1−x2
(
1 − x2 − y 2 + 1 −
p
2 − x2 − y 2 ) dydx.
SOLUTIONS to Review Problems for Chapter Sixteen
We simplify the integral by converting to polar coordinates
Volume =
=
Z
Z
2π
0
2π
0
= 2π
Z
1
0
p
1 − r2 + 1 −
p
2 − r 2 r drdθ
(2 − r 2 )3/2
(1 − r 2 )3/2
r2
+
+
−
3
2
3
1
1
+ −
2
3
1
23/2
− +
3
3
= 2π
1
dθ
0
√ 7
2 2
−
= 1.41.
6
3
50.
2
2
6
10
f (x, y) = √ e−50(x−5) √ e−18(y−15)
2π
2π
30 −50(x−5)2 −18(y−15)2
e
=
π
51. Suppose the brick is set up as shown in Figure 16.123.
z
5
3
6
1
y
?
x
Figure 16.123
The brick has m/v = density = 1. The moment of inertia about the z-axis is
Iz =
=
=
Z
Z
Z
5/2
−5/2
5/2
−5/2
5/2
Z
Z
3/2
−3/2
3/2
Z
1/2
1(x2 + y 2 ) dz dy dx
−1/2
(x2 + y 2 ) dy dx
−3/2
(3x2 +
−5/2
9
) dx
4
45
85
125
+
=
=
4
4
2
The moment of inertia about the y-axis is
Iy =
=
=
Z
Z
Z
5/2
−5/2
5/2
−5/2
5/2
Z
Z
3/2
−3/2
3/2
Z
1(x2 + z 2 ) dz dy dx
−1/2
(x2 +
−3/2
(3x2 +
−5/2
1/2
1
) dy dx
12
1
) dx
4
125
5
65
=
+ =
4
4
2
1189
1190
Chapter Sixteen /SOLUTIONS
The moment of inertia about the x-axis is
Ix =
=
=
Z
Z
Z
5/2
−5/2
5/2
−5/2
5/2
Z
Z
(
−5/2
= 5·
3/2
−3/2
3/2
Z
1/2
1(y 2 + z 2 ) dz dy dx
−1/2
(y 2 +
−3/2
1
) dy dx
12
9
1
+ ) dx
4
4
25
10
=
4
2
52. Let the ball be centered at the origin. Since a ball looks the same from all directions, we can choose the axis of rotation;
in this case, let it be the z-axis. It is best to use spherical coordinates, so then
x2 + y 2 = (ρ sin φ cos θ)2 + (ρ sin φ sin θ)2
= ρ2 sin2 φ
Then m/v = Density = 1, so the moment of inertia is
Iz =
=
=
=
=
Z
Z
R
0
R
Z
0
Z
0
Z
R
0
R
R
0
Z
Z
Z
Z
2π
0
2π
0
2π
Z
Z
π
1(ρ2 sin2 φ)ρ2 sin φ dφ dθ dρ
0
π
ρ4 (sin φ)(1 − cos2 φ) dφ dθ dρ
0
π
1
ρ (− cos φ + cos3 φ) dθ dρ
3
0
4
0
2π
0
4 4
ρ dθ dρ
3
8
8π 4
ρ dρ =
πR5
3
15
53. Set up the cylinder with the base centered at the origin on the xy plane, facing up. (See Figure 16.124.) Newton’s Law of
Gravitation states that the force exerted between two particles is
F =G
m1 m2
ρ2
where G is the gravitational constant, m1 and m2 are the masses, and ρ is the distance between the particles. We take a
small volume element, so m1 = m, and m2 = δdV . In cylindrical coordinates,
if m is at (0,0,0) and δdV is at (r, θ, z),
√
(see Figure 16.124), then the distance from m to δdV is given by ρ = r2 + z 2 for r1 ≤ r ≤ r2 and 0 ≤ z ≤ h.
z
r2
-
r1
-
φ
m
r
Figure 16.124
6
z
?
δdV
SOLUTIONS to Review Problems for Chapter Sixteen
1191
Due to the symmetry of the cylinder the sum of all the horizontal forces is zero; the net force on m is vertical. The force
acting on the particle as a result of the small piece dV makes an angle φ with the vertical and therefore has vertical
component
Vertical force on
GmδdV
GmδdV
Gmδz
z
particle from small = √
· cos φ = 2
=
·√
3 dV.
