Chapt_09_2

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99 Vector Calculus
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504
9.13 Surface Integrals
EXERCISES 9.13
Surface Integrals
1. Letting z = 0, we have 2x + 3y = 12. Using f(x, y) = z = 3 − 1
2
x − 3
4
y we have fx = −12 ,
fy = −34 , 1 + f
2
x + f
2
y =
29
16
. Then
A =
∫ 6
0
∫ 4−2x/3
0
√
29/16 dy dx =
√
29
4
∫ 6
0
(
4− 2
3
x
)
dx =
√
29
4
(
4x− 1
3
x2
) ∣∣∣∣6
0
=
√
29
4
(24− 12) = 3
√
29 .
2. We see from the graph in Problem 1 that the plane is entirely above the region bounded by
r = sin 2θ in the first octant. Using f(x, y) = z = 3− 1
2
x− 3
4
y we have fx = −12 , fy = −
3
4
,
1 + f2x + f
2
y =
29
16
. Then
A =
∫ π/2
0
∫ sin 2θ
0
√
29/16 r dr dθ =
√
29
4
∫ π/2
0
1
2
r2
∣∣∣∣sin 2θ
0
dθ =
√
29
8
∫ π/2
0
sin2 2θ dθ
=
√
29
8
(
1
2
θ − 1
8
sin 4θ
) ∣∣∣∣π/2
0
=
√
29π
32
.
3. Using f(x, y) = z =
√
16− x2 we see that for 0 ≤ x ≤ 2 and 0 ≤ y ≤ 5, z > 0.
Thus, the surface is entirely above the region. Now fx = − x√
16− x2 , fy = 0,
1 + f2x + f
2
y = 1 +
x2
16− x2 =
16
16− x2 and
A =
∫ 5
0
∫ 2
0
4√
16− x2 dx dy = 4
∫ 5
0
sin−1
x
4
∣∣∣∣2
0
dy = 4
∫ 5
0
π
6
dy =
10π
3
.
4. The region in the xy-plane beneath the surface is bounded by the graph of x2 + y2 = 2.
Using f(x, y) = z = x2 + y2 we have fx = 2x, fy = 2y, 1 + f2x + f
2
y = 1 + 4(x
2 + y2).
Then,
A =
∫ 2π
0
∫ √2
0
√
1 + 4r2 r dr dθ =
∫ 2π
0
1
12
(1 + 4r2)3/2
∣∣∣∣
√
2
0
dθ =
1
12
∫ 2π
0
(27− 1)dθ = 13π
3
.
5. Letting z = 0 we have x2 + y2 = 4. Using f(x, y) = z = 4 − (x2 + y2) we have fx = −2x,
fy = −2y, 1 + f2x + f2y = 1 + 4(x2 + y2). Then
A =
∫ 2π
0
∫ 2
0
√
1 + 4r2 r dr dθ =
∫ 2π
0
1
3
(1 + 4r2)3/2
∣∣∣∣2
0
dθ
=
1
12
∫ 2π
0
(173/2 − 1)dθ = π
6
(173/2 − 1).
505
9.13 Surface Integrals
6. The surfaces x2 + y2 + z2 = 2 and z2 = x2 + y2 intersect on the cylinder 2x2 + 2y2 = 2
or x2 + y2 = 1. There are portions of the sphere within the cone both above and
below the xy-plane. Using f(x, y) =
√
2− x2 − y2 we have fx = − x√
2− x2 − y2 ,
fy = − y√
2− x2 − y2 , 1 + f
2
x + f
2
y =
2
2− x2 − y2 . Then
A = 2
[∫ 2π
0
∫ 1
0
√
2√
2− r2 r dr dθ
]
= 2
√
2
∫ 2π
0
−
√
2− r2
∣∣∣1
0
dθ
= 2
√
2
∫ 2π
0
(
√
2− 1)dθ = 4π
√
2(
√
2− 1).
7. Using f(x, y) = z =
√
25− x2 − y2 we have fx = − x√
25− x2 − y2 ,
fy = − y√
25− x2 − y2 , 1 + f
2
x + f
2
y =
25
25− x2 − y2 . Then
A =
∫ 5
0
∫ √25−y2/2
0
5√
25− x2 − y2 dx dy = 5
∫ 5
0
sin−1
x√
25− y2
∣∣∣∣
√
25−y2/2
0
dy
= 5
∫ 5
0
π
6
dy =
25π
6
.
8. In the first octant, the graph of z = x2 − y2 intersects the xy-plane in the line y = x. The
surface is in the firt octant for x > y. Using f(x, y) = z = x2 − y2 we have fx = 2x,
fy = −2y, 1 + f2x + f2y = 1 + 4x2 + 4y2. Then
A =
∫ π/4
0
∫ 2
0
√
1 + 4r2 r dr dθ =
∫ π/4
0
1
12
(1 + 4r2)3/2
∣∣∣∣2
0
dθ
=
1
12
∫ π/4
0
(173/2 − 1)dθ = π
48
(173/2 − 1).
9. There are portions of the sphere within the cylinder both above and below the xy-plane.
Using f(x, y) = z =
√
a2 − x2 − y2 we have fx = − x√
12 − x2 − y2 , fy = −
y√
a2 − x2 − y2 ,
1 + f2x + f
2
y =
a2
a2 − x2 − y2 . Then, using symmetry,
A = 2
[
2
∫ π/2
0
∫ a sin θ
0
a√
a2 − r2 r dr dθ
]
= 4a
∫ π/2
0
−
√
a2 − r2
∣∣∣a sin θ
0
dθ
= 4a
∫ π/2
0
(a− a
√
1− sin2 θ )dθ = 4a2
∫ π/2
0
(1− cos θ)dθ
= 4a2(θ − sin θ)
∣∣∣π/2
0
= 4a2
(π
2
− 1
)
= 2a2(π − 2).
10. There are portions of the cone within the cylinder both above and below the xy-plane. Using
f(x, y) = 12
√
x2 + y2 , we have fx =
x
2
√
x2 + y2
, fy =
y
2
√
x2 + y2
, 1 + f2x + f
2
y =
5
4 .
Then, using symmetry,
506
9.13 Surface Integrals
A = 2
[
2
∫ π/2
0
∫ 2 cos θ
0
√
5
4
r dr dθ
]
= 2
√
5
∫ π/2
0
1
2
r2
∣∣∣∣2 cos θ
0
dθ
= 4
√
5
∫ π/2
0
cos2 θ dθ = 4
√
5
(
1
2
θ +
1
4
sin 2θ
) ∣∣∣∣π/2
0
=
√
5π.
11. There are portions of the surface in each octant with areas equal to the area of the portion
in the first octant. Using f(x, y) = z =
√
a2 − y2 we have fx = 0, fy = y√
a2 − y2 ,
1 + f2x + f
2
y =
a2
a2 − y2 . Then
A = 8
∫ a
0
∫ √a2−y2
0
a√
a2 − y2 dx dy = 8a
∫ a
0
x√
a2 − y2
∣∣∣∣
√
a2−y2
0
dy = 8a
∫ a
0
dy = 8a2.
12. From Example 1, the area of the portion of the hemisphere within x2 + y2 = b2 is 2πa(a − √a2 − b2 ). Thus,
the area of the sphere is A = 2 lim
b→a
2πa(a−
√
a2 − b2 ) = 2(2πa2) = 4πa2.
13. The projection of the surface onto the xz-plane is shown in the graph. Using f(x, z) =
y =
√
a2 − x2 − z2 we have fx = − x√
a2 − x2 − z2 ,
fz = − z√
a2 − x2 − z2 , 1 + f
2
x + f
2
z =
a2
a2 − x2 − z2 . Then
A =
∫ 2π
0
∫ √a2−c21
√
a2−c22
a√
a2 − r2 r dr dθ = a
∫ 2π
0
−
√
a2 − r2
∣∣∣∣
√
a2−c21
√
a2−c22
dθ
= a
∫ 2π
0
(c2 − c1) dθ = 2πa(c2 − c1).
14. The surface area of the cylinder x2 + z2 = a2 from y = c1 to y = c2 is the area of a cylinder of radius a and
height c2 − c1. This is 2πa(c2 − c1).
15. zx = −2x, zy = 0; dS =
√
1 + 4x2 dA∫∫
S
x dS =
∫ 4
0
∫ √2
0
x
√
1 + 4x2 dx dy =
∫ 4
0
1
12
(1 + 4x2)3/2
∣∣∣∣
√
2
0
dy
=
∫ 4
0
13
6
dy =
26
3
16. See Problem 15.∫∫
S
xy(9− 4z) dS =
∫∫
S
xy(1 + 4x2) dS =
∫ 4
0
∫ √2
0
xy(1 + 4x2)3/2 dx dy
=
∫ 4
0
y
20
(1 + 4x2)5/2
∣∣∣∣
√
2
0
dy =
∫ 4
0
242
20
y dy =
121
10
∫ 4
0
y dy =
121
10
(
1
2
y2
) ∣∣∣∣4
0
=
484
5
507
9.13 Surface Integrals
17. zx =
x√
x2 + y2
, zy =
y√
x2 + y2
; dS =
√
2 dA.
Using polar coordinates,∫∫
S
xz3 dS =
∫∫
R
x(x2 + y2)3/2
√
2 dA =
√
2
∫ 2π
0
∫ 1
0
(r cos θ)r3/2r dr dθ
=
√
2
∫ 2π
0
∫ 1
0
r7/2 cos θ dr dθ =
√
2
∫ 2π
0
2
9
r9/2 cos θ
∣∣∣∣1
0
dθ
=
√
2
∫ 2π
0
2
9
cos θ dθ =
2
√
2
9
sin θ
∣∣∣∣2π
0
= 0.
18. zx =
x√
x2 + y2
, zy =
y√
x2 + y2
; dS =
√
2 dA.
Using polar coordinates,∫∫
S
(x + y + z) dS =
∫∫
R
(x + y +
√
x2 + y2 )
√
2 dA
=
√
2
∫ 2π
0
∫ 4
1
(r cos θ + r sin θ + r)r dr dθ
=
√
2
∫ 2π
0
∫ 4
1
r2(1 + cos θ + sin θ) dr dθ =
√
2
∫ 2π
0
1
3
r3(1 + cos θ + sin θ)
∣∣∣∣4
1
dθ
=
63
√
2
3
∫ 2π
0
(1 + cos θ + sin θ) dθ = 21
√
2(θ + sin θ − cos θ)
∣∣∣2π
0
= 42
√
2π.
19. z =
√
36− x2 − y2 , zx = − x√
36− x2 − y2 , zy = −
y√
36− x2 − y2 ;
dS =
√
1 +
x2
36− x2 − y2 +
y2
36− x2 − y2 dA =
6√
36− x2 − y2 dA.
Using polar coordinates,∫∫
S
(x2 + y2)z dS =
∫∫
R
(x2 + y2)
√
36− x2 − y2 6√
36− x2 − y2 dA
= 6
∫ 2π
0
∫ 6
0
r2r dr dθ = 6
∫ 2π
0
1
4
r4
∣∣∣∣6
0
dθ = 6
∫ 2π
0
324 dθ = 972π.
20. zx = 1, zy = 0; dS =
√
2 dA∫∫
S
z2 dS =
∫ 1
−1
∫ 1−x2
0
(x + 1)2
√
2 dy dx =
√
2
∫ 1
−1
y(x + 1)2
∣∣∣1−x2
0
dx
=
√
2
∫ 1
−1
(1− x2)(x + 1)2 dx =
√
2
∫ 1
−1
(1 + 2x− 2x3 − x4) dx
=
√
2
(
x + x2 − 1
2
x4 − 1
5
x5
) ∣∣∣∣1
−1
=
8
√
2
5
508
9.13 Surface Integrals
21. zx = −x, zy = −y; dS =
√
1 + x2 + y2 dA∫∫
S
xy dS =
∫ 1
0
∫ 1
0
xy
√
1 + x2 + y2 dx dy =
∫ 1
0
1
3
y(1 + x2 + y2)3/2
∣∣∣∣1
0
dy
=
∫ 1
0
[
1
3
y(2 + y2)3/2 − 1
3
y(1 + y2)3/2
]
dy
=
[1
15
(2 + y2)5/2 − 1
15
(1 + y2)5/2
] ∣∣∣∣1
0
=
1
15
(35/2 − 27/2 + 1)
22. z =
1
2
+
1
2
x2 +
1
2
y2, zx = x, zy = y; dS =
√
1 + x2 + y2 dA
Using polar coordinates,∫∫
S
2z dS =
∫∫
R
(1 + x2 + y2)
√
1 + x2 + y2 dA
=
∫ π/2
π/3
∫ 1
0
(1 + r2)
√
1 + r2 r dr dθ
=
∫ π/2
π/3
∫ 1
0
(1 + r2)3/2r dr dθ =
∫ π/2
π/3
1
5
(1 + r2)5/2
∣∣∣∣1
0
dθ =
1
5
∫ π/2
π/3
(25/2 − 1) dθ
=
4
√
2− 1
5
(π
2
− π
3
)
=
(4
√
2− 1)π
30
.
23. yx = 2x, yz = 0; dS =
√
1 + 4x2 dA∫∫
S
24
√
y z dS =
∫ 3
0
∫ 2
0
24xz
√
1 + 4x2 dx dz =
∫ 3
0
2z(1 + 4x2)3/2
∣∣∣2
0
dz
= 2(173/2 − 1)
∫ 3
0
z dz = 2(173/2 − 1)
(
1
2
z2
) ∣∣∣∣3
0
= 9(173/2 − 1)
24. xy = −2y, xz = −2z; dS =
√
1 + 4y2 + 4z2 dA
Using polar coordinates,∫∫
S
(1 + 4y2 + 4z2)1/2 dS =
∫ π/2
0
∫ 2
1
(1 + 4r2)r dr dθ
=
∫ π/2
0
1
16
(1 + 4r2)2
∣∣∣2
1
dθ =
1
16
∫ π/2
0
12 dθ =
3π
8
.
25. Write the equation of the surface as y =
1
2
(6−x−3z). Then yx = −12 , yz = −
3
2
; dS =
√
1 + 1/4 + 9/4 =
√
14
2
.∫∫
S
(3z2 + 4yz) dS =
∫ 2
0
∫ 6−3z
0
[
3z2 + 4z
1
2
(6− x− 3z)
] √
14
2
dx dz
=
√
14
2
∫ 2
0
[3z2x− z(6− x− 3z)2]
∣∣∣6−3z
0
dz
=
√
14
2
∫ 2
0
(
[3z2(6− 3z)− 0]− [0− z(6− 3z)2]) dz
=
√
14
2
∫ 2
0
(36z − 18z2) dz =
√
14
2
(18z2 − 6z3)
∣∣∣2
0
=
√
14
2
(72− 48) = 12
√
14
509
9.13 Surface Integrals
26. Write the equation of the surface as x = 6 − 2y − 3z. Then xy = −2, xz = −3; dS =
√
1 + 4 + 9 =
√
14 .∫∫
S
(3z2 + 4yz) dS =
∫ 2
0
∫ 3−3z/2
0
(3z2 + 4yz)
√
14 dy dz =
√
14
∫ 2
0
(3yz + 2y2z)
∣∣∣3−3z/2
0
dz
=
√
14
∫ 2
0
[
9z
(
1− z
2
)
+ 18z
(
1− z
2
)2]
dz =
√
14
∫ 2
0
(
27z − 45
2
z2 +
9
2
z3
)
dz
=
√
14
(
27
2
z2 − 15
2
z3 +
9
8
z4
) ∣∣∣∣2
0
=
√
14(54− 60 + 18) = 2
√
14
27. The density is ρ = kx2. The surface is z = 1−x−y. Then zx = −1, zy = −1; dS =
√
3 dA.
m =
∫∫
S
kx2 dS = k
∫ 1
0
∫ 1−x
0
x2
√
3 dy dx =
√
3 k
∫ 1
0
1
3
x3
∣∣∣∣1−x
0
dx
=
√
3
3
k
∫ 1
0
(1− x)3 dx =
√
3
3
k
[
−1
4
(1− x)4
] ∣∣∣∣1
0
=
√
3
12
k
28. zx = − x√
4− x2 − y2 , zy = −
y√
4− x2 − y2 ;
dS =
√
1 +
x2
4− x2 − y2 +
y2
4− x2 − y2 dA =
2√
4− x2 − y2 dA.
