MATH 52, FALL 2007 - SOLUTIONS TO HOMEWORK 6.
Problem 25, page 343.
Determine
Z Z Z
x2 + y 2 + 2z 2 dV,
W
where W is the solid cylinder defined by the inequalities x2 + y 2 ≤ 4, −1 ≤ z ≤ 2.
Solution: Cylindrical coordinates are most convenient. The cylinder is defined by
r ≤ 2 and −1 ≤ z ≤ 2:
r=2
Z 2 Z 2π Z 2
Z 2 Z 2π Z 2 Z 2π
1 4
2
2
2 2
(r + 2z )r dr dθ dz =
r +z r
dθ dz =
4 + 4z 2 dθ dz
4
−1 0
0
−1 0
−1 0
r=0
Z 2
8 3 2
2
=
8π + 8πz dz = 8πz + πz
= 48π .
3
−1
−1
Problem 26, page 343.
Determine
Z Z Z
dV
p
,
2
x + y2 + z2
W
where W is the region bounded by the two spheres x2 + y 2 + z 2 = a2 and x2 + y 2 + z 2 = b2 ,
for 0 < a < b.
Solution: Spherical coordinates are most convenient. The region is defined by a ≤
ρ ≤ b:
Z 2π Z π Z b
Z 2π Z π Z b
Z 2π Z π
1 2 2
1 2
ρ sin ϕ dρ dϕ dθ =
(b −a ) sin ϕ dϕ dθ
ρ sin ϕ dρ dϕ dθ =
0
0
a ρ
0
0
a
0
0 2
π
Z 2π Z 2π
1 2
2
(b2 − a2 ) dθ = 2π(b2 − a2 ) .
=
− (b − a ) cos ϕ dθ =
2
0
0
0
Problem 27, page 343.
Determine
Z Z Z
p
2
2
2
x2 + y 2 + z 2 ex +y +z dV,
W
where W is the region bounded by the two spheres x2 + y 2 + z 2 = a2 and x2 + y 2 + z 2 = b2 ,
for 0 < a < b.
Solution: Spherical coordinates are most convenient. The region is defined by a ≤
ρ ≤ b:
Z 2π Z π
Z b
Z 2π Z π Z b
Z b
2
2
ρ2 2
ρe ρ sin ϕ dρ dϕ dθ =
sin ϕ dϕ dθ
ρ3 eρ dρ = 4π
ρ3 eρ dρ
0
0
a
0
0
1
a
a
2
MATH 52, FALL 2007 - SOLUTIONS TO HOMEWORK 6.
2
2
The integrand is equal to u dv, where u = ρ2 , dv = ρeρ dρ, and v = 12 eρ . We apply
integration by parts:
ρ=b
Z b
Z ρ=b
1 2 ρ2 b
2
ρeρ dρ
− 4π
= 4π uv
v du = 4π ρ e
− 4π
2
a
ρ=a
a
ρ=a
2 b2
2 a2
ρ2
b
2
2
2
2
2
2
= 2π(b2 eb −eb −a2 ea +ea ) = 2π((b2 − 1)eb − (a − 1)ea ) .
= 2π(b e −a e )−2π e
a
Problem 29, page 343.
Find the volume of the region W that represents the intersection of the solid cylinder x2 +y 2 ≤
1 and the solid ellipsoid 2(x2 + y 2 ) + z 2 ≤ 10.
Solution: The cylinder and ellipsoid are symmetric around the z axis,
√ so cylindrical
coordinates
are convenient. The region is defined by r ≤ 1 and − 10 − 2r2 ≤ z ≤
√
2
10 − 2r :
r=1
Z 2π Z 1 Z √10−2r2
Z 2π Z 1 p
Z 2π 1
2 3/2
2
dθ
r dz dr dθ =
2r 10 − 2r dr dθ =
− (10−2r )
√
3
0
0
0
0
0
− 10−2r2
r=0
Z
=
0
2π
1 3/2 1 3/2
10 − 8 dθ =
3
3
2π
Z
0
√
√
√
√
4 2π √
1 √
2π
(10 10−8 8) dθ =
(10 10−16 2) =
(5 5 − 8) .
3
3
3
Problem 30, page 343.
Find the volume of the solid W that is bounded by the paraboloid z = 9 − x2 − y 2 , the xy
plane, and the cylinder x2 + y 2 = 4.