2 + z 2 )2
2 + z2
2 + z2) 2
r + z2
(
r
r
(r
piece of cylinder
Thus, since dV = rdzdrdθ,
Total force =
Z
2π
0
Z
r2
r1
= 2πGmδ
= 2πGmδ
Z
Z
Z
h
0
r2
r1
r2
r1
Z
Gmδz
r dzdrdθ
(r2 + z 2 )3/2
h
zr
drdz
(r2 + z 2 )3/2
0
r
1−
1
(r2 + h2 ) 2
r2
= 2πGmδ(r − (r + h ) )
r1
p
dr
1
2 2
2
r22 + h2 +
= 2πGmδ(r2 − r1 −
p
r12 + h2 ).
54. The outer circle is a semicircle of radius 4. This is shown in Figure 16.125, with center at D. Thus, CE = 2 and DC = 2,
while AD = 4. Notice that angle ADO is a right angle.
A
r
4
r−2
O
2
D
2
C
E
B
Figure 16.125
Suppose the large circle has center O and radius r. Then OA = r and OD = OC − DC = r − 2. Applying
Pythagoras’ Theorem to triangle OAD gives
r2 = 42 + (r − 2)2
r2 = 16 + r 2 − 4r + 4
r = 5.
If we put the origin at O, the equation of the large circle is x2 + y 2 = 25. In the same coordinates, the equation of the
small circle, which has center at D = (3, 0), is (x − 3)2 + y 2 = 16. The right hand side of the two circles are given by
x=
p
and
25 − y 2
x=3+
p
16 − y 2 .
Since the y-coordinate of A is 4 and the y-coordinate of B is −4, we have
Z Z √
3+
4
Area =
=
Z
−4
1 dxdy
√
25−y 2
4
(3 +
−4
16−y 2
= 13.95.
p
16 − y 2 −
p
25 − y 2 ) dy
1192
Chapter Sixteen /SOLUTIONS
55. Let’s denote the (x, y) coordinates of the points in the lagoon by L. Since x and y are measured in kilometers and d is
measured in meters, and 1 km = 1000 m, the volume of a small piece of the lagoon is given by
∆V ≈ d(x, y)(1000∆x)(1000∆y)m3 .
Thus, the total volume of the lagoon is given by
V = 10002
Z
d(x, y) dxdy.
L
Changing coordinates using u = x/2 and v = y − f (x) converts the depth function to:
d(x(u, v), y(u, v)) = 40 − 160v 2 − 160u2 = 160(
1
− u2 − v 2 ) meters.
4
Thus, the points in the lagoon have (u, v) coordinates in the disk, D, given by u 2 + v 2 ≤ 1/4.
The Jacobian of the transformation is:
∂x
∂u
∂y
∂x
∂v
∂y
∂u ∂v
Thus, the integral in u, v coordinates is
V = 10002
Z
d(x, y) dxdy = 106
L
Converting to polar coordinates, we have
V = 320 · 10
6
Z
2π
0
Z
1/2
0
Z
2
0 = 0
2f (2u) 1 = 2.
160(
D
1
− u2 − v 2 )2 dudv = 320 · 106
4
Z
(
D
1/2
1 r2
r4 1
− )
( − r2 )r drdθ = 320 · 106 2π(
4
4 2
4 0
1
− u2 − v 2 ) dudv.
4
= 107 π m3 .
CAS Challenge Problems
56. The region is the triangle to the right of the y-axis, below the line y = 1, and above the line y = x. Thus the integral can
R1Ry 2
R1R1 2
be written as 0 x ey dydx or as 0 0 ey dxdy. The second of these integrals can be evaluated easily by hand:
Z
1
0
Z
x=y Z 1
2
e dxdy =
e x
dy =
yey dy
0
0
x=0
1
1
1 2
= ey = (e − 1)
2
2
y
Z
y2
0
1
y2
0
The other integral cannot be done by hand with the methods you have learned, but some computer algebra systems will
compute it and give the same answer.
57. In Cartesian coordinates the integral is
Z p
3
x2 + y 2 dA =
D
In polar coordinates it is
Z p
3
x2
+
y2
dA =
D
=
Z
Z
2π
0
Z
1
−1
1
Z √1−x2 p
3
−
√
x2 + y 2 dydx.
1−x2
√
3
r2 rdrdθ =
0
2π
0
Z
6π
3
dθ =
8
7
Z
2π
0
Z
1
r5/3 drdθ
0
The Cartesian coordinate version requires the use of a computer algebra system. Some CASs may be able to handle it
and may give the answer in terms of functions called hypergeometric functions. To compare the answers are the same you
may need to ask the CAS to give a numerical value for the answer. It’s possible your CAS will not be able to handle the
integral at all.