Using symmetry and polar coordinates,
m = 4
∫∫
S
|xy| dS = 4
∫ π/2
0
∫ 2
0
(r2 cos θ sin θ)
2√
4− r2 r dr dθ
= 4
∫ π/2
0
∫ 2
0
r2(4− r2)−1/2 sin 2θ(r dr) dθ u = 4− r2, du = −2r dr, r2 = 4− u
= 4
∫ π/2
0
∫ 0
4
(4− u)u−1/2 sin 2θ
(
−1
2
du
)
dθ = −2
∫ π/2
0
∫ 0
4
(4u−1/2 − u1/2) sin 2θ du dθ
= −2
∫ π/2
0
(
8u1/2 − 2
3
u3/2
) ∣∣∣∣0
4
sin 2θ dθ = −2
∫ π/2
0
(
−32
3
sin 2θ
)
dθ =
64
3
(
−1
2
cos 2θ
) ∣∣∣∣π/2
0
=
64
3
.
29. The surface is g(x, y, z) = y2 + z2 − 4 = 0. ∇g = 2yj + 2zk,
|∇g| = 2
√
y2 + z2 ; n =
yj + zk√
y2 + z2
;
F · n = 2yz√
y2 + z2
+
yz√
y2 + z2
=
3yz√
y2 + z2
; z =
√
4− y2 , zx = 0,
zy = − y√
4− y2 ; dS =
√
1 +
y2
4− y2 dA =
2√
4− y2 dA
Flux =
∫∫
S
F · n dS =
∫∫
R
3yz√
y2 + z2
2√
4− y2 dA =
∫∫
R
3y
√
4− y2√
y2 + 4− y2
2√
4− y2 dA
=
∫ 3
0
∫ 2
0
3y dy dx =
∫ 3
0
3
2
y2
∣∣∣∣2
0
dx =
∫ 3
0
6 dx = 18
510
9.13 Surface Integrals
30. The surface is g(x, y, z) = x2 + y2 + z − 5 = 0. ∇g = 2xi + 2yj + k,
|∇g| =
√
1 + 4x2 + 4y2 ; n =
2xi + 2yj + k√
1 + 4x2 + 4y2
; F · n = z√
1 + 4x2 + 4y2
;
zx = −2x, zy = −2y, dS =
√
1 + 4x2 + 4y2 dA. Using polar coordinates,
Flux =
∫∫
S
F · n dS =
∫∫
R
z√
1 + 4x2 + 4y2
√
1 + 4x2 + 4y2 dA =
∫∫
R
(5− x2 − y2) dA
=
∫ 2π
0
∫ 2
0
(5− r2)r dr dθ =
∫ 2π
0
(
5
2
r2 − 1
4
r4
) ∣∣∣∣2
0
dθ =
∫ 2π
0
6 dθ = 12π.
31. From Problem 30, n =
2xi + 2yj + k√
1 + 4x2 + 4y2
. Then F · n = 2x
2 + 2y2 + z√
1 + 4x2 + 4y2
. Also, from Problem 30,
dS =
√
1 + 4x2 + 4y2 dA. Using polar coordinates,
Flux =
∫∫
S
F · n dS =
∫∫
R
2x2 + 2y2 + z√
1 + 4x2 + 4y2
√
1 + 4x2 + 4y2 dA =
∫∫
R
(2x2 + 2y2 + 5− x2 − y2) dA
=
∫ 2π
0
∫ 2
0
(r2 + 5)r dr dθ =
∫ 2π
0
(
1
4
r4 +
5
2
r2
) ∣∣∣∣2
0
dθ =
∫ 2π
0
14 dθ = 28π.
32. The surface is g(x, y, z) = z − x− 3 = 0. ∇g = −i + k, |∇g| = √2 ; n = −i + k√
2
;
F · n = 1√
2
x3y +
1√
2
xy3; zx = 1, zy = 0, dS =
√
2 dA. Using polar coordinates,
Flux =
∫∫
S
F · n dS =
∫∫
R
1√
2
(x3y + xy3)
√
2 dA =
∫∫
R
xy(x2 + y2) dA
=
∫ π/2
0
∫ 2 cos θ
0
(r2 cos θ sin θ)r2r dr dθ =
∫ π/2
0
∫ 2 cos θ
0
r5 cos θ sin θ dr dθ
=
∫ π/2
0
1
6
r6 cos θ sin θ
∣∣∣∣2 cos θ
0
dθ =
1
6
∫ π/2
0
64 cos7 θ sin θ dθ =
32
3
(
−1
8
cos8 θ
) ∣∣∣∣π/2
0
=
4
3
.
33. The surface is g(x, y, z) = x2 + y2 + z − 4. ∇g = 2xi + 2yj + k,
|∇g| =
√
4x2 + 4y2 + 1 ; n =
2xi + 2yj + k√
4x2 + 4y2 + 1
; F · n = x
3 + y3 + z√
4x2 + 4y2 + 1
;
zx = −2x, zy = −2y, dS =
√
1 + 4x2 + 4y2 dA. Using polar coordinates,
Flux =
∫∫
S
F · n dS =
∫∫
R
(x3 + y3 + z) dA =
∫∫
R
(4− x2 − y2 + x3 + y3) dA
=
∫ 2π
0
∫ 2
0
(4− r2 + r3 cos3 θ + r3 sin3 θ) r dr dθ
=
∫ 2π
0
(
2r2 − 1
4
r4 +
1
5
r5 cos3 θ +
1
5
r5 sin3 θ
) ∣∣∣∣2
0
dθ
=
∫ 2π
0
(
4 +
32
5
cos3 θ +
32
5
sin3 θ
)
dθ = 4θ
∣∣∣2π
0
+ 0 + 0 = 8π.
511
9.13 Surface Integrals
34. The surface is g(x, y, z) = x+ y+ z−6. ∇g = i+ j+k, |∇g| = √3 ; n = (i+ j+k)/√3 ;
F · n = (ey + ex + 18y)/√3 ; zx = −1, zy = −1, dS =
√
1 + 1 + 1 dA =
√
3 dA.
Flux =
∫∫
S
F · n dS =
∫∫
r
(ey + ex + 18y) dA =
∫ 6
0
∫ 6−x
0
(ey + ex + 18y) dy dx
=
∫ 6
0
(ey + yex + 9y2)
∣∣∣6−x
0
dx =
∫ 6
0
[e6−x + (6− x)ex + 9(6− x)2 − 1] dx
= [−e6−x + 6ex − xex + ex − 3(6− x)3 − x]
∣∣∣6
0
= (−1 + 6e6 − 6e6 + e6 − 6)− (−e6 + 6 + 1− 648) = 2e6 + 634 ≈ 1440.86
35. For S1: g(x, y, z) = x2 + y2 − z, ∇g = 2xi + 2yj − k, |∇g| =
√
4x2 + 4y2 + 1 ; n1 =
2xi + 2yj− k√
4x2 + 4y2 + 1
;
F · n1 = 2xy
2 + 2x2y − 5z√
4x2 + 4y2 + 1
; zx = 2x, zy = 2y, dS1 =
√
1 + 4x2 + 4y2 dA. For S2: g(x, y, z) = z − 1,
∇g = k, |∇g| = 1; n2 = k; F ·n2 = 5z; zx = 0, zy = 0, dS2 = dA. Using polar coordinates and R: x2 +y2 ≤ 1
we have
Flux =
∫∫
S1
F · n1 dS1 +
∫∫
S2
F · n2 dS2 =
∫∫
R
(2xy2 + 2x2y − 5z) dA +
∫∫
R
5z dA
=
∫∫
R
[2xy2 + 2x2y − 5(x2 + y2) + 5(1)] dA
=
∫ 2π
0
∫ 1
0
(2r3 cos θ sin2 θ + 2r3 cos2 θ sin θ − 5r2 + 5)r dr dθ
=
∫ 2π
0
(
2
5
r5 cos θ sin2 θ +
2
5
r5 cos2 θ sin θ − 5
4
r4 +
5
2
r2
) ∣∣∣∣1
0
dθ
=
∫ 2π
0
[
2
5
(cos θ sin2 θ + cos2 θ sin θ) +
5
4
]
dθ =
2
5
(
1
3
sin3 θ − 1
3
cos3 θ
) ∣∣∣∣2π
0
+
5
4
θ
∣∣∣∣2π
0
=
2
5
[
−1
3
−
(
−1
3
)]
+
5
2
π =
5
2
π.
36. For S1: g(x, y, z) = x2 + y2 + z − 4, ∇g = 2xi + 2yj + k, |∇g| =
√
4x2 + 4y2 + 1 ;
n1 =
2xi + 2yj + k√
4x2 + 4y2 + 1
; F · n1 = 6z2/
√
4x2 + 4y2 + 1 ; zx = −2x, zy = −2y,
dS1 =
√
1 + 4x2 + 4y2 dA. For S2: g(x, y, z) = x2 + y2 − z, ∇g = 2xi + 2yj− k,
|∇g| =
√
4x2 + 4y2 + 1 ; n2 =
2xi + 2yj− k√
4x2 + y2 + 1
; F · n2 = −6z2/
√
4x2 + 4y2 + 1 ; zx = 2x, zy = 2y,
dS2 =
√
1 + 4x2 + 4y2 dA. Using polar coordinates and R: x2 + y2 ≤ 2 we have
Flux =
∫∫
S1
F · n1 dS1 +
∫∫
S1
F · n2 dS2 =∫∫
R
6z2 dA +
∫∫
−6z2 dA
=
∫∫
R
[6(4− x2 − y2)2 − 6(x2 + y2)2] dA = 6
∫ 2π
0
∫ √2
0
[(4− r2)2 − r4] r dr dθ
= 6
∫ 2π
0
[
−1
6
(4− r2)3 − 1
6
r6
] ∣∣∣∣
√
2
0
dθ = −
∫ 2π
0
[(23 − 43) + (
√
2 )6] dθ =
∫ 2π
0
48 dθ = 96π.
512
9.13 Surface Integrals
37. The surface is g(x, y, z) = x2 + y2 + z2 − a2 = 0. ∇g = 2xi + 2yj + 2zk,
|∇g| = 2
√
x2 + y2 + z2 ; n =
xi + yj + zk√
x2 + y2 + z2
;
F · n = −(2xi + 2yj + 2zk) · xi + yj + zk√
x2 + y2 + z2
= −2x
2 + 2y2 + 2z2√
x2 + y2 + z2
= −2
√
x2 + y2 + z2 = −2a.
Flux =
∫∫
S
−2a dS = −2a× area = −2a(4πa2) = −8πa3
38. n1 = k, n2 = −i, n3 = j, n4 = −k, n5 = i, n6 = −j; F · n1 = z = 1, F · n2 = −x = 0, F · n3 = y = 1,
F · n4 = −z = 0, F · n5 = x = 1, F · n6 = −y = 0; Flux =
∫∫
S1
1 dS +
∫∫
S3
1 dS +
∫∫
S5
1 dS = 3
39. Refering to the solution to Problem 37, we find n =
xi + yj + zk√
x2 + y2 + z2
and dS =
a√
a2 − x2 − y2 dA.
Now
F · n = kq r|r|3 ·
r
|r| =
kq
|r|4 |r|
2 =
kq
|r|2 =
kq
x2 + y2 + z2
=
kq
a2
and
Flux =
∫∫
S
F · n dS =
∫∫
S
kq
a2
dS =
kq
a2
× area = kq
a2
(4πa2) = 4πkq.
40. We are given σ = kz. Now zx − x√
16− x2 − y2 , zy = −
y√
16− x2 − y2 ;
dS =
√
1 +
x2
16− x2 − y2 +
y2
16− x2 − y2 dA =
4√
16− x2 − y2 dA
Using polar coordinates,
Q =
∫∫
S
kz dS = k
∫∫
R
√
16− x2 − y2 4√
16− x2 − y2 dA = 4k
∫ 2π
0
∫ 3
0
r dr dθ
= 4k
∫ 2π
0
1
2
r2
∣∣∣∣3
0
dθ = 4k
∫ 2π
0
9
2
dθ = 36πk.
41. The surface is z = 6− 2x− 3y. Then zx = −2, zy = −3, dS =
√
1 + 4 + 9 =
√
14 dA.
The area of the surface is
A(s) =
∫∫
S
dS =
∫ 3
0
∫ 2−2x/3
0
√
14 dy dx =
√
14
∫ 3
0
(
2− 2
3
x
)
dx
=
√
14
(
2x− 1
3
x2
) ∣∣∣∣3
0
= 3
√
14 .
x¯ =
1
3
√
14
∫∫
S
x dS =
1
3
√
14
∫ 3
0
∫ 2−2x/3
0
√
14x dy dx =
1
3
∫ 3
0
xy
∣∣∣∣2−2x/3
0
dx
=
1
3
∫ 3
0
(
2x− 2
3
x2
)
dx =
1
3
(
x2 − 2
9
x3
) ∣∣∣∣3
0
= 1
y¯ =
1
3
√
14
∫∫
S
y dS =
1
3
√
14
∫ 3
0
∫ 2−2x/3
0
√
14 y dy dx =
1
3
∫ 3
0
1
2
y2
∣∣∣∣2−2x/3
0
dx
=
1
6
∫ 3
0
(
2− 2
3
x
)2
dx =
1
6
[
−1
2
(
2− 2
3
x
)3] ∣∣∣∣3
0
=
2
3
513
z¯ =
1
3
√
14
∫∫
S
z dS =
1
3
√
14
∫ 3
0
∫ 2−2x/3
0
(6− 2x− 3y)
√
14 dy dx
=
1
3
∫ 3
0
(
6y − 2xy − 3
2
y2
) ∣∣∣∣2−2x/3
0
dx =
1
3
∫ 3
0
(
6− 4x + 2
3
x2
)
dx =
1
3
(
6x− 2x2 + 2
9
x3
) ∣∣∣∣3
0
= 2
The centroid is (1, 2/3, 2).
42. The area of the hemisphere is A(s) = 2πa2. By symmetry, x¯ = y¯ = 0.
zx = − x√
a2 − x2 − y2 , zy = −
y√
a2 − x2 − y2 ;
dS =
√
1 +
x2
a2 − x2 − y2 +
y2
a2 − x2 − y2 dA =
a√
a2 − x2 − y2 dA
Using polar coordinates,
z =
∫∫
S
z dS
2πa2
=
1
2πa2
∫∫
R
√
a2 − x2 − y2 a√
a2 − x2 − y2 dA =
1
2πa
∫ 2π
0
∫ a
0
r dr dθ
=
1
2πa
∫ 2π
0
1
2
r2
∣∣∣∣a
0
dθ =
1
2πa
∫ 2π
0
1
2
s2 dθ =
a
2
.
The centroid is (0, 0, a/2).
43. The surface is g(x, y, z) = z − f(x, y) = 0. ∇g = −fxi− fyj + k, |∇g| =
√
f2x + f2y + 1 ;
n =
−fxi− fyj + k√
1 + f2x + f2y
; F · n = −Pfx −Qfy + R√
1 + f2x + f2y
; dS =
√
1 + f2x + f2y dA
∫∫
S
F · n dS =
∫∫
R
−Pfx −Qfy + R√
1 + f2x + f2y
√
1 + f2x + f2y dA =
∫∫
R
(−Pfx −Qfy + R) dA
9.13 Surface Integrals
EXERCISES 9.14
Stokes’ Theorem
1. Surface Integral: curl F = −10k. Letting g(x, y, z) = z− 1, we have ∇g = k and n = k. Then∫∫
S
(curl F) · n dS =
∫∫
S
(−10) dS = −10× (area of S) = −10(4π) = −40π.
Line Integral: Parameterize the curve C by x = 2 cos t, y = 2 sin t, z = 1, for 0 ≤ t ≤ 2π. Then∮ˇ
C
F · dr =
∮ˇ
C
5y dx− 5x dy + 3 dz =
∫ 2π
0
[10 sin t(−2 sin t)− 10 cos t(2 cos t)] dt
=
∫ 2π
0
(−20 sin2 t− 20 cos2 t) dt =
∫ 2π
0
−20 dt = −40π.
514
9.14 Stokes’ Theorem
2. Surface Integral: curl F = 4i− 2j− 3k. Letting g(x, y, z) = x2 + y2 + z − 16,
∇g = 2xi + 2yj + k, and n = (2xi + 2yj + k)/
√
4x2 + 4y2 + 1 . Thus,∫∫
S
(curl F) · n dS =
∫∫
S
8x− 4y − 3√
4x2 + 4y2 + 1
dS.
