Solution to Section 5.4, assigned April 23
(1) (Section 5.4, Problem 10) Evaluate the iterated triple integral
Z 2 Z √4−x2 Z √4−x2 −y2
xydzdydx
0
0
Solution:
Z 2 Z √4−x2 Z √4−x2 −y2
0
=
0
Z 2Z
0
xydzdydx =
0
√
0
0
4−x2
Z
2
Z
√
4−x2
0
0
√
z= 4−x2 =y 2
dydx
[xyz]z=0
iy=√4−x2
h x
p
2
2
3/2
− (4 − x − y )
xy 4 − x2 − y 2dydx =
dx
3
0
0
2
Z 2
32
1
x
2 5/2
2 3/2
=
4−x
(4 − x ) dx = −
=
15
15
0 3
0
Z
2
(2) (Problem 5.4, Problem 14) Evaluate the volume integral (triple
integral) of f (x, y, z) = x2 over S, where S is the solid bounded
by the paraboloids z = x2 + y 2 and z = 8 − x2 − y 2 .
Solution:
z=8−x2 −y2
S
z=x2+y2
R
x 2 +y 2 =4
Figure 1. Region S bounded above by paraboloid z =
8−x2 −y 2 and below by paraboloid z = x2 +y 2. Surfaces
intersect on the curve x2 + y 2 = 4 = z. So boundary of
the projected region R in the x − y plane is x2 + y 2 = 4.
Where the two surfaces intersect z = x2 + y 2 = 8 − x2 − y 2.
So, 2x2 + 2y 2 = 8 or x2 + y 2 = 4 = z, this is the curve at
the intersection of the two surfaces. Therefore, the boundary
of projected region R in the x − y plane is given by the circle
x2 + y 2 = 4. So R can be treated as a y simple region in the
1
2
√
x − y plane, with upper and lower curves y = ± 4 − x2 for
−2 ≤ x ≤ 2. Therefore,
#
Z Z "Z 8−x2 −y2
Z Z Z
x2 dV =
x2 dz dA
=
Z
2
−2
Z
S
√
R
4−x2
√
− 4−x2
x2 z
z=8−x2 −y2
z=x2 +y 2
x2 +y 2
dydx =
Z
2
−2
Z
√
4−x2
√
− 4−x2
x2 8 − 2x2 − 2y 2 dydx
y=√4−x2
2
2
dx
=
x2 8y − 2x2 y − y 3
√
3
−2
y=− 4−x2
Z 2 √
√
3/2
4
2
2
2
4−x
=
dx
x 16 4 − x2 − 4x 4 − x2 −
3
−2
Z
Substituting x = 2 sin θ and noting that dx = 2 cos θdθ we get
Z Z Z
Z π
2
256
2
2
2
3
sin θ 256 cos θ − 256 sin θ cos θ −
x dV =
cos θ cos θdθ
3
S
− π2
Z π/2 512
4
2
2
2
4
2
cos θ sin θ
=
dθ 512 cos θ sin θ − 512 sin θ cos θ −
3
0
Using 512 cos2 θ sin2 θ = 128 sin2 (2θ) = 64 (1 − cos[4θ]),
−512 sin4 θ cos2 θ = −128 sin2 (2θ) sin2 θ = −32 [1 − cos 4θ] [1 − cos 2θ]
= −32 + 32 cos 2θ + 32 cos 4θ − 16 cos 2θ − 16 cos 6θ
−
512
128 2
32
cos4 θ sin2 θ = −
sin (2θ) cos2 θ = − [1 − cos 4θ] [1 + cos 2θ]
3
3
3
32
1
1
=−
1 + cos 2θ − cos 4θ − cos 2θ − cos 6θ
3
2
2
Since the integral of cos [2mθ] for m = 1, 2, 3 is a multiple of
sin[2mθ] which is zero at θ = π/2, it follows that
Z Z Z
π
32
32
2
[θ]02 = π
x dV = 64 − 32 −
3
3
S
(3) (Section 5.4, Problem 18) Find the volume of the indicated solid
region S inside the cylinders x2 + y 2 = a2 and x2 + z 2 = a2 .
Solution: Consider only the part of S that lies in the region
x ≥ 0, y ≥ 0, z ≥ 0 From symmetry of the region under the
transformation x → −x, y → −y and z → −z, it follows that
the volume of this region S1 is V8 , where V is the volume of S.
We treat S1 as an x-simple region in 3-D.
3
O plot3d({sqrt(9-x^2), sqrt(9-y^2)},x=-3..3,y=-3..3);
Figure 2. Part of the region S bounded by x2 + z 2 = a2
and x2 + y 2 = a2 for x ≥ 0
Note that the projection of region S1 on the y − z plane,
call it R is a a square 0 ≤ y ≤ a, 0 ≤ z ≤ a. We break
up R into two region R1 = {(y, z) : a ≥ y ≥ z ≥ 0, } and R2 =
{(y, z) : a ≥ z >py ≥ 0}. In region (y, z) ∈ R1 , x-ranges
√ from
2
2
x = 0 to x = a − y (since this is smaller than √a2 − z 2 .