CHECK YOUR UNDERSTANDING
58.
Z
Z
1193
R 0 R 1 x+y
x+y
dydx = 1/2 and −1 0 (x−y)
3 dxdy = −1/2. This does not contradict the theorem because the
3
(x − y)
0
−1
function is not continuous everywhere inside the region of integration. In fact, it is not even defined at the origin.
1
0
59.
Average value for F =
=
1
Area
1
4h2
= a+
Z
−h
h
(a + bx4 + cy 4 + dx2 y 2 + ex3 y 3 ) dx dy
−h
4ah2 +
4ch6
4dh6
4bh6
+
+
5
5
9
1
(9b + 9c + 5d) h4
45
The limit is
lim (a +
h→0
Z
h
1
(9b + 9c + 5d) h4 ) = a.
45
Notice that F (0, 0) = a.
1
Average value for G =
4h2
Z
h
−h
1
=
4h2
= c+
The limit is
lim
h→0
Notice that G(0, 0) = b + c.
Finally,
c+
Average value for H =
Z
h
(a sin(kx) + b cos(ky) + c) dx dy
−h
4 ch2 k + bh sin(hk)
k
!
b sin(hk)
.
hk
b sin(hk)
hk
= c + b lim
h→0
(a + b) −1 + e2h
sin(hk)
= b + c.
hk
−2 − 2h − h2 + e2h 2 − 2h + h2
4e2h h2
.
You may need to simplify the answer given by your CAS to get this form. The limit of this as h → 0 (calculated with a
CAS) is 0. This is equal to H(0, 0).
In each case the limit of the average values over smaller and smaller squares centered at the origin is equal to the value
of the function at the origin. We conjecture that this is true in general for a continuous function. This makes sense because
when the square is small, the function is approximately constant on the square with value equal to its value at the origin.
Therefore the integral is approximately the area times the value of the function, so the average value is approximately the
value of the function. This approximation gets better and better as h → 0.
CHECK YOUR UNDERSTANDING
1. False. For example, if f (x, y) < 0 for all (x, y) in the region R, then
2. True. The double integral is the limit of the sum
X
f (x, y)∆A =
X
R
R
k∆A = k
f dA is negative.
X
∆A
over rectangles that lie inside the region R. As the area ∆A → 0, this sum approaches k · Area(R).
3. False. The function f (x, y) = exy is largest at the (1, 1) corner of R, so for any (x, y) in R we have exy ≤ e1·1 = e.
Then
Z
exy dA = lim
So
R
R
R
exy dA ≤ e ≈ 2.7.
∆A→0
X
exy ∆A ≤ lim
∆A→0
X
e∆A = e lim
∆A→0
X
∆A = e · Area(R) = e.
1194
Chapter Sixteen /SOLUTIONS
4. False. For example, if f = 1, then
R
R
1 dA = Area(R) = 6 and
5. True. The double integral is the limit of the sum
X
R
S
1 dA = Area(S) = 6.
ρ(x, y)∆A. Each of the terms ρ(x, y)∆A is an approximation of
∆A→0
the total population inside a small rectangle of area ∆A. Thus the limit of the sum of all of these numbers as ∆A → ∞
gives the total population of the region R.
6. False. If the graph of f has equal volumes above and below the xy-plane over the region R, the double integral is zero
without having f (x, y) = 0 everywhere.
7. True. Writing the definition of the integral of g, we have
Z
X
g dA = lim
∆A→0
R
g(x, y)∆A = lim
∆A→0
X
kf (x, y)∆A = k lim
∆A→0
X
f (x, y)∆A = k
Z
8. False. As a counterexample, let R be a rectangle with area 2 and take f (x, y) = g(x, y) = 1. Then
R
R
R
1 dA = Area(R) = 2, but R f dA = R g dA = Area(R) · Area(R) = 4.
R
f dA.
R
R
R
f · g dA =
9. False. There is no reason to expect this to be true, since the behavior of f on one half of R can be completely unrelated
to the behavior of f on the other half. As a counterexample, suppose that f is defined so that f (x,
R y) = 0 for points
(x, y) lying in S, and f (x, y) = 1 for points (x, y) lying in the part of R that is not in S. Then S f dA = 0, since
R
f = 0 on all of S. To evaluate R f dA, note that f = 1 on the square S1 which is 0 ≤ x ≤ 1, 1 ≤ y ≤ 2. Then
R
R
f dA = S f dA = Area(S1 ) = 1, since f = 0 on S.