Letting the surface be z = 16− x2 − y2, we have zx = −2x, zy = −2y, and
dS =
√
1 + 4x2 + 4y2 dA. Then, using polar coordinates,∫∫
S
(curl F) · n dS =
∫∫
R
(8x− 4y − 3) dA =
∫ 2π
0
∫ 4
0
(8r cos θ − 4r sin θ − 3) r dr dθ
=
∫ 2π
0
(
8
3
r3 cos θ − 4
3
r3 sin θ − 3
2
r2
) ∣∣∣∣4
0
dθ =
∫ 2π
0
(
512
3
cos θ − 256
3
sin θ − 24
)
dθ
=
(
512
3
sin θ +
256
3
cos θ − 24θ
) ∣∣∣∣2π
0
= −48π.
Line Integral: Parameterize the curve C by x = 4 cos t, y = 4 sin t, z = 0, for 0 ≤ t ≤ 2π. Then,∮ˇ
C
F · dr =
∮ˇ
C
2z dx− 3x dy + 4y dz =
∫ 2π
0
[−12 cos t(4 cos t)] dt
=
∫ 2π
0
−48 cos2 t dt = (−24t− 12 sin 2t)
∣∣∣2π
0
= −48π.
3. Surface Integral: curl F = i + j + k. Letting g(x, y, z) = 2x + y + 2z − 6, we have
∇g = 2i+ j+2k and n = (2i+ j+2k)/3. Then
∫∫
S
(curl F) ·n dS =
∫∫
S
5
3
dS. Letting
the surface be z = 3− 12y − x we have zx = −1, zy = − 12 , and
dS =
√
1 + (−1)2 + (− 12 )2 dA = 32 dA. Then∫∫
S
(curl F) · n dS =
∫∫
R
5
3
(
3
2
)
dA =
5
2
× (area of R) = 5
2
(9) =
45
2
.
Line Integral: C1: z = 3−x, 0 ≤ x ≤ 3, y = 0; C2: y = 6−2x, 3 ≥ x ≥ 0, z = 0; C3: z = 3−y/2, 6 ≥ y ≥ 0,
x = 0.∮ˇ
C
z dx + x dy + y dz =
∫∫
C1
z dx +
∫
C2
x dy +
∫
C3
y dz
=
∫ 3
0
(3− x) dx +
∫ 0
3
x(−2 dx) +
∫ 0
6
y(−dy/2)
=
(
3x− 1
2
x2
) ∣∣∣∣3
0
−x2
∣∣∣∣0
3
−1
4
y2
∣∣∣∣0
6
=
9
2
− (0− 9)− 1
4
(0− 36) = 45
2
4. Surface Integral: curl F = 0 and
∫∫
S
(curl F) · n dS = 0.
Line Integral: the curve is x = cos t, y = sin t, z = 0, 0 ≤ t ≤ 2π.∮ˇ
C
x dx + y dy + z dz =
∫ 2π
0
[cos t(− sin t) + sin t(cos t)]dt = 0.
515
9.14 Stokes’ Theorem
5. curl F = 2i + j. A unit vector normal to the plane is n = (i + j + k)/
√
3 . Taking the
equation of the plane to be z = 1− x− y, we have zx = zy = −1. Thus,
dS =
√
1 + 1 + 1 dA =
√
3 dA and∮ˇ
C
F · dr =
∫∫
S
(curl F) · n dS =
∫∫
S
√
3 dS =
√
3
∫∫
R
√
3 dA
= 3× (area of R) = 3(1/2) = 3/2.
6. curl F = −2xzi + z2k. A unit vector normal to the plane is n = (j + k)/
√
2 . From z = 1− y, we have zx = 0
and zy = −1. Thus, dS =
√
1 + 1 dA =
√
2 dA and
∮ˇ
C
F · dr =
∫∫
S
(curl F) · n dS =
∫∫
R
1√
2
z2
√
2 dA =
∫∫
R
(1− y)2 dA
=
∫ 2
0
∫ 1
0
(1− y)2 dy dx =
∫ 2
0
−1
3
(1− y)3
∣∣∣∣1
0
dx =
∫ 2
0
1
3
dx =
2
3
.
7. curl F = −2yi− zj− xk. A unit vector normal to the plane is n = (j + k)/√2 . From z = 1− y we have zx = 0
and zy = −1. Then dS =
√
1 + 1 dA =
√
2 dA and
∮ˇ
C
F · dr =
∫∫
S
(curl F) · n dS =
∫∫
R
[
− 1√
2
(z + x)
]√
2 dA =
∫∫
R
(y − x− 1) dA
=
∫ 2
0
∫ 1
0
(y − x− 1) dy dx =
∫ 2
0
(
1
2
y2 − xy − y
) ∣∣∣∣1
0
dx =
∫ 2
0
(
−x− 1
2
)
dx
=
(
−1
2
x2 − 1
2
x
) ∣∣∣∣2
0
= −3.
8. curl F = 2i + 2j + 3k. Letting g(x, y, z) = x + 2y + z − 4, we have ∇g = i + 2j + k
and n = (i + 2j + k)/
√
6 . From z = 4 − x − 2y we have zx = −1 and zy = −2. Then
dS =
√
6 dA and∮ˇ
C
F · dr =
∫∫
S
(curl F) · n dS =
∫∫
R
1√
6
(9)
√
6 dA =
∫∫
R
9 dA = 9(4) = 36.
9. curl F = (−3x2− 3y2)k. A unit vector normal to the plane is n = (i+ j+k)/√3 . From
z = 1− x− y, we have zx = zy = −1 and dS =
√
3 dA. Then,using polar coordinates,∮ˇ
C
F · dr =
∫∫
S
(curl F) · n dS =
∫∫
R
(−
√
3x2 −
√
3 y2)
√
3 dA
= 3
∫∫
R
(−x2 − y2) dA = 3
∫ 2π
0
∫ 1
0
(−r2)r dr dθ
= 3
∫ 2π
0
−1
4
r4
∣∣∣∣1
0
dθ = 3
∫ 2π
0
−1
4
dθ = −3π
2
.
10. curl F = 2xyzi− y2zj + (1− x2)k. A unit vector normal to the surface is n = 2yj + k√
4y2 + 1
. From z = 9− y2 we
have zx = 0, zy = −2y and dS =
√
1 + 4y2 dA. Then
516
9.14 Stokes’ Theorem
∮ˇ
C
F · dr =
∫∫
S
(curl F) · n dS =
∫∫
R
(−2y3z + 1− x2) dA =
∫ 3
0
∫ y/2
0
[−2y3(9− y2) + 1− x2] dx dy
=
∫ 3
0
(
−18y3x + 2y5x + x− 1
3
x3
) ∣∣∣∣y/2
0
dy =
∫ 3
0
(
−9y4 + y6 + 1
2
y − 1
24
y3
)
dy
=
(
−9
5
y5 +
1
7
y7 +
1
4
y2 − 1
96
y4
) ∣∣∣∣3
0
≈ 123.57.
11. curl F = 3x2y2k. A unit vector normal to the surface is
n =
8xi + 2yj + 2zk√
64x2 + 4y2 + 4z2
=
4xi + yj + zk√
16x2 + y2 + z2
.
From zx = − 4x√
4− 4x2 − y2 , zy = −
y√
4− 4x2 − y2 we obtain dS = 2
√
1 + 3x2
4− 4x2 − y2 dA. Then
∮ˇ
C
F · dr =
∫∫
S
(curl F) · n dS =
∫∫
R
3x2y2z√
16x2 + y2 + z2
(
2
√
1 + 3x2
4− 4x2 − y2
)
dA
=
∫∫
R
3x2y2 dA Using symmetry
= 12
∫ 1
0
∫ 2√1−x2
0
x2y2 dy dx = 12
∫ 1
0
(
1
3
x2y3
) ∣∣∣∣2
√
1−x2
0
dx
= 32
∫ 1
0
x2(1− x2)3/2dx x = sin t, dx = cos t dt
= 32
∫ π/2
0
sin2 t cos4 t dt = π.
12. curl F = i + j + k. Taking the surface S bounded by C to be the portion of the plane
x + y + z = 0 inside C, we have n = (i + j + k)/
√
3 and dS =
√
3 dA.∮ˇ
C
F · dr =
∫∫
S
(curl F) · n dS =
∫∫
S
√
3 dS =
√
3
∫∫
R
√
3 dA = 3× (area of R)
The region R is obtained by eliminating z from the equations of the plane and the sphere.
This gives x2 + xy + y2 = 12 . Rotating axes, we see that R is enclosed by the ellipse
X2/(1/3) + Y 2/1 = 1 in a rotated coordinate system. Thus,∮ˇ
C
F · dr = 3× (area of R) = 3
(
π
1√
3
1
)
=
√
3π.
13. Parameterize C by x = 4 cos t, y = 2 sin t, z = 4, for 0 ≤ t ≤ 2π. Then∫∫
S
(curl F) · n dS =
∮ˇ
C
F · dr =
∮ˇ
C
6yz dx + 5x dy + yzex
2
dz
=
∫ 2π
0
[6(2 sin t)(4)(−4 sin t) + 5(4 cos t)(2 cos t) + 0] dt
= 8
∫ 2π
0
(−24 sin2 t + 5 cos2 t) dt = 8
∫ 2π
0
(5− 29 sin2 t) dt = −152π.
517
9.14 Stokes’ Theorem
14. Parameterize C by x = 5 cos t, y = 5 sin t, z = 4, for 0 ≤ t ≤ 2π. Then,∫∫
S
(curl F) · n dS =
∮ˇ
C
F · r =
∮ˇ
C
y dx + (y − x) dy + z2 dz
=
∫ 2π
0
[(5 sin t)(−5 sin t) + (5 sin t− 5 cos t)(5 cos t)] dt
=
∫ 2π
0
(25 sin t cos t− 25) dt =
(
25
2
sin2 t− 25t
) ∣∣∣∣2π
0
= −50π.
15. Parameterize C by C1: x = 0, z = 0, 2 ≥ y ≥ 0; C2: z = x, y = 0, 0 ≤ x ≤ 2;
C3: x = 2, z = 2, 0 ≤ y ≤ 2; C4: z = x, y = 2, 2 ≥ x ≥ 0. Then∫∫
S
(curl F) · n dS =
∮ˇ
C
F · r =
∮ˇ
C
3x2 dx + 8x3y dy + 3x2y dz
=
∫
C1
0 dx + 0 dy + 0 dz +
∫
C2
3x2 dx +
∫
C3
64 dy +
∫
C4
3x2 dx + 6x2 dx
=
∫ 2
0
3x2 dx +
∫ 2
0
64 dy +
∫ 0
2
9x2 dx = x3
∣∣∣2
0
+ 64y
∣∣∣2
0
+ 3x3
∣∣∣0
2
= 112.
16. Parameterize C by x = cos t, y = sin t, z = sin t, 0 ≤ t ≤ 2π. Then∫∫
S
(curl F) · n dS =
∮ˇ
C
F · r =
∮ˇ
C
2xy2z dx + 2x2yz dy + (x2y2 − 6x) dz
=
∫ 2π
0
[2 cos t sin2 t sin t(− sin t) + 2 cos2 t sin t sin t cos t
+ (cos2 t sin2 t− 6 cos t) cos t] dt
=
∫ 2π
0
(−2 cos t sin4 t + 3 cos3 t sin2 t− 6 cos2 t) dt = −6π.
17. We take the surface to be z = 0. Then n = k and dS = dA. Since curl F =
1
1 + y2
i + 2zex
2
j + y2k,∮ˇ
C
z2ex
2
dx + xy dy + tan−1 y dz =
∫∫
S
(curl F) · n dS =
∫∫
S
y2 dS =
∫∫
R
y2 dA
=
∫ 2π
0
∫ 3
0
r2 sin2 θ r dr dθ =
∫ 2π
0
1
4
r4 sin2 θ
∣∣∣∣3
0
dθ
=
81
4
∫ 2π
0
sin2 θ dθ =
81π
4
.
18. (a) curl F = xzi− yzj. A unit vector normal to the surface is n = 2xi + 2yj + k√
4x2 + 4y2 + 1
and
dS =
√
1 + 4x2 + 4y2 dA. Then, using x = cos t, y = sin t, 0 ≤ t ≤ 2π, we have∫∫
S
(curl F) · n dS =
∫∫
R
(2x2z − 2y2z) dA =
∫∫
R
(2x2 − 2y2)(1− x2 − y2) dA
=
∫∫
R
(2x2 − 2y2 − 2x4 + 2y4) dA
=
∫ 2π
0
∫ 1
0
(2r2 cos2 θ − 2r2 sin2 θ − 2r4 cos4 θ + 2r4 cos4 θ) r dr dθ
= 2
∫ 2π
0
∫ 1
0
[r3 cos 2θ − r5(cos2 θ − sin2 θ)(cos2 θ + sin2 θ)] dr dθ
518
9.15 Triple Integrals
= 2
∫ 2π
0
∫ 1
0
(r3 cos 2θ − r5 cos 2θ) dr dθ = 2
∫ 2π
0
cos 2θ
(
1
4
r4 − 1
6
r6
) ∣∣∣∣1
0
dθ
=
1
6
∫ 2π
0
cos 2θ dθ = 0.
(b) We take the surface to be z = 0. Then n = k, curl F · n = curl F · k = 0 and
∫∫
S
(curl F) · n dS = 0.
(c) By Stokes’ Theorem, using z = 0, we have∫∫
S
(curl F) · n dS =
∮ˇ
C
F · dr =
∮ˇ
C
xyz dz =
∮ˇ
C
xy(0) dz = 0.
EXERCISES 9.15
Triple Integrals
1.
∫ 4
2
∫ 2
−2
∫ 1
−1
(x + y + z)dx dy dz =
∫ 4
2
∫ 2
−2
(
1
2
x2 + xy + xz
) ∣∣∣∣1
−1
dy dz
=
∫ 4
2
∫ 2
−2
(2y + 2z) dy dz =
∫ 4
2
(y2 + 2yz)
∣∣∣2
−2
dz =
∫ 4
2
8z dz = 4z2
∣∣∣4
2
= 48
2.
∫ 3
1
∫ x
1
∫ xy
2
24xy dz dy dx =
∫ 3
1
∫ x
1
24xyz
∣∣∣xy
2
dy dx =
∫ 3
1
∫ x
1
(24x2y2 − 48xy)dy dx
=
∫ 3
1
(8x2y3 − 24xy2)
∣∣∣x
1
dx =
∫ 3
1
(8x5 − 24x3 − 8x2 + 24x) dx
=
(
4
3
x6 − 6x4 − 8
3
x3 + 12x2
) ∣∣∣∣3
1
= 522− 14
3
=
1552
3
3.
∫ 6
0
∫ 6−x
0
∫ 6−x−z
0
dy dz dx =
∫ 6
0
∫ 6−x
0
(6− x− z)dz dx =
∫ 6
0
(
6z − xz − 1
2
z2
) ∣∣∣∣6−x
0
dx
=
∫ 6
0
[
6(6− x)− x(6− x)− 1
2
(6− x)2
]
dx =
∫ 6
0
(
18− 6x + 1
2
x2
)
dx
=
(
18x− 3x2 + 1
6
x3
) ∣∣∣∣6
0
= 36
4.
∫ 1
0
∫ 1−x
0
∫ √y
0
4x2z3 dz dy dx =
∫ 1
0
∫ 1−x
0
x2z4
∣∣∣√y
0
dy dx =
∫ 1
0
∫ 1−x
0
x2y2 dy dx
=
∫ 1
0
1
3
x2y3
∣∣∣∣1−x
0
dx =
1
3
∫ 1
0
x2(1− x)3 dx = 1
3
∫ 1
0
(x2 − 3x3 + 3x4 − x5)dx
=
1
3
(
1
3
x3 − 3
4
x4 +
3
5
x5 − 1
6
x6
) ∣∣∣∣1
0
=
1
180
519
9.15 Triple Integrals
5.
∫ π/2
0
∫ y2
0
∫ y
0
cos
x
y
dz dx dy =
∫ π/2
0
∫ y2
0
y cos
x
y
dx dy =
∫ π/2
0
y2 sin
x
y
∣∣∣∣y2
0
dy
=
∫ π/2
0
y2 sin y dy Integration by parts
= (−y2 cos y + 2 cos y + 2y sin y)
∣∣∣π/2
0
= π − 2
6.
∫ √2
0
∫ 2
√
y
∫ ex2
0
x dz dx dy =
∫ √2
0
∫ 2
√
y
xex
2
dx dy =
∫ √2
0
1
2
ex
2
∣∣∣∣2√
y
dy =
1
2
∫ √2
0
(e4 − ey)dy
=
1
2
(ye4 − ey)
∣∣∣√2
0
=
1
2
[(e4
√
2− e
√
2 )− (−1)] = 1
2
(1 + e4
√
2− e
√
2 )
7.
