Thomas Calculus 13th [Solutions]

Chapter 16 Practice Exercises 1229
54. ( xz) z, ( yz) z, (1) 0 F 2z Flux = 2z r dr d dz
x z z
D
2
2 4 25 r
2 4 2
2
2z dz r dr d r 16 r dr d 64 d 128
0 0 3 0 0 0
55.
2 2 2 2 2
( 2 x) 2, ( 3 y) 3, ( z) 1 F 4; x y z 2 and x y z z 1
x y z
2 2 2 1 2 r
2 12 3
x y 1 Flux 4 dV 4 2 dz r dr d 4 r 2 r r dr d
0 0 r
0 0
D
4 2
7 2 2
0 12 3 2 d
3
7 8 2
2
56.
2 2
(6 x y) 6, ( x z) 0, (4 yz) 4y F 6 4 y;
z x y r
x y z
/2 1 r
/2 1 2 3
Flux (6 4 y) dV (6 4r sin ) dz r dr d 6r 4r sin dr d
0 0 0 0 0
D
/2
(2 sin ) d 1
0
57. F yi zj xk F 0 Flux F n d F dV 0
S
D
58.
2 3 2 2
F 3xz i yj z k F 3z 1 3z 1 Flux F n d F dV
S
D
2
4 16 x /2 y/2 4
1 dz dy dx 16 x dx x x 8
0 0 0 0 16 48
0
3
2 3 4
59.
2 2 2 2
F xy i x yj yk F y x 0 Flux F n d F dV
S
D
2 2 2 1 1 2 2 1 3 2
x y dV r dz r dr d 2r dr d 1 d
0 0 1 0 0 0 2
D
60. (a) F (3z
1) k F 3 Flux across the hemisphere F n d F dV 3 dV
(b)
1 4 3 3
3 a 2 a
2 3
S D D
2 2 2 2 2 2 2 2
f ( x, y, z) x y z a 0 f 2xi 2yj 2zk f 4x 4y 4z 4a 2a
since
0 2xi 2yj 2zk xi yj zk
a n F n
2
(3 z 1) z
a a a
; p k f p f k 2 z f p 2 z
f
since z 0 d 2a dA a dA F n d (3z 1) z a dA (3z 1) dx dy
f p 2z z a z
S R R
R
xy
3 2 2 2 1 2 a 2 2 2 2
3 2 3
a x y dx dy
0 0 3 a r 1 r dr d a a d a
0 2
2 a ,
which is the flux across the hemisphere. Across the base we find F 3(0) 1 k k since z 0 in the
xy
xy
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