R
1
10. True. Since all points in the region R satisfy x < y, it is true that at every point in R, f (x, y) = x + x < x + y = g(x, y).
Since all of the values of f in R are less than those of g, the average of the values of f is less than the average of the
values of g.
11. False. Since the inside integral is performed with respect to x and the outside integral with respect to y, the region of
integration is the rectangle 5 ≤ x ≤ 12, 0 ≤ y ≤ 1.
12. False. The iterated integral
R2R1
0
0
f dxdy is over a rectangle. The correct limits are
R 1 R 2−x
f dydx.
x
0
13. True. For any point in the region of integration we have 1 ≤ x ≤ 2, and so y is between the positive numbers 1 and 8.
14. False. The sign of
R2R2
1
RbRd
a
c
f dydx depends on the behavior of the function f on the region of integration. For example,
(−x)dydx = − 32 .
1
15. False. The integrals
R 1 R yThe region of
R 1 R x are over different regions, so there is no reason to expect their values to be equal.
integration of 0 0 f dydx is the triangle with vertices (0, 0), (1, 1) and (1, 0), while the region of 0 0 f dxdy is the
triangle with vertices (0, 0), (1, 1) and (0, 1).
16. True. Since f does not depend on x, the inside integral (which is with respect to x) evaluates to
(f − 0) = f . Thus
Z bZ
a
1
f dxdy =
0
Z
b
R1
0
f dy.
x=1
f dx = xf =
x=0
a
17. False. The given limits describe only the upper half disk where y ≥ 0. The correct limits are
R a R √a2 −x2
−a
−
√
a2 −x2
f dydx.
18. True. As an example, take the region inside the triangle with vertices (0, 0), (1, 1) and (2, 3). Iterated integration over this
region using either the dxdy or the dydx orders requires breaking the region into two pieces:
Z
1
0
Z
3x/2
f dydx +
x
Z
2
1
Z
Z
3x/2
f dydx
or
2x−1
1
0
Z
y
f dxdy +
2y/3
Z
3
1
Z
1 (y+1)
2
f dxdy
2y/3
19. True. In the inner integral with respect to y, the function g(x) can be treated as a constant, so
The result of the integral
Rd
c
Z
a
Z bZ
a
d
g(x)h(y) dxdy =
c
Z
b
g(x)
a
Z
d
h(y) dy
c
dx.
h(y) dy is a constant, so may be factored out of the integral with respect to x. Thus we have
b
g(x)
Z
d
h(y) dy
c
dx =
Z
d
h(y) dy
c
Z
·
b
g(x) dx .
a
PROJECTS FOR CHAPTER SIXTEEN
20. False. As a counterexample, consider
Z
2
0
Z
2
R2R2
0
0
(x + y) dxdy. We have
(x + y) dxdy =
0
and
Z
Z
2
x dx +
0
1195
Z
2
0
2
x=2
x2
dy =
+ yx 2
x=0
x=2
Z
2
(2 + 2y) dy = 8
0
y=2
x2 y 2 +
= 2 + 2 = 4.
2 x=0
2 y=0
y dy =
0
21. False. The integral gives the total mass of the material contained in W.
22. True. The region lies above the square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 and below the plane z = x.
23. False. The given limits only cover the part of the unit ball in the first octant where x ≥ 0, y ≥ 0, and z ≥ 0. To cover the
entire unit ball the limits are
√
√
Z
1
−1
Z
1−x2
−
√
1−x2
Z
1−x2 −y 2
−
√
f dzdydx.
1−x2 −y 2
24. True. Both sets of limits describe the solid region lying above the triangle x + y ≤ 1, x ≥ 0, y ≥ 0, z = 0 and below the
plane x + y + z = 1.
25. True. Both sets of limits describe the solid region lying above the rectangle −1 ≤ x ≤ 1, 0 ≤ y ≤ 1, z = 0 and below
the parabolic cylinder z = 1 − x2 .
RbRdRk
26. False. The iterated integral is of the form like a c e f dz dy dx only if the rectangular region has faces parallel to the
coordinate axes. More general rectangular regions, such as a cube with one corner at the origin and the opposite corner at
(0, 0, 1) will need to be written as the sum of iterated integrals where the limits are not constant.
27. False. As a counterexample, consider f (x, y, z) = 12 − x. Then f is positive on half the cube and negative on the other
R1R1R1
half. Symmetry can be used to show that 0 0 0 ( 21 − x)dz dy dx = 0.