∫ 1
0
∫ 1
0
∫ 2−x2−y2
0
xyez dz dx dy =
∫ 1
0
∫ 1
0
xyez
∣∣∣2−x2−y2
0
dx dy =
∫ 1
0
∫ 1
0
(xye2−x
2−y2 − xy)dx dy
=
∫ 1
0
(
−1
2
ye2−x
2−y2 − 1
2
x2y
) ∣∣∣∣1
0
dy =
∫ 1
0
(
−1
2
ye1−y
2 − 1
2
y +
1
2
ye2−y
2
)
dy
=
(
1
4
e1−y
2 − 1
4
y2 − 1
4
e2−y
2
) ∣∣∣∣1
0
=
(
1
4
− 1
4
− 1
4
e
)
−
(
1
4
e− 1
4
e2
)
=
1
4
e2 − 1
2
e
8.
∫ 4
0
∫ 1/2
0
∫ x2
0
1√
x2 − y2 dy dx dz =
∫ 4
0
∫ 1/2
0
sin−1
y
x
∣∣∣∣x2
0
dx dz =
∫ 4
0
∫ 1/2
0
sin−1 x dx dz
Integration by parts
=
∫ 4
0
(x sin−1 x +
√
1− x2 )
∣∣∣1/2
0
dz =
∫ 4
0
(
1
2
π
6
+
√
3
2
− 1
)
dz =
π
3
+ 2
√
3− 4
9.
∫∫∫
D
z dV =
∫ 5
0
∫ 3
1
∫ y+2
y
z dx dy dz =
∫ 5
0
∫ 3
1
xz
∣∣∣y+2
y
dy dz =
∫ 5
0
∫ 3
1
2z dy dz
=
∫ 5
0
2yz
∣∣∣3
1
dz =
∫ 5
0
4z dz = 2z2
∣∣∣5
0
= 50
10. Using symmetry,∫∫∫D
(x2 + y2) dV = 2
∫ 2
0
∫ 4
x2
∫ 4−y
0
(x2 + y2) dz dy dx = 2
∫ 2
0
∫ 4
x2
(x2 + y2)z
∣∣∣4−y
0
dy dx
= 2
∫ 2
0
∫ 4
x2
(4x2 − x2y + 4y2 − y3) dy dx
= 2
∫ 2
0
(
4x2y − 1
2
x2y2 +
4
3
y3 − 1
4
y4
) ∣∣∣∣4
x2
dx
= 2
∫ 2
0
[(
8x2 +
64
3
)
−
(
4x4 +
5
6
x6 − 1
4
x8
)]
dx
= 2
(
8
3
x3 +
64
3
x− 4
5
x5 − 5
42
x7 +
1
36
x9
) ∣∣∣∣2
0
=
23,552
315
.
520
9.15 Triple Integrals
11. The other five integrals are
∫ 4
0
∫ 2−x/2
0
∫ 4
x+2y
F (x, y, z) dz dy dx,∫ 4
0
∫ z
0
∫ (z−x)/2
0
F (x, y, z) dy dx dz,
∫ 4
0
∫ 4
x
∫ (z−x)/2
0
F (x, y, z) dy dz dx,∫ 4
0
∫ z/2
0
∫ z−2y
0
F (x, y, z) dx dy dz,
∫ 2
0
∫ 4
2y
∫ z−2y
0
F (x, y, z) dx dz dy.
12. The other five integrals are
∫ 3
0
∫ √36−4y2/3
0
∫ 3
1
F (x, y, z) dz dx dy,
∫ 3
1
∫ 2
0
∫ √36−9x2/2
0
F (x, y, z) dy dx dz,
∫ 3
1
∫ 3
0
∫ √36−4y2/3
0
F (x, y, z) dx dy dz,
∫ 3
0
∫ 3
1
∫ √36−4y2/3
0
F (x, y, z) dx dz dy,
∫ 2
0
∫ 3
1
∫ √36−9x2/2
0
F (x, y, z) dy dz dx.
13. (a) V =
∫ 2
0
∫ 8
x3
∫ 4
0
dz dy dx (b) V =
∫ 8
0
∫ 4
0
∫ y1/3
0
dx dz dy (c) V =
∫ 4
0
∫ 2
0
∫ 8
x2
dy dx dz
14. Solving z =
√
x and x + z = 2, we obtain x = 1, z = 1. (a) V =
∫ 3
0
∫ 1
0
∫ 2−z
z2
dx dz dy
(b) V =
∫ 1
0
∫ 2−z
z2
∫ 3
0
dy dx dz (c) V =
∫ 3
0
∫ 1
0
∫ √x
0
dz dx dy +
∫ 3
0
∫ 2
1
∫ 2−x
0
dz dx dy
15. 16. The region in the first
octant is shown.
17. 18.
19. 20.
521
9.15 Triple Integrals
21. Solving x = y2 and 4− x = y2, we obtain x = 2, y = ±√2 . Using symmetry,
V = 2
∫ 3
0
∫ √2
0
∫ 4−y2
y2
dx dy dz = 2
∫ 3
0
∫ √2
0
(4− 2y2)dy dz
= 2
∫ 3
0
(
4y − 2
3
y3
) ∣∣∣∣
√
2
0
dz = 2
∫ 3
0
8
√
2
3
dz = 16
√
2 .
22. V =
∫ 2
0
∫ √4−x2
0
∫ x+y
0
dz dy dx =
∫ 2
0
∫ √4−x2
0
z
∣∣∣x+y
0
dy dx
=
∫ 2
0
∫ √4−x2
0
(x + y) dy dx =
∫ 2
0
(
xy +
1
2
y2
) ∣∣∣∣
√
4−x2
0
dx
=
∫ 2
0
[
x
√
4− x2 + 1
2
(4− x2)
]
dx =
[
−1
3
(4− x2)3/2 + 2x− 1
6
x3
] ∣∣∣∣2
0
=
(
4− 4
3
)
−
(
−8
3
)
=
16
3
23. Adding the two equations, we obtain 2y = 8. Thus, the paraboloids intersect
in the plane y = 4. Their intersection is a circle of radius 2. Using symmetry,
V = 4
∫ 2
0
∫ √4−x2
0
∫ 8−x2−z2
x2+z2
dy dz dx = 4
∫ 2
0
∫ √4−x2
0
(8− 2x2 − 2z2) dz dx
= 4
∫ 2
0
[
2(4− x2)z − 2
3
z3
] ∣∣∣∣
√
4−x2
0
dx = 4
∫ 2
0
4
3
(4− x2)3/2dx Trig substitution
=
16
3
[
−x
8
(2x2 − 20)
√
4− x2 + 6 sin−1 x
2
] ∣∣∣∣2
0
= 16π.
24. Solving x = 2, y = x, and z = x2 + y2, we obtain the point (2, 2, 8).
V =
∫ 2
0
∫ x
0
∫ x2+y2
0
dz dy dx =
∫ 2
0
∫ x
0
(x2 + y2) dy dx =
∫ 2
0
(
x2y +
1
3
y3
) ∣∣∣∣x
0
dx
=
∫ 2
0
4
3
x3 dx =
1
3
x4
∣∣∣∣2
0
=
16
3
.
25. We are given ρ(x, y, z) = kz.
m =
∫ 8
0
∫ 4
0
∫ y1/3
0
kz dx dz dy = k
∫ 8
0
∫ 4
0
xz
∣∣∣y1/3
0
dz dy = k
∫ 8
0
∫ 4
0
y1/3z dz dy
= k
∫ 8
0
1
2
y1/3z2
∣∣∣∣4
0
dy = 8k
∫ 8
0
y1/3dy = 8k
(
3
4
y4/3
) ∣∣∣∣8
0
= 96k
Mxy =
∫ 8
0
∫ 4
0
∫ y1/3
0
kz2 dx dz dy = k
∫ 8
0
∫ 4
0
xz2
∣∣∣y1/3
0
dz dy = k
∫ 8
0
∫ 4
0
y1/3z2 dz dy
= k
∫ 8
0
1
3
y1/3z3
∣∣∣∣4
0
dy =
64
3
k
∫ 8
0
y1/3dy =
64
3
k
(
3
4
y4/3
) ∣∣∣∣8
0
= 256k
522
9.15 Triple Integrals
Mxz =
∫ 8
0
∫ 4
0
∫ y1/3
0
kyz dx dz dy = k
∫ 8
0
∫ 4
0
xyz
∣∣∣y1/3
0
dz dy = k
∫ 8
0
∫ 4
0
y4/3z dz dy
= k
∫ 8
0
1
2
y4/3z2
∣∣∣∣4
0
dy = 8k
∫ 8
0
y4/3dy = 8k
(
3
7
y7/3
) ∣∣∣∣8
0
=
3072
7
k
Myz =
∫ 8
0
∫ 4
0
∫ y1/3
0
kxz dx dz dy = k
∫ 8
0
∫ 4
0
1
2
x2z
∣∣∣∣y1/3
0
dz dy =
1
2
k
∫ 8
0
∫ 4
0
y2/3z dz dy
=
1
2
k
∫ 8
0
1
2
y2/3z2
∣∣∣∣4
0
dy = 4k
∫ 8
0
y2/3dy = 4k
(
3
5
y5/3
) ∣∣∣∣8
0
=
384
5
k
x¯ = Myz/m =
384k/5
96k
= 4/5; y¯ = Mxz/m =
3072k/7
96k
= 32/7; z¯ = Mxy/m =
256k
96k
= 8/3
The center of mass is (4/5, 32/7, 8/3).
26. We use the form of the integral in Problem 14(b) of this section. Without loss of generality, we take ρ = 1.
m =
∫ 1
0
∫ 2−z
z2
∫ 3
0
dy dx dz =
∫ 1
0
∫ 2−z
z2
3 dx dz = 3
∫ 1
0
(2− z − z2) dz = 3
(
2z − 1
2
z2 − 1
3
z3
) ∣∣∣∣1
0
=
7
2
Mxy =
∫ 1
0
∫ 2−z
z2
∫ 3
0
z dy dx dz =
∫ 1
0
∫ 2−z
z2
yz
∣∣∣3
0
dx dz =
∫ 1
0
∫ 2−z
z2
3z dx dz
= 3
∫ 1
0
xz
∣∣∣2−z
z2
dz = 3
∫ 1
0
(2z − z2 − z3) dz = 3
(
z2 − 1
3
z3 − 1
4
z4
) ∣∣∣∣1
0
=
5
4
Mxz =
∫ 1
0
∫ 2−z
z2
∫ 3
0
y dy dx dz =
∫ 1
0
∫ 2−z
z2
1
2
y2
∣∣∣∣3
0
dx dz =
9
2
∫ 1
0
∫ 2−z
z2
dx dz
=
9
2
∫ 1
0
(2− z − z2) dz = 9
2
(
2z − 1
2
z2 − 1
3
z3
) ∣∣∣∣1
0
=
21
4
Myz =
∫ 1
0
∫ 2−z
z2
∫ 3
0
x dy dx dz =
∫ 1
0
∫ 2−z
z2
xy
∣∣∣3
0
dx dz =
∫ 1
0
∫ 2−x
z2
3x dx dz
= 3
∫ 1
0
1
2
x2
∣∣∣∣2−z
z2
dz =
3
2
∫ 1
0
(4− 4z + z2 − z4) dz = 3
2
(
4z − 2z2 + 1
3
z3 − 1
5
z5
) ∣∣∣∣1
0
=
16
5
x¯ = Myz/m =
16/5
7/2
= 32/35, y¯ = Mxz/m =
21/4
7/2
= 3/2, z¯ = Mxy/m =
5/4
7/2
= 5/14.
The centroid is (32/35, 3/2, 5/14).
27. The density is ρ(x, y, z) = ky. Since both the region and the density function are symmetric
with respect to the xy-and yz-planes, x¯ = z¯ = 0. Using symmetry,
m = 4
∫ 3
0
∫ 2
0
∫ √4−x2
0
ky dz dx dy = 4k
∫ 3
0
∫ 2
0
yz
∣∣∣√4−x2
0
dx dy = 4k
∫ 3
0
∫ 2
0
y
√
4− x2 dx dy
= 4k
∫ 3
0
y
(x
2
√
4− x2 + 2 sin−1 x
2
) ∣∣∣∣2
0
dy = 4k
∫ 3
0
πy dy = 4πk
(
1
2
y2
) ∣∣∣∣3
0
= 18πk
Mxz = 4
∫ 3
0
∫ 2
0
∫ √4−x2
0
ky2 dz dx dy = 4k
∫ 3
0
∫ 2
0
y2z
∣∣∣√4−x2
0
dx dy = 4k
∫ 3
0
∫ 2
0
y2
√
4− x2 dx dy
= 4k
∫ 3
0
y2
(x
2
√
4− x2 + 2 sin−1 x
2
) ∣∣∣∣2
0
dy = 4k
∫ 3
0
πy2 dy = 4πk
(
1
3
y3
) ∣∣∣∣3
0
= 36πk.
y¯ = Mxz/m =
36πk
18πk
= 2. The center of mass is (0, 2, 0).
523
9.15 Triple Integrals
28. The density is ρ(x, y, z) = kz.
m =
∫ 1
0
∫ x
x2
∫ y+2
0
kz dz dy dx = k
∫ 1
0
∫ x
x2
1
2
z2
∣∣∣∣y+2
0
dy dx
=
1
2
k
∫ 1
0
∫ x
x2
(y + 2)2dy dx =
1
2
k
∫ 1
0
1
3
(y + 2)3
∣∣∣∣x
x2
dx
=
1
6
k
∫ 1
0
[(x + 2)3 − (x2 + 2)3] dx = 1
6
k
∫ 1
0
[(x + 2)3 − (x6 + 6x4 + 12x2 + 8)]dx
=
1
6
k
[
1
4
(x + 2)4 − 1
7
x7 − 6
5
x5 − 4x3 − 8x
] ∣∣∣∣1
0
=
407
840
k
Mxy =
∫ 1
0
∫ x
x2
∫ y+2
0
kz2 dz dy dx = k
∫ 1
0
∫ x
x2
1
3
z3
∣∣∣∣y+2
0
dy dx =
1
3
k
∫ 1
0
∫ x
x2
(y + 2)3dy dx
=
1
3
k
∫ 1
0
1
4
(y + 2)4
∣∣∣∣x
x2
dx =
1
12
k
∫ 1
0
[(x + 2)4 − (x2 + 2)4] dx
=
1
12
k
∫ 1
0
[(x + 2)4 − (x8 + 8x6 + 24x4 + 32x2 + 16)] dx
=
1
12
k
[
1
5
(x + 2)5 − 1
9
x9 − 8
7
x7 − 24
5
− 32
3
x3 − 16x
] ∣∣∣∣1
0
=
1493
1890
k
Mxz =
∫ 1
0
∫ x
x2
∫ y+2
0
kyz dz dy dx = k
∫ 1
0
∫ x
x2
1
2
yz2
∣∣∣∣y+2
0
dy dx =
1
2
k
∫ 1
0
∫ x
x2
y(y + 2)2 dy dx
=
1
2
k
∫ 1
0
∫ x
x2
(y3 + 4y2 + 4y) dy dx =
1
2
k
∫ 1
0
(
1
4
y4+
4
3
y3 + 2y2
) ∣∣∣∣x
x2
dx
=
1
2
k
∫ 1
0
(
−1
4
x8 − 4
3
x6 − 74x4 + 4
3
x3 + 2x2
)
dx
=
1
2
k
(
− 1
36
x9 − 4
21
x7 − 7
20
x5 +
1
3
x4 +
2
3
x3
) ∣∣∣∣1
0
=
68
315
k
Myz =
∫ 1
0
∫ x
x2
∫ y+2
0
kxz dz dy dx = k
∫ 1
0
∫ x
x2
1
2
xz2
∣∣∣∣y+2
0
dy dx =
1
2
k
∫ 1
0
∫ x
x2
x(y + 2)2 dy dx
=
1
2
k
∫ 1
0
1
3
x(y + 2)3
∣∣∣∣x
x2
dx =
1
6
k
∫ 1
0
[x(x + 2)3 − x(x2 + 2)3] dx
=
1
6
k
∫ 1
0
[x4 + 6x3 + 12x2 + 8x− x(x2 + 2)3] dx
=
1
6
k
[
1
5
x5 +
3
2
x4 + 4x3 + 4x2 − 1
8
(x2 + 2)4
] ∣∣∣∣1
0
=
21
80
k
x¯ = Myz/m =
21k/80
407k/840
= 441/814, y¯ = Mxz/m =
68k/315
407k/840
= 544/1221,
z¯ = Mxy/m =
1493k/1890
407k/840
= 5972/3663. The center of mass is (441/814, 544/1221, 5972/3663).
