Thomas Calculus 13th [Solutions]

Section 16.6 Surface Integrals 1201
i j k
2 2 2 2 2
r r a cos cos a cos sin a sin a sin cos i a sin sin j a sin cos k
a sin sin a sin cos 0
F n d F r r
3 3 2 3 3 2 3 2
| | sin cos sin sin sin cos
| r r | r r d d a a a d d
a
3 sin d d since F xi yj zk ( asin cos ) i ( a sin sin ) j ( a cos ) k
2 3 3
F n d a sin d d 4 a
0 0
S
23. Let the parametrization be r( x, y) xi yj (2 a x y) k, 0 x a, 0 y a rx
i k and ry
j k
i j k
rx
ry
r x r y 1 0 1 i j k F n d F | r |
| x y|
x r y dy dx
r r
0 1 1
2xy 2 y(2 a x y) 2 x(2 a x y) dy dx since F 2xyi 2yzj 2xzk
2xyi 2 y(2 a x y) j 2 x(2 a x y)
k F n d
S
a a a a
2xy 2 y(2 a x y) 2 x(2 a x y) dy dx
0 0 0 0
4ay 2
2y 4ax 2
2x 2xy dy dx
a 3 2 2 4
4
4 a 3a x 2ax dx 4 3 2 a 13a
0 3 3 2 3 6
24. Let the parametrization be r( , z) (cos ) i (sin ) j zk,
0 z a , 0 2 (where r
2
x
2
y 1 on
i j k
the cylinder) r ( sin ) i (cos ) j and rz
k r rz
sin cos 0 (cos ) i (sin ) j
0 0 1
F n d F r rz
2 2
| | cos sin
| z| z dz d
r r
dz d dz d , since (cos ) (sin ) z 2 a
F n d
0 0 1 dz d
S
2 a
25. Let the parametrization be r( r, ) ( r cos ) i ( r sin ) j rk, 0 r 1 (since 0 z 1) 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
3 2 2
| | sin cos
| r|
r d dr r r
r r
d dr since
F
2 2 1 3 2 2 2
1 2
r sin cos i rk F n d r sin cos r dr d sin cos 1
0 0 0 4 3
S
d
3
2
1 cos
2
12 3 0 3
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