Thomas Calculus 13th [Solutions]

1202 Chapter 16 Integrals and Vector Fields
26. Let the parametrization be r( r, ) ( r cos ) i ( r sin ) j 2 rk, 0 r 1 (since 0 z 2) and 0 2
i j k
rr
(cos ) i (sin ) j 2k and r ( r sin ) i ( r cos ) j r rr
r sin r cos 0
cos sin 2
(2r cos ) (2r sin ) r F n d F r rr
|
| r rr
|
r r | d dr
3 2
2r sin cos
3
4r cos sin r d dr since F
2 2
r sin i
2
2r
cos j k
2 1 3 2 3 2
1 2
F n d 2r sin cos 4r cos sin r dr d sin cos cos sin 1 d
0 0 0 2 2
S
3 2
2
1 sin 1 sin 1
6 2 2 0
27. Let the parametrization be r( r, ) ( r cos ) i ( r sin ) j rk, 1 r 2 (since 1 z 2) and 0 2
i j k
rr
(cos ) i (sin ) j k and r ( r sin ) i ( r cos ) j r rr
r sin r cos 0
cos sin 1
( r cos ) ( r sin ) r F n d F r rr
2 2 2 2 3
| | cos sin
| r | r d dr r r r
r r
d dr
2 3 2 2 2 2 3
r r d dr since F ( r cos ) i ( r sin ) j r k F n d r r dr d 73
0 1
6
S
28. Let the parametrization be r( r, ) ( r cos ) i ( r sin ) j
2
r k, 0 r 1 (since 0 z 1) and 0 2
i j k
rr
(cos ) i (sin ) j 2rk and r ( r sin ) i ( r cos ) j r rr
r sin r cos 0
cos sin 2r
2 2 r
3 2 3 2
2r cos 2r sin r d r r
| | 8 cos 8 sin
| r|
r d dr r r
r r
2r d dr
3 3 2 1
8r 2r d dr since F (4r cos ) i (4r sin ) j 2k F n d 8r 0 0
S
2r dr d 2
29. g( x, y, z) z, p k g k | g | 1 and | g p | 1 Flux = F n d ( F k)
dA
S
R
2 3
0 0 3 dy dx 18
30. g( x, y, z) y, p j g j | g | 1 and | g p | 1 Flux = F n d ( F j)
dA
S
R
2 7 2
2 dz dx 2(7 2) dx 10(2 1) 30
1 2 1
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