524
9.15 Triple Integrals
29. m =
∫ 1
−1
∫ √1−x2
−√1−x2
∫ 8−y
2+2y
(x + y + 4) dz dy dx
30. Both the region and the density function are symmetric with respect to the xz- and
yz-planes. Thus, m = 4
∫ 2
−1
∫ √1+z2
0
∫ √1+z2−y2
0
z2 dx dy dz.
31. We are given ρ(x, y, z) = kz.
Iy =
∫ 8
0
∫ 4
0
∫ y1/3
0
kz(x2 + z2)dx dz dy = k
∫ 8
0
∫ 4
0
(
1
3
x3z + xz3
) ∣∣∣∣y1/3
0
dz dy
= k
∫ 8
0
∫ 4
0
(
1
3
yz + y1/3z3
)
dz dy = k
∫ 8
0
(
1
6
yz2 +
1
4
y1/3z4
) ∣∣∣∣4
0
dy
= k
∫ 8
0
(
8
3
y + 64y1/3
)
dy = k
(
4
3
y2 + 48y4/3
) ∣∣∣∣8
0
=
2560
3
k
From Problem 25, m = 96k. Thus, Rg =
√
Iy/m =
√
2560k/3
96k
=
4
√
5
3
.
32. We are given ρ(x, y, z) = k.
Ix =
∫ 1
0
∫ 2−z
z2
∫ 3
0
k(y2 + z2)dy dx dz = k
∫ 1
0
∫ 2−z
z2
(
1
3
y3 + yz2
) ∣∣∣∣3
0
dx dz = k
∫ 1
0
∫ 2−z
z2
(9 + 3z2) dx dz
= k
∫ 1
0
(9x + 3xz2)
∣∣∣2−z
z2
dz = k
∫ 1
0
(18− 9z − 3z2 − 3z3 − 3z4) dz
= k
(
18z − 9
2
z2 − z3 − 3
4
z4 − 3
5
z5
) ∣∣∣∣1
0
=
223
20
k
m =
∫ 1
0
∫ 2−z
z2
∫ 3
0
k dy dx dz = k
∫ 1
0
∫ 2−z
z2
3 dx dz = 3k
∫ 1
0
(2− z − z2) dz = 3k
(
2z − 1
2
z2 − 1
3
z3
) ∣∣∣∣1
0
=
7
2
k
Rg =
√
Ix
m
=
√
223k/20
7k/2
=
√
223
70
33. Iz = k
∫ 1
0
∫ 1−x
0
∫ 1−x−y
0
(x2 + y2) dz dy dx = k
∫ 1
0
∫ 1−x
0
(x2 + y2)(1− x− y) dy dx
= k
∫ 1
0
∫ 1−x
0
(x2 − x3 − x2y + y2 − xy2 − y3) dy dx
= k
∫ 1
0
[
(x2 − x3)y − 1
2
x2y2 +
1
3
(1− x)y3 − 1
4
y4
] ∣∣∣∣1−x
0
dx
= k
∫ 1
0
[
1
2
x2 − x3 + 1
2
x4 +
1
12
(1− x)4
]
dx = k
[
1
6
x6 − 1
4
x4 +
1
10
x5 − 1
60
(1− x)5
] ∣∣∣∣1
0
=
k
30
525
9.15 Triple Integrals
34. We are given ρ(x, y, z) = kx.
Iy =
∫ 1
0
∫ 2
0
∫ 4−z
z
kx(x2 + z2) dy dx dz = k
∫ 1
0
∫ 2
0
(x3 + xz2)y
∣∣∣4−z
z
dx dz
= k
∫ 1
0
∫ 2
0
(x3 + xz2)(4− 2z) dx dz = k
∫ 1
0
(
1
4
x4 +
1
2
x2z2
)
(4− 2z)
∣∣∣∣2
0
dz
= k
∫ 1
0
(4 + 2z2)(4− 2z) dz = 4k
∫ 1
0
(4− 2z + 2z2 − z3) dz = 4k
(
4z − z2 + 2
3
z3 − 1
4
z4
) ∣∣∣∣1
0
=
41
3
k
35. x = 10 cos 3π/4 = −5√2 ; y = 10 sin 3π/4 = 5√2 ; (−5√2 , 5√2 , 5)
36. x = 2 cos 5π/6 = −√3 ; y = 2 sin 5π/6 = 1; (−√3 , 1,−3)
37. x =
√
3 cosπ/3 =
√
3/2; y =
√
3 sinπ/3 = 3/2; (
√
3/2, 3/2,−4)
38. x = 4 cos 7π/4 = 2
√
2 ; y = 4 sin 7π/4 = −2√2 ; (2√2 ,−2√2 , 0)
39. With x = 1 and y = −1 we have r2 = 2 and tan θ = −1. The point is (√2 ,−π/4,−9).
40. With x = 2
√
3 and y = 2 we have r2 = 16 and tan θ = 1/
√
3 . The point is (4, π/6, 17).
41. With x = −√2 and y = √6 we have r2 = 8 and tan θ = −√3 . The point is (2√2 , 2π/3, 2).
42. With x = 1 and y = 2 we have r2 = 5 and tan θ = 2. The point is (
√
5 , tan−1 2, 7).
43. r2 + z2 = 25 44. r cos θ + r sin θ − z = 1
45. r2 − z2 = 1 46. r2 cos2 θ + z2 = 16
47. z = x2 + y2 48. z = 2y
49. r cos θ = 5, x = 5 50. tan θ = 1/
√
3 , y/x = 1/
√
3 , x =
√
3 y, x > 0
51. The equations are r2 = 4, r2 + z2 = 16, and z = 0.
V =
∫ 2π
0
∫ 2
0
∫ √16−r2
0
r dz dr dθ =
∫ 2π
0
∫ 2
0
r
√
16− r2 dr dθ
=
∫ 2π
0
−1
3
(16− r2)3/2
∣∣∣∣2
0
dθ =
∫ 2π
0
1
3
(64− 24
√
3 ) dθ =
2π
3
(64− 24
√
3 )
52. The equation is z = 10− r2.
V =
∫ 2π
0
∫ 3
0
∫ 10−r2
1
r dz dr dθ =
∫ 2π
0
∫ 3
0
r(9− r2) dr dθ =
∫ 2π
0
(
9
2
r2 − 1
4
r4
) ∣∣∣∣3
0
dθ
=
∫ 2π
0
81
4
dθ =
81π
2
.
53. The equations are z = r2, r = 5, and z = 0.
V =
∫ 2π
0
∫ 5
0
∫ r2
0
r dz dr dθ =
∫ 2π
0
∫ 5
0
r3 dr dθ =
∫ 2π
0
1
4
r4
∣∣∣∣5
0
dθ
=
∫ 2π
0
625
4
dθ =
625π
2
526
9.15 Triple Integrals
54. Substituting the first equation into the second, we see that the surfaces intersect
in the plane y = 4. Using polar coordinates in the xz-plane, the equations of the
surfaces become y = r2 and y = 12r
2 + 2.
V =
∫ 2π
0
∫ 2
0
∫ r2/2+2
r2
r dy dr dθ =
∫ 2π
0
∫ 2
0
r
(
r2
2
+ 2− r2
)
dr dθ
=
∫ 2π
0
∫ 2
0
(
2r − 1
2
r3
)
dr dθ =
∫ 2π
0
(
r2 − 1
8
r4
) ∣∣∣∣2
0
dθ =
∫ 2π
0
2 dθ = 4π
55. The equation is z =
√
a2 − r2 . By symmetry, x¯ = y¯ = 0.
m =
∫ 2π
0
∫ a
0
∫ √a2−r2
0
r dz dr θ =
∫ 2π
0
∫ a
0
r
√
a2 − r2 dr dθ
=
∫ 2π
0
−1
3
(a2 − r2)3/2
∣∣∣∣a
0
dθ =
∫ 2π
0
1
3
a3 dθ =
2
3
πa3
Mxy =
∫ 2π
0
∫ a
0
∫ √a2−r2
0
zr dz dr dθ =
∫ 2π
0
∫ a
0
1
2
rz2
∣∣∣∣
√
a2−r2
0
dr dθ =
1
2
∫ 2π
0
∫ a
0
r(a2 − r2) dr dθ
=
1
2
∫ 2π
0
(
1
2
a2r2 − 1
4
r4
) ∣∣∣∣a
0
dθ =
1
2
∫ 2π
0
1
4
a4 dθ =
1
4
πa4
z¯ = Mxy/m =
πa4/4
2πa3/3
= 3a/8. The centroid is (0, 0, 3a/8).
56. We use polar coordinates in the yz-plane. The density is ρ(x, y, z) = kz. By symmetry,
y¯ = z¯ = 0.
m =
∫ 2π
0
∫ 4
0
∫ 5
0
kxr dx dr dθ = k
∫ 2π
0
∫ 4
0
1
2
rz2
∣∣∣∣5
0
dr dθ =
k
2
∫ 2π
0
∫ 4
0
25r dr dθ
=
25k
2
∫ 2π
0
1
2
r2
∣∣∣∣4
0
dθ =
25k
2
∫ 2π
0
8 dθ = 200kπ
Myz =
∫ 2π
0
∫ 4
0
∫ 5
0
kx2r dx dr dθ = k
∫ 2π
0
∫ 4
0
1
3
rx3
∣∣∣∣5
0
dr dθ =
1
3
k
∫ 2π
0
∫ 4
0
125r dr dθ
=
1
3
k
∫ 2π
0
125
2
r2
∣∣∣∣4
0
dθ =
1
3
k
∫ 2π
0
1000 dθ =
2000
3
kπ
x¯ = Myz/m =
2000kπ/3
200kπ
= 10/3. The center of mass of the given solid is (10/3, 0, 0).
57. The equation is z =
√
9− r2 and the density is ρ = k/r2. When z = 2, r = √5 .
Iz =
∫ 2π
0
∫ √5
0
∫ √9−r2
2
r2(k/r2)r dz dr dθ = k
∫ 2π
0
∫ √5
0
rz
∣∣∣√9−r2
2
dr dθ
= k
∫ 2π
0
∫ √5
0
(r
√
9− r2 − 2r) dr dθ = k
∫ 2π
0
[
−1
3
(9− r2)3/2 − r2
] ∣∣∣∣
√
5
0
dθ
= k
∫ 2π
0
4
3
dθ =
8
3
πk
527
9.15 Triple Integrals
58. The equation is z = r and the density is ρ = kr.
Ix =
∫ 2π
0
∫ 1
0
∫ 1
r
(y2 + z2)(kr)r dz dr dθ = k
∫ 2π
0
∫ 1
0
∫ 1
r
(r4 sin2 θ + r2z2) dz dr dθ
= k
∫ 2π
0
∫ 1
0
[
(r4 sin2 θ)z +
1
3
r2z3
] ∣∣∣∣1
r
dr dθ
= k
∫ 2π
0
∫ 1
0
(
r4 sin2 θ +
1
3
r2 − r5 sin2 θ − 1
3
r5
)
dr dθ
= k
∫ 2π
0
(
1
5
r5 sin2 θ +
1
9
r3 − 1
6
r6 sin2 θ − 1
18
r6
) ∣∣∣∣1
0
dθ = k
∫ 2π
0
(
1
30
sin2 θ +
1
18
)
dθ
= k
(
1
60
θ − 1
120
sin 2θ +
1
18
θ
) ∣∣∣∣2π
0
=
13
90
πk
59. (a) x = (2/3) sin(π/2) cos(π/6) =
√
3/3; y = (2/3) sin(π/2) sin(π/6) = 1/3;
z = (2/3) cos(π/2) = 0; (
√
3/3, 1/3, 0)
(b) Withx =
√
3/3 and y = 1/3 we have r2 = 4/9 and tan θ =
√
3/3. The point is (2/3, π/6, 0).
60. (a) x = 5 sin(5π/4) cos(2π/3) = 5
√
2/4; y = 5 sin(5π/4) sin(2π/3) = −5√6/4;
z = 5 cos(5π/4) = −5√2/2; (5√2/4,−5√6/4,−5√2/2)
(b) With x = 5
√
2/4 and y = −5√6/4 we have r2 = 25/2 and tan θ = −√3 .
The point is (5/
√
2 , 2π/3,−5√2/2).
61. (a) x = 8 sin(π/4) cos(3π/4) = −4; y = 8 sin(π/4) sin(3π/4) = 4; z = 8 cos(π/4) = 4√2 ;
(−4, 4, 4√2)
(b) With x = −4 and y = 4 we have r2 = 32 and tan θ = −1. The point is (4√2 , 3π/4, 4√2 ).
62. (a) x = (1/3) sin(5π/3) cos(π/6) = −1/4; y = (1/3) sin(5π/3) sin(π/6) = −√3/12;
z = (1/3) cos(5π/3) = 1/6; (−1/4,−√3/12, 1/6)
(b) With x = −1/4 and y = −√3/12 we have r2 = 1/12 and tan θ = √3/3. The point is (1/2√3 , π/6, 1/6).
63. With x = −5, y = −5, and z = 0, we have ρ2 = 50, tan θ = 1, and cosφ = 0. The point is (5√2 , π/2, 5π/4).
64. With x = 1, y = −√3 , and z = 1, we have ρ2 = 5, tan θ = −√3 , and cosφ = 1/√5 . The point is
(
√
5 , cos−1 1/
√
5 ,−π/3).
65. With x =
√
3/2, y = 1/2, and z = 1, we have ρ2 = 2, tan θ = 1/
√
3 , and cosφ = 1/
√
2 . The point is
(
√
2 , π/4, π/6).
66. With x = −√3/2, y = 0, and z = −1/2, we have ρ2 = 1, tan θ = 0, and cosφ = −1/2. The point is (1, 2π/3, 0).
67. ρ = 8
68. ρ2 = 4ρ cosφ; ρ = 4 cosφ
69. 4z2 = 3x2 + 3y2 + 3z2; 4ρ2 cos2 φ = 3ρ2; cosφ = ±√3/2; φ = π/6 or equivalently, φ = 5π/6
70. −x2 − y2 − z2 = 1− 2z2; −ρ2 = 1− 2ρ2 cos2 φ; ρ2(2 cos2 φ− 1) = 1
71. x2 + y2 + z2 = 100
72. cosφ = 1/2; ρ2 cos2 φ = ρ2/4; 4z2 = x2 + y2 + z2; x2 + y2 = 3z2
73. ρ cosφ = 2; z = 2
74. ρ(1− cos2 φ) = cosφ; ρ2 − ρ2 cos2 φ = ρ cosφ; x2 + y2 + z2 − z2 = z; z = x2 + y2
528
9.15 Triple Integrals
75. The equations are φ = π/4 and ρ = 3.
V =
∫ 2π
0
∫ π/4
0
∫ 3
0
ρ2 sinφdρ dφ dθ =
∫ 2π
0
∫ π/4
0
1
3
ρ3 sinφ
∣∣∣∣3
0
dφ dθ =
∫ 2π
0
∫ π/4
0
9 sinφdφ dθ
=
∫ 2π
0
−9 cosφ
∣∣∣π/4
0
dθ = −9
∫ 2π
0
(√
2
2
− 1
)
dθ = 9π(2−
√
2 )
76. The equations are ρ = 2, θ = π/4, and θ = π/3.∫ π/3
π/4
∫ π/2
0
∫ 2
0
ρ2 sinφdρ dφ dθ =
∫ π/3
π/4
∫ π/2
0
1
3
ρ3 sinφ
∣∣∣∣2
0
dφ dθ
=
∫ π/3
π/4
∫ π/2
0
8
3
sinφdφ dθ =
8
3
∫ π/3
π/4
− cosφ
∣∣∣π/2
0
dθ
=
8
3
∫ π/3
π/4
(0 + 1) dθ =
2π
9
77. From Problem 69, we have φ = π/6. Since the figure is in the first octant and z = 2 we also
have θ = 0, θ = π/2, and ρ cosφ = 2.