28. True. Since
R
W
f dV = lim
and since f > g, we have
X
f (xi , yj , zk )∆V, where (xi , yj , zk ) is a point inside the ijk-th subbox of volume ∆V,
i,j,k
lim
X
f (xi , yj , zk )∆V > lim
i,j,k
X
g(xi , yj , zk )∆V =
i,j,k
Z
g dV.
W
29. False. As a counterexample, let W1 be the solid cube 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ z ≤ 1, and let W2 be the solid cube
1
1
1
1
−
1 ) = 1 and volume(W2 ) = 8 . Now if f (x, y, z) = −1, then
R 2 ≤ x ≤ 0, − 2 ≤ y ≤ 0, − 2 ≤ z ≤R 0. Then volume(W
1
f dV = 1 · −1 which is less than W f dV = 8 · −1.
W
1
2
30. True. If W is the solid region lying under theRgraph of f and above the region RRin the xy-plane, we can compute the
volume of W either using the double integral R f dA, or using the triple integral W 1 dV.
PROJECTS FOR CHAPTER SIXTEEN
1. (a) We are integrating over the whole plane, so converting to polar coordinates gives
∞
Z 2π
Z 2π Z ∞
Z 2π
Z ∞Z ∞
1 −r2 1
−r 2
−(x2 +y 2 )
− e dθ =
e rdrdθ =
e
dxdy =
dθ = π.
2
2
0
0
0
0
−∞ −∞
0
(b) Rewriting the integrand as a product gives
Z
Z ∞Z ∞
2
2
e−(x +y ) dxdy =
−∞
−∞
∞
−∞
Z
∞
2
2
e−x e−y dxdy.
−∞
2
Now e−y is a constant as far as the integral with respect to x is concerned, so
Z ∞
Z ∞Z ∞
Z ∞
−x2 −y 2
−y 2
−x2
e
e dxdy =
e
e
dx dy.
−∞
−∞
−∞
−∞
1196
Chapter Sixteen /SOLUTIONS
We assume that the integral with respect to x converges, and so is a constant as far as the integral with
respect to y is concerned. Thus, we have
Z ∞
Z ∞
Z ∞
Z ∞
−y 2
−x2
−x2
−y 2
e dy .
e
dx
e
dx dy =
e
But
R∞
−∞
e−x dx and
Z
∞
−∞
Z
∞
e
R∞
−∞
−∞
−∞
−∞
−∞
2
2
e−y dy are the same number, so we can write
−(x2 +y 2 )
dxdy =
−∞
Z
∞
e
−x2
dx
−∞
(c) Using the results of parts (a) and (b), we have
2 Z
Z ∞
2
e−x dx =
−∞
∞
−∞
Z
Z
∞
∞
e
−y 2
−∞
e−(x
2
+y 2 )
dy
=
Z
∞
e
−x2
dx
−∞
2
.
dxdy = π.
−∞
Taking square roots and observing that the integral we are looking for is positive, we have
Z ∞
√
2
e−x dx = π.
−∞
2. (a) We want to find the average value of |x − y| over the square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1:
Z 1Z 1
|x − y| dy dx.
Average distance between gates =
0
0
Let’s fix x, with 0 ≤ x ≤ 1. Then |x − y| =
Z
1
0
|x − y| dy =
Z
x
0
(x − y) dy +
y − x for y ≥ x
. Therefore
x − y for y ≤ x
Z
1
x
(y − x) dy
x 2
1
2
2
y
= x2 − x + 1 − x − x + x2
=
+
−
xy
2
2
2
2
0
x
1
= x2 − x + .
2
y2
xy −
2
So,
Average distance between gates =
Z
1
Z
1
|x − y| dy dx
Z 1
Z 1 Z 1
1
(x2 − x + ) dx
|x − y| dy dx =
=
2
0
0
0
1
3
2
1
x
x
1 =
−
+ x = .
3
2
2
3
0
0
0
(b) There are (n + 1)2 possible pairs (i, j) of gates, i = 0, . . . , n, j = 0, . . . , n, so the sum given represents
the average distances apart of all such gates. The Riemann sum with ∆x = ∆y = 1/n, if we choose the
least x and y-values in each subdivision is
n−1
X
X n−1
i=0
i
− j 1 ,
n n n2
j=0
which for large n is just about the same as the other sum. For n = 5 the sum is about 0.389; for n = 10
the sum is about 0.364.