2
2
In region (y, z) ∈ R1 , x-ranges
p from x = 0 to x = a − z
2
2
(since this is smaller than a − y . So, it follows that the
total volume of S1 is
Z "Z √a2 −y2 #
Z "Z √a2 −z 2 #
V
dx dA
=
dx dA +
8
0
R1
0
R2
Z aZ yp
Z aZ z√
Z a p
2
2
2
2
=
a − y dzdy +
a − z dydz = 2
y a2 − y 2dy
0
0
0
0
0
a 2
2
= − (a2 − y 2 )3/2 0 = a3
3
3
16 3
Therefore, volume of S is V = 3 a .
4
(4) (Section 5.4, Problem 24). Find the centroid of the given solid
bounded by the paraboloids z = 1 + x2 + y 2 and z = 5 − x2 − y 2
with density proportional to the distnace from the z = 5 plane.
Solution: From the problem statement, density ρ = k|z − 5| =
k(5−z) since region is below plane z = 5. The plot of the region
S between the two paraboloids is similar to (Secion 5.4, Problem
14 ) we have solved above, whose projection R in the x−y plane
is bounded by the curve given by 1 + x2 + y 2 = 5 − x2 − y 2, or
x2 + y 2 = 2. So, we have mass
M=
√
Z
√
2
√
− 2
Z
√
√
2−x2
√
− 2−x2
Z
5−x2 −y 2
1+x2 +y 2
k(5 − z)dzdydx
√
√
√
8
2
3/2
2
16 − 8x − 8y dydx = k √ 16 2 − x2 − 8x 2 − x2 − (2 − x )
dx
=k √
3
− 2 0
− 2
Z √2
Z π/2
Z π/2 128
32
32
1 1
2 3/2
4
(2−x ) dx =
= k
k
cos θdθ = k
1 + + cos(4θ) + 2 cos(2θ) dθ
3
3
3
2 2
0
0
0
= 8πk
Z
2
Z
2−x2
2
2
Z
2
Now, from symmetry of the shape, it follows that xc = 0 = yc .
So, we only need to calculate
Z
√
2
√
− 2
Z
√
2−x2
√
− 2−x2
Z
5−x2 −y 2
= 2k
√
2
√
kz(5 − z)dzdydx
1+x2 +y 2
Z √2 Z √2−x2
2−x2
√
− 2
0
5 2 z3
z −
2
3
z=5−x2 −y2
dydx
z=1+x2 +y 2
56 2 6 2 6
4
4
= 2k √
+ x + y − 4x − 4y dydx
−4x − 4y − 8x y + 2x y + 2x y +
3
3
3
− 2 0
√
Z 2
√
√
√
152
64
32
1424 √
2
4
6
= 4k
x 2 − x2 − x 2 − x2 +
x 2 − x2 dx
2 − x2 −
105
35
35
105
0
x 152
32
76 3
16
= 4k
x(2 − x2 )1/2 +
arcsin 1/2 +
x(2 − x2 )3/2 +
x (2 − x2 )3/2
3
3
2
105
315
√2
64π
4 5
=
x (2 − x2 )3/2
k
−
105
3
0
Z
Z
So, zc =
64πk
3(8πk)
2
=
8
3
2
2 2
and xc = 0, 0, 83 .
4 2
2 4
5
(5) (Section 5.4, Problem 27) Reverse the order of integration appropriate for a z-simple and x-simple regions.
Z 2 Z √1−z 2 /4 Z 3√1−x2 −z 2 /4
f (x, y, z)dydxdz
0
0
0
p
Solution: Since y ranges from 0 to y = 3 1 − x2 − z 4 /4, we
2
2
have the upper surface y9 + x2 + z4 = 1, which is an ellipsoid.
We also note that the projectedpregion R in the x − z plane has
goes between x = 0 and x = 1 − z 2 /4, the latter being the
boundary of an ellipse, while z ranges from 0 to 2. Therefore,
it is clear that the region S is the first octant of an ellipsoid
2
2
bounded by x2 + y9 + z4 = 1.
Treating S as a z-simple
q region, we have lower surface z = 0
and upper-surface z = 2 1 − x2 −
y2
.
9
The projected region in
2
the x − y is the the inside of the ellipse x2 + y9 = 1 in the first
quadrant, which may be described as a y-simple region in the
2-D x − y plane:
n
o
√
(x, y) : 0 ≤ y ≤ 3 1 − x2 , 0 ≤ x ≤ 1
So, the integral above is the same as
Z 1Z
0
√
3 1−x2
0
Z
q
2
2 1−x2 − y9
f (x, y, z)dzdydx
0
Treating S asqa x simple region, we have for fixed y − z, x
2
2
going from 0 to 1 − y9 − z4 . The projected region in the y −z
plane can be described as a z-simple region in the y − z plane
and described by
)
(
r
y2
(y, z) : 0 ≤ z ≤ 2 1 − , 0 ≤ y ≤ 3
9
So, the above integral is the same as
Z 3Z
0
q
2
2 1− y9
0
Z
q
2
2
1− y9 − z4
f (x, y, z)dxdzdy
0
(6) (Section 5.4,RProblem
30) Using Theorem 5.4.3, determine whether
RR
the integral
zdV is positive, negative or 0, where S is the
S
solid bounded by the paraboloid z = −x2 − y 2 and the plane
z = −4.
6
Solution: Note from the description of the region
R R R that f (x, y, z) =
z < 0 in S. Therefore, from theorem 4.5.3,
zdV < 0.
S