V =
∫ π/2
0
∫ π/6
0
∫ 2 secφ
0
ρ2 sinφdρ dφ dθ =
∫ π/2
0
∫ π/6
0
1
3
ρ3 sinφ
∣∣∣∣2 secφ
0
dφ dθ
=
8
3
∫ π/2
0
∫ π/6
0
sec3 φ sinφdφ dθ =
8
3
∫ π/2
0
∫ π/6
0
sec2 φ tanφdφ dθ
=
8
3
∫ π/2
0
1
2
tan2 φ
∣∣∣∣π/6
0
dθ =
4
3
∫ π/2
0
1
3
dθ =
2
9
π
78. The equations are ρ = 1 and φ = π/4. We find the volume above the xy-plane and double.
V = 2
∫ 2π
0
∫ π/2
π/4
∫ 1
0
ρ2 sinφdρ dφ dθ = 2
∫ 2π
0
∫ π/2
π/4
1
3
ρ3 sinφ
∣∣∣∣1
0
dφ dθ
=
2
3
∫ 2π
0
∫ π/2
π/4
sinφdφ dθ =
2
3
∫ 2π
0
− cosφ
∣∣∣π/2
π/4
dθ =
2
3
∫ 2π
0
√
2
2
dθ =
2π
√
2
3
79. By symmetry, x¯ = y¯ = 0. The equations are φ = π/4 and ρ = 2 cosφ.
m =
∫ 2π
0
∫ π/4
0
∫ 2 cosφ
0
ρ2 sinφdρ dφ dθ =
∫ 2π
0
∫ π/4
0
1
3
ρ3 sinφ
∣∣∣∣2 cosφ
0
dφ dθ
=
8
3
∫ 2π
0
∫ π/4
0
sinφ cos3 φdφ dθ =
8
3
∫ 2π
0
−1
4
cos4 φ
∣∣∣∣π/4
0
dθ
= −2
3
∫ 2π
0
(
1
4
− 1
)
dθ = π
Mxy =
∫ 2π
0
∫ π/4
0
∫ 2 cosφ
0
zρ2 sinφdρ dφ dθ =
∫ 2π
0
∫ π/4
0
∫ 2 cosφ
0
ρ3 sinφ cosφdρ dφ dθ
=
∫ 2π
0
∫ π/4
0
1
4
ρ4 sinφ cosφ
∣∣∣∣2 cosφ
0
dφ dθ = 4
∫ 2π
0
∫ π/4
0
cos5 φ sinφdφ dθ
= 4
∫ 2π
0
−1
6
cos6 φ
∣∣∣∣π/4
0
dθ = −2
3
∫ 2π
0
(
1
8
− 1
)
dθ =
7
6
π
z¯ = Mxy/m =
7π/6
π
= 7/6. The centroid is (0, 0, 7/6).
529
9.15 Triple Integrals
80. We are given density = kz. By symmetry, x¯ = y¯ = 0. The equation is ρ = 1.
m =
∫ 2π
0
∫ π/2
0
∫ 1
0
kzρ2 sinφdρ dφ dθ = k
∫ 2π
0
∫ π/2
0
∫ 1
0
ρ3 sinφ cosφdρ dφ dθ
= k
∫ 2π
0
∫ π/2
0
1
4
ρ4 sinφ cosφ
∣∣∣∣1
0
dφ dθ =
1
4
k
∫ 2π
0
∫ π/2
0
sinφ cosφdφ dθ
=
1
4
k
∫ 2π
0
1
2
sin2 φ
∣∣∣∣π/2
0
dθ =
1
8
k
∫ 2π
0
dθ =
kπ
4
Mxy =
∫ 2π
0
∫ π/2
0
∫ 1
0
kz2ρ2 sinφdρ dφ dθ = k
∫ 2π
0
∫ π/2
0
∫ 1
0
ρ4 cos2 φ sinφdρ dφ dθ
= k
∫ 2π
0
∫ π/2
0
1
5
ρ5 cos2 φ sinφ
∣∣∣∣1
0
dφ dθ =
1
5
k
∫ 2π
0
∫ π/2
0
cos2 φ sinφdφ dθ
=
1
5
k
∫ 2π
0
−1
3
cos3 φ
∣∣∣∣π/2
0
dθ = − 1
15
k
∫ 2π
0
(0− 1) dθ = 2
15
kπ
z¯ = Mxy/m =
2kπ/15
kπ/4
= 8/15. The center of mass is (0, 0, 8/15).
81. We are given density = k/ρ.
m =
∫ 2π
0
∫ cos−1 4/5
0
∫ 5
4 secφ
k
ρ
ρ2 sinφdρ dφ dθ = k
∫ 2π
0
∫ cos−1 4/5
0
1
2
ρ2 sinφ
∣∣∣∣5
4 secφ
dφ dθ
=
1
2
k
∫ 2π
0
∫ cos−1 4/5
0
(25 sinφ− 16 tanφ secφ) dφ dθ
=
1
2
k
∫ 2π
0
(−25 cosφ− 16 secφ)
∣∣∣cos−1 4/5
0
dθ =
1
2
k
∫ 2π
0
[−25(4/5)− 16(5/4)− (−25− 16)] dθ
=
1
2
k
∫ 2π
0
dθ = kπ
82. We are given density = kρ.
Iz =
∫ 2π
0
∫ π
0
∫ a
0
(x2 + y2)(kρ)ρ2 sinφdρ dφ dθ
= k
∫ 2π
0
∫ π
0
∫ a
0
(ρ2 sin2 φ cos2 θ + ρ2 sin2 φ sin2 θ)ρ3 sinφdρ dφ dθ
= k
∫ 2π
0
∫ π
0
∫ a
0
ρ5 sin3 φdρ dφ dθ = k
∫ 2π
0
∫ π
0
1
6
ρ6 sin3 φ
∣∣∣∣a
0
dφ dθ =
1
6
ka6
∫ 2π
0
∫ π
0
sin3 φdφ dθ
=
1
6
ka3
∫ 2π
0
∫ π
0
(1− cos2 φ) sinφdφ dθ = 1
6
ka3
∫ 2π
0
(
− cosφ + 1
3
cos3 φ
) ∣∣∣∣π
0
dθ =
1
6
ka3
∫ 2π
0
4
3
dθ =
4π
9
ka6
530
9.16 Divergence Theorem
EXERCISES 9.16
Divergence Theorem
1. div F = y + z + x
The Triple Integral:∫∫∫
D
div F dV =
∫ 1
0
∫ 1
0
∫ 1
0
(x + y + z) dx dy dz
=
∫ 1
0
∫ 1
0
(
1
2
x2 + xy + xz
) ∣∣∣∣1
0
dy dz
=
∫ 1
0
∫ 1
0
(
1
2
+ y + z
)
dy dz =
∫ 1
0
(
1
2
y +
1
2
y2 + yz
) ∣∣∣∣1
0
dz
=
∫ 1
0
(1 + z) dz =
1
2
(1 + z2)
∣∣∣1
0
= 2− 1
2
=
3
2
The Surface Integral: Let the surfaces be S1 in z = 0, S2 in z = 1, S3 in y = 0, S4 in y = 1, S5 in x = 0,
and S6 in x = 1. The unit outward normal vectors are −k, k, −j, j, −i and i, respectively. Then∫∫
S
F · n dS =
∫∫
S1
F · (−k) dS1 +
∫∫
S2
F · k dS2 +
∫∫
S3
F · (−j) dS3 +
∫∫
S4
F · j dS4
+
∫∫
S5
F · (−i) dS5 +
∫∫
S6
F · i dS6
=
∫∫
S1
(−xz) dS1 +
∫∫
S2
xz dS2 +
∫∫
S3
(−yz) dS3 +
∫∫
S4
yz dS4
+
∫∫
S5
(−xy) dS5 +
∫∫
S6
xy dS6
=
∫∫
S2
x dS2 +
∫∫
S4
z dS4 +
∫∫
S6
y dS6
=
∫ 1
0
∫ 1
0
x dx dy +
∫ 1
0
∫ 1
0
z dz dx +
∫ 1
0
∫ 1
0
y dy dz
=
∫ 1
0
1
2
dy +
∫ 1
0
1
2
dx +
∫ 1
0
1
2
dz =
3
2
.
2. div F = 6y + 4z
The Triple Integral:∫∫∫
D
div F dV =
∫ 1
0
∫ 1−x
0
∫ 1−x−y
0
(6y + 4z) dz dy dx
=
∫ 1
0
∫ 1−x
0
(6yz + 2z2)
∣∣∣1−x−y
0
dy dx
=
∫ 1
0
∫ 1−x
0
(−4y2 + 2y − 2xy + 2x2 − 4x + 2) dy dx
531
9.16 Divergence Theorem
=
∫ 1
0
(
−4
3
y3 + y2 − xy2 + 2x2y − 4xy + 2y
) ∣∣∣∣1−x
0
dx
=
∫ 1
0
(
−5
3
x3 + 5x2 − 5x + 5
3
)
dx =
(
− 5
12
x4 +
5
3
x3 − 5
2
x2 +
5
3
x
) ∣∣∣∣1
0
=
5
12
The Surface Integral: Let the surfaces be S1 in the plane x + y + z = 1, S2 in z = 0, S3 in x = 0, and S4 in
y = 0. The unit outward normal vectors are n1 = (i+ j+k)/
√
3 , n2 = −k, n3 = −i, and n4 = −j, respectively.
Now on S1, dS1 =
√
3 dA1, on S3, x = 0, and on S4, y = 0, so∫∫
SF · n dS =
∫∫
S1
F · n1 dS1 +
∫∫
S2
F · (−k) dS2 +
∫∫
S3
F · (−j) dS3 +
∫∫
S4
F · (−i) dS4
=
∫ 1
0
∫ 1−x
0
(6xy + 4y(1− x− y) + xe−y) dy dx +
∫ 1
0
∫ 1−x
0
(−xe−y) dy dx
+
∫∫
S3
(−6xy) dS3 +
∫∫
S4
(−4yz) dS4
=
∫ 1
0
(
xy2 + 2y2 − 4
3
y3 − xe−y
) ∣∣∣∣1−x
0
dx +
∫ 1
0
xe−y
∣∣∣1−x
0
dx + 0 + 0
=
∫ 1
0
[x(1− x)2 + 2(1− x)2 − 4
3
(1− x)3 − xex−1 + x] dx +
∫ 1
0
(xex−1 − x) dx
=
[
1
2
x2 − 2
3
x3 +
1
4
x4 − 2
3
(1− x)3 + 1
3
(1− x)4
] ∣∣∣∣1
0
=
5
12
.
3. div F = 3x2 + 3y2 + 3z2. Using spherical coordinates,∫∫
S
F · n dS =
∫∫∫
D
3(x2 + y2 + z2) dV =
∫ 2π
0
∫ π
0
∫ a
0
3ρ2ρ2 sinφdρ dφ dθ
=
∫ 2π
0
∫ π
0
3
5
ρ5 sinφ
∣∣∣∣a
0
dφ dθ =
3a5
5
∫ 2π
0
∫ π
0
sinφdφ dθ
=
3a5
5
∫ 2π
0
− cosφ
∣∣∣π
0
dθ =
6a5
5
∫ 2π
0
dθ =
12πa5
5
.
4. div F = 4 + 1 + 4 = 9. Using the formula for the volume of a sphere,∫∫
S
F · n dS =
∫∫∫
D
9 dV = 9
(
4
3
π23
)
= 96π.
5. div F = 2(z − 1). Using cylindrical coordinates,∫∫
S
F · n dS =
∫∫∫
D
2(z − 1)V =
∫ 2π
0
∫ 4
0
∫ 5
1
2(z − 1) dz r dr dθ =
∫ 2π
0
∫ 4
0
(z − 1)2
∣∣∣5
1
r dr dθ
=
∫ 2π
0
∫ 4
0
16r dr dθ =
∫ 2π
0
8r2
∣∣∣4
0
dθ = 128
∫ 2π
0
dθ = 256π.
6. div F = 2x + 2z + 12z2.∫∫
S
F · n dS =
∫∫∫
D
div F dV =
∫ 3
0
∫ 2
0
∫ 1
0
(2x + 2z + 12z2) dx dy dz
=
∫ 3
0
∫ 2
0
(x2 + 2xz + 12xz2)
∣∣∣1
0
dy dz =
∫ 3
0
∫ 2
0
(1 + 2z + 12z2) dy dz
=
∫ 3
0
2(1 + 2z + 12z2) dz = (2z + 2z2 + 8z3)
∣∣∣3
0
= 240
532
9.16 Divergence Theorem
7. div F = 3z2. Using cylindrical coordinates,∫∫
S
F · n dS =
∫∫∫
D
div F dV =
∫ 2π
0
∫ √3
0
∫ √4−r2
0
3z2r dz dr dθ
=
∫ 2π
0
∫ √3
0
rz3
∣∣∣√4−r2
0
dr dθ =
∫ 2π
0
∫ √3
0
r(4− r2)3/2 dr dθ
=
∫ 2π
0
−1
5
(4− r2)5/2
∣∣∣∣
√
3
0
dθ =
∫ 2π
0
−1
5
(1− 32) dθ =
∫ 2π
0
31
5
dθ =
62π
5
.
8. div F = 2x.∫∫
S
F · n dS =
∫∫∫
D
div F dV =
∫ 3
0
∫ 9
x2
∫ 9−y
0
2x dz dy dx
=
∫ 3
0
∫ 9
x2
2x(9− y) dy dx =
∫ 3
0
−x(9− y)2
∣∣∣9
x2
dx =
∫ 3
0
x(9− x)2 dx
=
∫ 3
0
(x3 − 18x2 + 81x) dx =
(
1
4
x4 − 6x3 + 81
2
x2
) ∣∣∣∣3
0
=
891
4
9. div F =
1
x2 + y2 + z2
. Using spherical coordinates,
∫∫
S
F · n dS =
∫∫∫
D
div F dV =
∫ 2π
0
∫ π
0
∫ b
a
1
ρ2
ρ2 sinφdρ dφ dθ
=
∫ 2π
0
∫ π
0
(b− a) sinφdφ dθ = (b− a)
∫ 2π
0
− cosφ
∣∣∣π
0
dθ
= (b− a)
∫ 2π
0
2 dθ = 4π(b− a).
10. Since div F = 0,
∫∫
S
F · n dS =
∫∫∫
D
0 dV = 0.
11. div F = 2z + 10y − 2z = 10y.∫∫
S
F·n dS =
∫∫∫
D
10y dV =
∫ 2
0
∫ 2−x2/2
0
∫ 4−z
z
10y dy dz dx =
∫ 2
0
∫ 2−x2/2
0
5y2
∣∣∣4−z
z
dz dx
=
∫ 2
0
∫ 2−x2/2
0
(80− 40z) dz dx =
∫ 2
0
(80z − 20z2)
∣∣∣2−x2/2
0
dx =
∫ 2
0
(80− 5x4) dx = (80x− x5)
∣∣∣2
0
= 128
12. div F = 30xy.∫∫
S
F · n dS =
∫∫∫
D
30xy dV =
∫ 2
0
∫ 2−x
0
∫ 3
x+y
30xy dz dy dx
=
∫ 2
0
∫ 2−x
0
30xyz
∣∣∣3
x+y
dy dx
=
∫ 2
0
∫ 2−x
0
(90xy − 30x2y − 30xy2) dy dx
=
∫ 2
0
(45xy2 − 15x2y2 − 10xy3)
∣∣∣2−x
0
dx
=
∫ 2
0
(−5x4 + 45x3 − 120x2 + 100x) dx =
(
−x5 + 45
4
x4 − 40x3 + 50x2
) ∣∣∣∣2
0
= 28
533
9.16 Divergence Theorem
13. div F = 6xy2 + 1− 6xy2 = 1. Using cylindrical coordinates,∫∫
S
F · n dS =
∫∫∫
D
dV =
∫ π
0
∫ 2 sin θ
0
∫ 2r sin θ
r2
dz r dr dθ =
∫ π
0
∫ 2 sin θ
0
(2r sin θ − r2)r dr dθ
=
∫ π
0
(
2
3
r3 sin θ − 1
4
r4
) ∣∣∣∣2 sin θ
0
dθ =
∫ π
0
(
16
3
sin4 θ − 4 sin4 θ
)
dθ
=
4
3
∫ π
0
sin4 θ dθ =
4
3
(
3
8
θ − 1
4
sin 2θ +
1
32
sin 4θ
) ∣∣∣∣π
0
=
π
2
14. div F = y2 + x2. Using spherical coordinates, we have x2 + y2 = ρ2 sin2 φ and z = ρ cosφ or ρ = z secφ. Then∫∫
S
F · n dS =
∫∫∫
D
(x2 + y2) dS =
∫ 2π
0
∫ π/4
0
∫ 4 secφ
2 secφ
ρ2 sin2 φρ2 sinφdρ dφ dθ
=
∫ 2π
0
∫ π/4
0
1
5
ρ5 sin3 φ
∣∣∣∣4 secφ
2 secφ
dφ dθ =
∫ 2π
0
∫ π/4
0
992
5
sec5 φ sin3 φdφ dθ
=
∫ 2π
0
∫ π/4
0
992
5
tan3 φ sec2 φdφ dθ =
992
5
∫ 2π
0
1
4
tan4 φ
∣∣∣∣π/4
0
dθ =
992
5
∫ 2π
0
1
4
dθ =
496π
5
.