Solution: Cylindrical coordinates are most convenient. The region is defined by
r ≤ 2 and 0 ≤ z ≤ 9 − r2 :
Z 2π Z 2π Z 2
Z 2π Z 2 Z 9−r2
9 2 1 4 r=2
2
r dz dr dθ =
(9 − r )r dr dθ =
r − r
dθ
2
4
0
0
0
0
0
0
r=0
Z
2π
18 − 4 dθ = 2π(14) = 28π .
=
0
Problem 22, page 356
Find
pthe centroid of the region bounded above by the sphere of radius 5 and below by the cone
z = 2 x2 + y 2
Solution:
The region is symmetric about the z-axis, so that x̂ and ŷ lie on that axis (i.e. both are
zero). Thus we need only find
ZZZ
1
ẑ =
z dV
V ol(W )
MATH 52, FALL 2007 - SOLUTIONS TO HOMEWORK 6.
In spherical coordinates
Z 2π Z tan−1 1 Z 5
Z
2
3
=
ρ cos φ sin φ dρdφdθ = 2π
0
0
0
625π
=
2
tan−1
0
Z
tan−1
0
1
2
1
2
3
ρ4
cos φ sin φ
4
5
dφ
0
tan−1 1
2
625π
2
sin φ
cos φ sin φ dφ =
4
0
125π
4
=
Problem 3, Handout
x2
Find the volume of the region that lies inside the sphere x2 + y 2 + z 2 = 4 and the cylinder
+ y 2 − 2x = 0
Solution: Note that x2 + y 2 − 2x = 0 can be rewritten as (x − 1)2 + y 2 = 1. The
integral is
2 cos θ
Z π Z 2 cos θ p
Z π
Z π Z 2 cos θ Z √4−r2
2
2
2 −2
2 3/2
2
2
(4 − r )
dzr dr dθ =
2r 4 − r =
√
3
0
0
0
0
0
− 4−r2
0
π
Z
16 2
16 π 2
=
−
.
1 − sin3 θdθ =
3 0
3 2 3
Problem 4, Handout
Using spherical coordinates, find the volume of the solid inside the sphere of radius a and
above the cone z 2 = x2 + y 2 .
Solution: We have
Z 2π Z a Z π
Z
4
2
ρ sin φ dφdρdθ = 2π
0
0
0
a
π
Z
√
4
ρ − cos φ dρ = (2 − 2)π
2
0
0
a
ρ2 dρ =
0
√
=
2− 2 3
a π
3
Problem 5, Handout
Using cylindrical coordinates, find the moment of inertia of a solid hemisphere of radius 1
about the central axis perpendicular to the base.
Solution: The moment of inertia is given by
ZZZ
x2 + y 2 dV
H
In cylindrical coordinates,
Z
0
2π
Z
0
1Z
0
√
1−r2
r3 dzdrdθ
4
MATH 52, FALL 2007 - SOLUTIONS TO HOMEWORK 6.
1
Z
= 2π
r
3
p
3
1−
0
5
2(1 − r2 ) 2
2(1 − r2 ) 2
= −π
−
3
5
r2 dr
=
1
0
4π
15
Problem 6, Handout
Find the gravitational attraction of the region bounded above by the plane z = 1 and below
by the cone z 2 = x2 + y 2 on a unit mass placed at the origin. Assume that δ = 1.
Solution: In spherical coordinates,
Z
2π
π
4
Z
G
0
0
Z
1
cos φ
Z
cos φ sin φ dρdφdθ = 2πG
0
π
4
1
cos φ
ρ cos φ sin φ
dφ =
0
Z
= 2πG
π
4
0
√
2
).
2
sin φdφ = 2πG(1 −
0
Problem 7, Handout
Find the gravitational attraction of a solid sphere of radius 1 on a unit point mass Q on its
1
surface, if the density of the sphere at a point |P Q|− 2
Solution: We center the point mass at the origin, and the north pole of the sphere at
(2, 0, 0). In spherical coordinates, the gravitational attraction is
Z
2π
Z
G
0
0
π
2
Z
2 cos φ
−1/2
ρ
Z
sin φ cos φ dρdφdθ = 4πG
0
0
Z
√
= 2 2πG
0
π
2
(cos φ)
3/2
π
2
2 cos φ
1/2 ρ
sin φ cos φ dφ
0
√
√
π
2
8 2
4 2
5/2 cos φ =
πG .
sin φdφ = −
5
5
0