15. (a) div E = q
[ −2x2 + y2 + z2
(x2 + y2 + z2)5/2
+
x2 − 2y2 + z2
(x2 + y2 + z2)5/2
+
x2 + y2 − 2z2
(x2 + y2 + z2)5/2
]
= 0∫∫
S∪Sa
(E · n) dS =
∫∫∫
D
div E dV =
∫∫∫
D
0 dV = 0
(b) From (a),
∫∫
S
(E · n) dS +
∫∫
Sa
(E · n) dS = 0 and
∫∫
S
(E · n) dS = −
∫∫
Sa
(E · n) dS. On Sa,
|r| = a, n = −(xi + yj + zk)/a = −r/a and E · n = (qr/a3) · (−r/a) = −qa2/a4 = −q/a2. Thus∫∫
S
(E · n) dS = −
∫∫
Sa
(
− q
a2
)
dS =
q
a2
∫∫
Sa
dS =
q
a2
× (area of Sa) = q
a2
(4πa2) = 4πq.
16. (a) By Gauss’ Law
∫∫
(E · n) dS =
∫∫∫
D
4πρ dV , and by the Divergence Theorem∫∫
S
(E · n) dS =
∫∫∫
D
div E dV . Thus
∫∫∫
D
4πρ dV =
∫∫∫
D
div E dV and
∫∫∫
D
(4πρ− div E) dV = 0.
Since this holds for all regions D, 4πρ− div E = 0 and div E = 4πρ.
(b) Since E is irrotational, E = ∇φ and ∇2φ = ∇ · ∇φ = ∇E = div E = 4πρ.
17. Since div a = 0, by the Divergence Theorem∫∫
S
(a · n) dS =
∫∫∫
D
div a dV =
∫∫∫
D
0 dV = 0.
18. By the Divergence Theorem and Problem 30 in Section 9.7,∫∫
S
(curl F · n) dS =
∫∫∫
D
div (curl F) dV =
∫∫∫
D
0 dV = 0.
19. By the Divergence Theorem and Problem 27 in Section 9.7,∫∫
S
(f∇g) · n dS =
∫∫∫
D
div (f∇g) dV =
∫∫∫
D
∇ · (f∇g) dV =
∫∫∫
D
[f(∇ · ∇g) +∇g · ∇f ] dV
=
∫∫∫
D
(f∇2g +∇g · ∇f) dV.
534
9.17 Change of Variables in Multiple Integrals
20. By the Divergence Theorem and Problems 25 and 27 in Section 9.7,∫∫
S
(f∇g − g∇f) · n dS =
∫∫∫
D
div (f∇g − g∇f) dV =
∫∫∫
D
∇ · (f∇g − g∇f) dV
=
∫∫∫
D
[f(∇ · ∇g) +∇g · ∇f − g(∇ · ∇f)−∇f · ∇g] dV
=
∫∫∫
D
(f∇2g − g∇2f) dV.
21. If G(x, y, z) is a vector valued function then we define surface integrals and triple integrals of G component-wise.
In this case, if a is a constant vector it is easily shown that∫∫
S
a ·G dS = a ·
∫∫
S
G dS and
∫∫∫
D
a ·G dV = a ·
∫∫∫
D
G dV.
Now let F = fa. Then ∫∫
S
F · n dS =
∫∫
S
(fa) · n dS =
∫∫
S
a · (fn) dS
and, using Problem 27 in Section 9.7 and the fact that ∇ · a = 0, we have∫∫∫
D
div F dV =
∫∫∫
D
∇ · (fa) dV =
∫∫∫
D
[f(∇ · a) + a · ∇f ] dV =
∫∫∫
D
a · ∇f dV.
By the Divergence Theorem,∫∫
S
a · (fn) dS =
∫∫
S
F · n dS =
∫∫∫
D
div F dV =
∫∫∫
D
a · ∇f dV
and
a ·
(∫∫
S
fn dS
)
= a ·
(∫∫∫
D
∇f dV
)
or a ·
(∫∫
S
fn dS −
∫∫∫
D
∇f dV
)
= 0.
Since a is arbitrary, ∫∫
S
fn dS −
∫∫∫
D
∇f dV = 0 and
∫∫
S
fn dS =
∫∫∫
D
∇f dV.
22. B + W = −
∫∫
S
pn dS + mg = mg −
∫∫∫
D
∇p dV = mg −
∫∫∫
D
ρg dV = mg −
(∫∫∫
D
ρ dV
)
g
= mg −mg = 0
EXERCISES 9.17
Change of Variables in Multiple Integrals
1. T : (0, 0) → (0, 0); (0, 2) → (−2, 8); (4, 0) → (16, 20); (4, 2) → (14, 28)
2. Writing x2 = v − u and y = v + u and solving for u and v, we obtain u = (y − x2)/2 and v = (x2 + y)/2.
Then the images under T−1 are (1, 1) → (0, 1); (1, 3) → (1, 2); (√2 , 2) → (0, 2).
535
1 2
1
2
v=0
u=2
v=u
S
u
v
3 6
-4
-2
2
y=14-3x
y=x/2
y=-2x/3
x
y
-2 2 4
3
6
v=1
u=4
v=5
u=-1
S
u
v
-2 3
3
6
x+2y=1
x+2y=5
x-y=-1x-y=4x
y
-4 -2 2
2
v=0
u=1
v=2
u=0 S
u
v
-4 -2 2
2
y=0
x=1-y
x=y /4-42
2
x
y
2 4
2
4
v=1
u=2
v=2
u=1 S
u
v
2 4
2
4
y=1
y=4
y=x /4
y=x2
2
x
y
9.17 Change of Variables in Multiple Integrals
3. The uv-corner points (0, 0), (2, 0), (2, 2) corre-
spond to xy-points (0, 0), (4, 2), (6,−4).
v = 0: x = 2u, y = u =⇒ y = x/2
u = 2: x = 4 + v, y = 2− 3v =⇒
y = 2− 3(x− 4) = −3x + 14
v = u: x = 3u, y = −2u =⇒ y = −2x/3
4. Solving for x and y we see that the transformation
is x = 2u/3+v/3, y = −u/3+v/3. The uv-corner
points (−1, 1), (4, 1), (4, 5), (−1, 5) correspond
to the xy-points (−1/3, 2/3), (3,−1), (13/3, 1/3),
(1, 2).
v = 1: x + 2y = 1; v = 5: x + 2y = 5;
u = −1: x− y = −1; u = 4: x− y = 4
5. The uv-corner points (0, 0), (1, 0), (1, 2),
(0, 2) correspond to the xy-points (0, 0),
(1, 0), (−3, 2), (−4, 0).
v = 0: x = u2, y = 0 =⇒ y = 0 and 0 ≤ x ≤ 1
u = 1: x = 1− v2, y = v =⇒ x = 1− y2
v = 2: x = u2 − 4, y = 2u =⇒ x = y2/4− 4
u = 0: x = −v2, y = 0 =⇒ y = 0 and −4 ≤ x ≤ 0
6. The uv-corner points (1, 1), (2, 1), (2, 2), (1, 2)
correspond to the xy-points (1, 1), (2, 1), (4, 4), (2, 4).
v = 1: x = u, y = 1 =⇒ y = 1, 1 ≤ x ≤ 2
u = 2: x = 2v, y = v2 =⇒ y = x2/4
v = 2: x = 2u, y = 4 =⇒ y = 4, 2 ≤ x ≤ 4
u = 1: x = v, y = v2 =⇒ y = x2
7.
∂(x, y)
∂(u, v)
=
∣∣∣∣−ve−u e−uveu eu
∣∣∣∣ = −2v
8.
∂(x, y)
∂(u, v)
=
∣∣∣∣ 3e3u sin v e3u cos v3e3u cos v −e3u sin v
∣∣∣∣ = −3e6u
9.
∂(u, v)
∂(x, y)
=
∣∣∣∣−2y/x3 1/x2−y2/x2 2y/x
∣∣∣∣ = −3y2x4 = −3 ( yx2 )2 = −3u2; ∂(x, y)∂(u, v) = 1−3u2 = − 13u2
10.
∂(u, v)
∂(x, y)
=
∣∣∣∣∣∣∣∣
2(y2 − x2)
(x2 + y2)2
−4xy
(x2 + y2)2
4xy
(x2 + y2)2
2(y2 − x2)
(x2 + y2)2
∣∣∣∣∣∣∣∣ =
4
(x2 + y2)2
536
v=0
u=1
v=1
u=0 S
u
v
y=0
y=1-x
x=0
x
y
-6 -3 3 6
-3
3
R1
R2
R3
R4
R
x
y
-3 3
-2
2
S
u
v
3
-2
2
R1
R2
R3
R4
R
x
y
2 4
3
6
S
u
v
2
R1
R2
R3
R4
R
x
y
2
S
u
v
9.17 Change of Variables in Multiple Integrals
From u = 2x/(x2 + y2) and v = −2y(x2 + y2) we obtain u2 + v2 = 4/(x2 + y2). Then x2 + y2 = 4/(u2 + v2) and
∂(x, y)/∂(u, v) = (x2 + y2)2/4 = 4/(u2 + v2)2.
11. (a) The uv-corner points (0, 0), (1, 0), (1, 1), (0, 1) corre-
spond to the xy-points (0, 0), (1, 0), (0, 1), (0, 0).
v = 0: x = u, y = 0 =⇒ y = 0, 0 ≤ x ≤ 1
u = 1: x = 1− v, y = v =⇒ y = 1− x
v = 1: x = 0, y = u =⇒ x = 0, 0 ≤ y ≤ 1
u = 0: x = 0, y = 0
(b) Since the segment u = 0, 0 ≤ v ≤ 1 in the uv-plane maps to the origin in the xy-plane, the transformation
is not one-to-one.
12.
∂(x, y)
∂(u, v)
=
∣∣∣∣ 1− v v−u u
∣∣∣∣ = u. The transformation is 0 when u is 0, for 0 ≤ v ≤ 1.
13. R1: x + y = −1 =⇒ v = −1
R2: x− 2y = 6 =⇒ u = 6
R3: x + y = 3 =⇒ v = 3
R4: x− 2y = −6 =⇒ u = −6
∂(u, v)
∂(x, y)
=
∣∣∣∣ 1 −21 1
∣∣∣∣ = 3 =⇒ ∂(x, y)∂(u, v) = 13∫∫
R
(x + y) dA =
∫∫
S
v
(
1
3
)
dA′ =
1
3
∫ 3
−1
∫ 6
−6
v du dv =
1
3
(12)
∫ 3
−1
v dv = 4
(
1
2
)
v2
∣∣∣∣3
−1
= 16
14. R1: y = −3x + 3 =⇒ v = 3
R2: y = x− π =⇒ u = π
R3: y = −3x + 6 =⇒ v = 6
R4: y = x =⇒ u = 0
∂(u, v)
∂(x, y)
=
∣∣∣∣ 1 −13 1
∣∣∣∣ = 4 =⇒ ∂(x, y)∂(u, v) = 14
∫∫
R
cos 12 (x− y)
3x + y
dA =
∫∫
S
cosu/2
v
(
1
4
)
dA′ =
1
4
∫ 6
3
∫ π
0
cosu/2
v
du dv =
1
4
∫ 6
3
2 sinu/2
v
∣∣∣∣π
0
dv
=
1
2
∫ 6
3
dv
v
=
1
2
ln v
∣∣∣∣6
3
=
1
2
ln 2
15. R1: y = x2 =⇒ u = 1
R2: x = y2 =⇒ v = 1
R3: y = 12x
2 =⇒ u = 2
R4: x = 12y
2 =⇒ v = 2
∂(u, v)
∂(x, y)
=
∣∣∣∣∣ 2x/y −x2/y2−y2/x2 2y/x
∣∣∣∣∣ = 3 =⇒ ∂(x, y)∂(u, v) = 13
537
3
3
R1
R2
R3
R4
R
x
y
2
2
S
u
v
a b
c
d
R1
R2
R3
R4 R
u
v
a b
c
d
S
u
v
2 4
-2
2
R1
R2
R3
R4
R
x
y
5 10-2
2
S
u
v
9.17 Change of Variables in Multiple Integrals
∫∫
R
y2
x
dA =
∫∫
S
v
(
1
3
)
dA′ =
1
3
∫ 2
1
∫ 2
1
v du dv =
1
3
∫ 2
1
v dv =
1
6
v2
∣∣∣∣2
1
=
1
2
16. R1: x2 + y2 = 2y =⇒ v = 1
R2: x2 + y2 = 2x =⇒ u = 1
R3: x2 + y2 = 6y =⇒ v = 1/3
R4: x2 + y2 = 4x =⇒ u = 1/2
∂(u, v)
∂(x, y)
=
∣∣∣∣∣∣∣∣
2(y2 − x2)
(x2 + y2)2
−4xy
(x2 + y2)2
−4xy
(x2 + y2)2
2(x2 − y2)
(x2 + y2)2
∣∣∣∣∣∣∣∣ =
−4
(x2 + y2)2
Using u2 + v2 = 4/(x2 + y2) we see that ∂(x, y)/∂(u, v) = −4/(u2 + v2)2.
∫∫
R
(x2 + y2)−3dA =
∫∫
S
(
4
u2 + v2
)−3 ∣∣∣∣ −4(u2 + v2)2
∣∣∣∣ dA′ = 116
∫ 1
1/3
∫ 1
1/2
(u2 + v2) du dv =
115
5184
17. R1: 2xy = c =⇒ v = c
R2: x2 − y2 = b =⇒ u = b
R3: 2xy = d =⇒ v = d
R4: x2 − y2 = a =⇒ u = a
∂(u, v)
∂(x, y)
=
∣∣∣∣ 2x −2y2y 2x
∣∣∣∣ = 4(x2 + y2)
=⇒ ∂(x, y)
∂(u, v)
=
1
4(x2 + y2)
∫∫
R
(x2 + y2) dA =
∫∫
S
(x2 + y2)
1
4(x2 + y2)
dA′ =
1
4
∫ d
c
∫ b
a
du dv =
1
4
(b− a)(d− c)
18. R1: xy = −2 =⇒ v = −2
R2: x2 − y2 = 9 =⇒ u = 9
R3: xy = 2 =⇒ v = 2
R4: x2 − y2 = 1 =⇒ u = 1
∂(u, v)
∂(x, y)
=
∣∣∣∣ 2x −2yy x
∣∣∣∣ = 2(x2 + y2)
=⇒ ∂(x, y)
∂(u, v)
=
1
2(x2 + y2)∫∫
R
(x2 + y2) sinxy dA =
∫∫
S
(x2 + y2) sin v
(
1
2(x2 + y2)
)
dA′ =
1
2
∫ 2
−2
∫ 9
1
sin v du dv =
1
2
∫ 2
−2
8 sin v dv = 0
538
1
2
R1
R2
R3 R
x
y
1
2
S
u
v
-4 -2 2
3
R1
R2
R3
R
x
y
2
2
S
u
v
2 4
2
4
R1 R2
R3R4
R
x
y
2 4
2
4
S
u
v
9.17 Change of Variables in Multiple Integrals
19. R1: y = x2 =⇒ v + u = v − u =⇒ u = 0
R2 : y = 4− x2 =⇒ v + u = 4− (v − u)
=⇒ v + u = 4− v + u =⇒ v = 2
R3: x = 1 =⇒ v − u = 1 =⇒ v = 1 + u
∂(x, y)
∂(u, v)
=
∣∣∣∣∣∣
− 1
2
√
v − u
1
2
√
v − u
1 1
∣∣∣∣∣∣ = − 1√v − u
∫∫
R
x
y + x2
dA =
∫∫
S
√
v − u
2v
∣∣∣∣− 1√v − u
∣∣∣∣ dA′ = 12
∫ 1
0
∫ 2
1+u
1
v
dv du =
1
2
∫ 1
0
[ln 2− ln(1 + u)] du
=
1
2
ln 2− 1
2
[(1 + u) ln(1 + u)− (1 + u)]
∣∣∣1
0
=
1
2
ln 2− 1
2
[2 ln 2− 2− (0− 1)] = 1
2
− 1
2
ln 2
20. Solving x = 2u − 4v, y = 3u + v for u and v we
obtain u = 114x +
2
7y, v = − 314x + 17y. The xy-
corner points (−4, 1), (0, 0), (2, 3) correspond to
the uv-points (0, 1), (0, 0), (1, 0).
∂(x, y)
∂(u, v)
=
∣∣∣∣ 2 −43 1
∣∣∣∣ = 14∫∫
R
y dA =
∫∫
S
(3u + v)(14) dA′ = 14
∫ 1
0
∫ 1−u
0
(3u + v) dv du = 14
∫ 1
0
(
3uv +
1
2
v2
) ∣∣∣∣1−u
0
du
= 14
∫ 1
0
(
1
2
+ 2u− 5
2
u2
)
du =
(
7u + 14u2 − 35
3
u3
) ∣∣∣∣1
0
=
28
3
21. R1: y = 1/x =⇒ u = 1
R2: y = x =⇒ v = 1
R3: y = 4/x =⇒ u = 4
R4: y = 4x =⇒ v = 4
∂(u, v)
∂(x, y)
=
∣∣∣∣ y x−y/x2 1/x
∣∣∣∣ = 2yx =⇒ ∂(x, y)∂(u, v) = x2y
8
∫∫
R
y4 dA =
∫∫
S
u2v2
(
1
2v
)
du dv =
1
2
∫ 4
1
u2v du dv =
1
2
∫ 4
1
1
3
u3v
∣∣∣∣4
1
dv =
1
6
∫ 4
1
63v dv =
21
4
v2
∣∣∣∣4
1
=
315
4
22. Under the transformation u = y+z, v = −y+z, w = x−y the parallelepiped D is mapped to the parallelepiped
E: 1 ≤ u ≤ 3, −1 ≤ v ≤ 1, 0 ≤ w ≤ 3.
∂(u, v, w)
∂(x, y, z)
=
∣∣∣∣∣∣∣
0 1 1
0 −1 1
1 −1 0
∣∣∣∣∣∣∣ = 2 =⇒
∂(x, y, z)
∂(u, v, w)
=
1
2
539
-1 1
1
R1
R2R3 R
x
y
-1 1
1
S
u
v
-2
2
R1
R2
R3
R
x
y
2
2
S
u
v
4
4
R1
R2
R3
R4
R
x
y
2 4
4
8
S
u
v
9.17 Change of Variables in Multiple Integrals
∫∫
D
(4z + 2x− 2y) dV =
∫∫
E
(2u + 2v + 2w)
1
2
dV ′ =
1
2
∫ 3
0
∫ 1
−1
∫ 3
1
(2u + 2v + 2w) du dv dw
=
1
2
∫ 3
0
∫ 1
−1
(u2 + 2uv + 2uw)
∣∣∣3
1
dv dw =
1
2
∫ 3
0
∫ 1−1
(8 + 4v + 4w) dv dw
=
∫ 3
0
(4v + v2 + 2vw)
∣∣∣1
−1
dw =
∫ 3
0
(8 + 4w) dw = (8w + 2w2)
∣∣∣3
0
= 42
23. We let u = y − x and v = y + x.
R1: y = 0 =⇒ u = −x, v = x =⇒ v = −u
R2: x + y = 1 =⇒ v = 1
R3: x = 0 =⇒ u = y, v = y =⇒ v = u
∂(u, v)
∂(x, y)
=
∣∣∣∣−1 11 1
∣∣∣∣ = −2 =⇒ ∂(x, y)∂(u, v) = −12∫∫
R
e(y−x)/(y+x)dA =
∫∫
S
eu/v
∣∣∣∣−12
∣∣∣∣ dA′ = 12
∫ 1
0
∫ v
−v
eu/v du dv =
1
2
∫ 1
0
veu/v
∣∣∣v
−v
dv
=
1
2
∫ 1
0
v(e− e−1) dv = 1
2
(e− e−1)1
2
v2
∣∣∣∣1
0
=
1
4
(e− e−1)
24. We let u = y − x and v = y.
R1: y = 0 =⇒ v = 0, u = −x =⇒ v = 0, 0 ≤ u ≤ 2
R2: x = 0 =⇒ v = u
R3: y = x + 2 =⇒ u = 2
∂(u, v)
∂(x, y)
=
∣∣∣∣−1 10 1
∣∣∣∣ = −1 =⇒ ∂(x, y)∂(u, v) = −1
∫∫
R
ey
2−2xy+x2dA =
∫∫
S
eu
2 | − 1| dA′ =
∫ 2
0
∫ u
0
eu
2
dv du =
∫ 2
0
ueu
2
du =
1
2
eu
2
∣∣∣∣2
0
=
1
2
(e4 − 1)
25. Noting that R2, R3, and R4 have equations y+2x = 8, y−2x = 0,
and y + 2x = 2, we let u = y/x and v = y + 2x.
R1: y = 0 =⇒ u = 0, v = 2x =⇒ u = 0, 2 ≤ v ≤ 8
R2: y + 2x = 8 =⇒ v = 8
R3: y − 2x = 0 =⇒ u = 2
R4: y + 2x = 2 =⇒ v = 2
∂(u, v)
∂(x, y)
=
∣∣∣∣−y/x2 1/x2 1
∣∣∣∣ = −y + 2xx2 =⇒ ∂(x, y)∂(u, v) = − x2y + 2x∫∫
R
(6x + 3y) dA = 3
∫∫
S
(y + 2x)
∣∣∣∣− x2y + 2x
∣∣∣∣ dA′ = 3∫∫
S
x2 dA′
From y = ux we see that v = ux + 2x and x = v/(u + 2). Then
3
∫∫
S
x2 dA′ = 3
∫ 2
0
∫ 8
2
v2(u + 2)2 dv du =
∫ 2
0
v3
(u + 2)2
∣∣∣∣8
2
du = 504
∫ 2
0
du
(u + 2)2
= − 504
u + 2
∣∣∣∣2
0
= 126.
540
4
-2
2
R1 R2
R3R4
R
x
y
4
-2
2
S
u
v
9.17 Change of Variables in Multiple Integrals
26. We let u = x + y and v = x− y.
R1: x + y = 1 =⇒ u = 1
R2: x− y = 1 =⇒ v = 1
R3: x + y = 3 =⇒ u = 3
R4: x− y = −1 =⇒ v = −1
∂(u, v)
∂(x, y)
=
∣∣∣∣ 1 11 −1
∣∣∣∣ = −2 =⇒ ∂(x, y)∂(u, v) = −12∫∫
R
(x + y)4ex−y dA =
∫∫
S
u4ev
∣∣∣∣−12
∣∣∣∣ dA′ = 12
∫ 3
1
∫ 1
−1
u4ev dv du =
1
2
∫ 3
1
u4ev
∣∣∣1
−1
du
=
e− e−1
2
∫ 3
1
u4 du =
e− e−1
10
u5
∣∣∣∣3
1
=
242(e− e−1)
10
=
121
5
(e− e−1)
27. Let u = xy and v = xy1.4. Then xy1.4 = c =⇒ v = c; xy = b =⇒ u = b; xy1.4 = d =⇒ v = d;
xy = a =⇒ u = a.
∂(u, v)
∂(x, y)
=
∣∣∣∣ y xy1.4 1.4xy0.4
∣∣∣∣ = 0.4xy1.4 = 0.4v =⇒ ∂(x, y)∂(u, v) = 52v∫∫
R
dA =
∫∫
S
5
2v
dA′ =
∫ d
c
∫ b
a
5
2v
du dv =
5
2
(b− a)
∫ d
c
dv
v
=
5
2
(b− a)(ln d− ln c)
28. The image of the ellipsoid x2/a2 + y2/b2 + z2/c2 = 1 under the transformation u = x/a, v = y/b, w = z/c, is
the unit sphere u2 + v2 + w2 = 1. The volume of this sphere is 43π. Now
∂(x, y, z)
∂(u, v, w)
=
∣∣∣∣∣∣∣
a 0 0
0 b 0
0 0 c
∣∣∣∣∣∣∣ = abc
and ∫∫∫
D
dV =
∫∫∫
E
abc dV ′ = abc
∫∫∫
E
dV ′ = abc
(
4
3
π
)
=
4
3
πabc.
29. The image of the ellipse is the unit circle x2 + y2 = 1. From
∂(x, y)
∂(u, v)
=
∣∣∣∣ 5 00 3
∣∣∣∣ = 15 we obtain
∫∫
R
(
x2
25
+
y2
9
)
dA =
∫∫
S
(u2 + v2)15 dA′ = 15
∫ 2π
0
∫ 1
0
r2r dr dθ =
15
4
∫ 2π
0
r4
∣∣∣1
0
dθ
=
15
4
∫ 2π
0
dθ =
15π
2
.
30.
∂(x, y, z)
∂(ρ, φ, θ)
=
∣∣∣∣∣∣∣
sinφ cos θ ρ cosφ cos θ −ρ sinφ sin θ
sinφ sin θ ρ cosφ sin θ ρ sinφ cos θ
cosφ −ρ sinφ 0
∣∣∣∣∣∣∣
= cosφ(ρ2 sinφ cosφ cos2 θ + ρ2 sinφ cosφ sin2 θ) + ρ sinφ(ρ sin2 φ cos2 θ + ρ sin2 φ sin2 θ)
= ρ2 sinφ cos2 φ(cos2 θ + sin2 θ) + ρ2 sin3 φ(cos2 θ + sin2 θ) = ρ2 sinφ(cos2 φ + sin2 φ) = ρ2 sinφ
541
9.17 Change of Variables in Multiple Integrals
CHAPTER 9 REVIEW EXERCISES
CHAPTER 9 REVIEW EXERCISES
1. True; |v(t)| = √2
2. True; for all t, y = 4.
3. True
4. False; consider r(t) = t2i. In this case, v(t) = 2ti and a(t) = 2i. Since v · a = 4t, the velocity and acceleration
vectors are not orthogonal for t �= 0.
5. False; ∇f is perpendicular to the level curve f(x, y) = c.
6. False; consider f(x, y) = xy at (0, 0).
7. True; the value is 4/3.
8. True; since 2xy dx− x2 dy is not exact.
9. False;
∫
C
x dx + x2 dy = 0 from (−1, 0) to (1, 0) along the x-axis and along the semicircle y = √1− x2 , but
since x dx + x2 dy is not exact, the integral is not independent of path.
10. True
11. False; unless the first partial derivatives are continuous.
12. True 13. True
14. True; since curl F = 0 when F is a conservative vector field.
15. True 16. True 17. True 18. True
19. F = ∇φ = −x(x2 + y2)−3/2i− y(x2 + y2)−3/2j
20. curl F =
∣∣∣∣∣∣∣
i j k
∂/∂x ∂/∂y ∂/∂z
f(x) g(y) h(z)
∣∣∣∣∣∣∣ = 0
21. v(t) = 6i + j + 2tk; a(t) = 2k. To find when the particle passes through the plane, we solve −6t+ t+ t2 = −4
or t2 − 5t + 4 = 0. This gives t = 1 and t = 4. v(1) = 6i + j + 2k, a(1) = 2k; v(4) = 6i + j + 8k, a(4) = 2k
22. We are given r(0) = i + 2j + 3k.
r(t) =
∫
v(t) dt =
∫
(−10ti + (3t2 − 4t)j + k) dt = −5t2i + (t3 − 2t2)j + tk + c
i + 2j + 3k = r(0) = c
r(t) = (1− 5t2)i + (t3 − 2t2 + 2)j + (t + 3)k
r(2) = −19i + 2j + 5k
23. v(t) =
∫
a(t) dt =
∫
(
√
2 sin ti +
√
2 cos tj) dt = −
√
2 cos ti +
√
2 sin tj + c;
−i + j + k = v(π/4) = −i + j + c, c = k; v(t) = −√2 cos ti +√2 sin tj + k;
542
CHAPTER 9 REVIEW EXERCISES
r(t) = −√2 sin ti−√2 cos tj + tk + b; i + 2j + (π/4)k = r(π/4) = −i− j + (π/4)k + b, b = 2i + 3j;
r(t) = (2−√2 sin t)i + (3−√2 cos t)j + tk; r(3π/4) = i + 4j + (3π/4)k
24. v(t) = ti + t2j− tk; |v| = t√t2 + 2 , t > 0; a(t) = i + 2tj− k; v · a = t + 2t3 + t = 2t + 2t3;
v × a = t2i + t2k, |v × a| = t2√2 ; aT = 2t + 2t
3
t
√
t2 + 2
=
2 + 2t2√
t2 + 2
, aN =
t2
√
2
t
√
t2 + 2
=
√
2 t√
t2 + 2
;
κ =
t2
√
2
t3(t2 + 2)3/2
=
√
2
t(t2 + 2)3/2
25.
26. r′(t) = sinh ti + cosh tj + k, r′(1) = sinh 1i + cosh 1j + k;
|r′(t)| =
√
sinh2 t + cosh2 t + 1 =
√
2 cosh2 t =
√
2 cosh t; |r′(1)| = √2 cosh 1;
T(t) =
1√
2
tanh ti +
1√
2
j +
1√
2
sech tk, T(1) =
1√
2
(tanh 1i + j + sech 1k);
dT
dt
=
1√
2
sech2 ti− 1√
2
sech t tanh tk;
d
dt
T(1) =
1√
2
sech2 1i− 1√
2
sech 1 tanh 1k,∣∣∣∣ ddtT(1)
∣∣∣∣ = sech 1√2 √sech2 1 + tanh2 1 = 1√2 sech 1; N(1) = sech 1i− tanh 1k;
B(1) = T(1)×N(1) = − 1√
2
tanh 1i +
1√
2
(tanh2 1 + sech2 1)j− 1√
2
sech 1k
=
1√
2
(− tanh 1i + j− sech 1k)
κ =
∣∣∣∣ ddtT(1)
∣∣∣∣ /|r′(1)| = (sech 1)/√2√2 cosh 1 = 12 sech2 1
27. ∇f = (2xy − y2)i + (x2 − 2xy)j; u = 2√
40
i +
6√
40
j =
1√
10
(i + 3j);
Duf =
1√
10
(2xy − y2 + 3x2 − 6xy) = 1√
10
(3x2 − 4xy − y2)
28. ∇F = 2x
x2 + y2 + z2
i +
2y
x2 + y2 + z2
j +
2z
x2 + y2 + z2
k; u = −2
3
i +
1
3
j +
2
3
k; DuF =
−4x + 2y + 4z
3(x2 + y2 + z2)
29. fx = 2xy4, fy = 4x2y3.
(a) u = i, Du(1, 1) = fx(1, 1) = 2
(b) u = (i− j)/√2 , Du(1, 1) = (2− 4)/
√
2 = −2/√2
(c) u = j, Du(1, 1) = fy(1, 1) = 4
30. (a)
dw
dt
=
∂w
∂x
dx
dt
+
∂w
∂y
dy
dt
+
∂w
∂z
dz
dt
=
x√
x2 + y2 + z2
6 cos 2t +
y√
x2 + y2 + z2
(−8 sin 2t) + z√
x2 + y2 + z2
15t2
=
(6x cos 2t− 8y sin 2t + 15zt2)√
x2 + y2 + z2
543
CHAPTER 9 REVIEW EXERCISES
(b)
∂w
∂t
=
∂w
∂x
∂x
∂t
+
∂w
∂y
∂y
∂t
+
∂w
∂z
∂z
∂t
=
x√
x2 + y2 + z2
6
r
cos
2t
r
+
y√
x2 + y2 + z2
(
8r
t2
sin
2r
t
)
+
z√
x2 + y2 + z2
15t2r3
=
(
6x
r
cos
2t
r
+
8yr
t2
sin
2r
t
+ 15zt2r3
)
√
x2 + y2 + z2
31. F (x, y, z) = sinxy − z; ∇F = y cosxyi + x cosxyj− k; ∇F (1/2, 2π/3,
√
3/2) =
π
3
i +
1
4
j− k. The equation of
the tangent plane is
π
3
(
x− 1
2
)
+
1
4
(
y − 2π
3
)
−
(
z −
√
3
2
)
= 0
or 4πx + 3y − 12z = 4π − 6√3 .
32. We want to find a normal to the surface that