Thomas Calculus 13th [Solutions]

1196 Chapter 16 Integrals and Vector Fields
16.6 SURFACE INTEGRALS
i j k
1. Let the parametrization be r( x, z) xi 2
x j zk rx
i 2xj and rz k rx rz 1 2x
0
0 0 1
2 xi j | rx
rz
|
3/2
2
2 3 2 2 3
1 2
4x 1 G( x, y, z) d x 4x 1 dx dz 4x 1 dz
0 0 0 12
S
0
3
1
17 17 1
17 17 1 dz
0 12 4
2. Let the parametrization be
2
r( x, y) xi yj 4 y k, 2 y 2 rx
i and
y
ry
j k
2
4 y
i j k
2
y
y
r 1 0 0 | | 1 2
x ry j k rx ry
4 y
4 y 4 y
y
0 1
2
4 y
2 2
2
4 2 2
G( x, y, z) d 4 y 2 dy dx 24
1 2 2
4 y
S
3. Let the parametrization be r( , ) (sin cos ) i (sin sin ) j (cos ) k (spherical coordinates with 1
on the sphere), 0 , 0 2 r (cos cos ) i (cos sin ) j (sin ) k and
i j k
r ( sin sin ) i (sin cos ) j r r cos cos cos sin sin
sin sin sin cos 0
2 2 4 2 4 2 2 2
sin cos i sin sin j sin cos k | r r | sin cos sin sin sin cos
2 2 2 2 2
sin ; x sin cos G( x, y, z) cos sin G( x, y, z) d cos sin (sin ) d d
0 0
S
2 2 2 u cos
2 1 2 2
cos 1 cos (sin ) d d ; cos u 1 du d
0 0 du sin d 0 1
2 3 1
2 2 2 sin 2
2
cos u u d 4 cos d 4
4
0 3
1
3 0 3 2 4 0 3
4. Let the parametrization be r( , ) ( a sin cos ) i ( a sin sin ) j ( a cos ) k (spherical coordinates with
a , a 0, on the sphere), 0 (since z 0 ), 0 2
2
r ( a cos cos ) i ( a cos sin ) j ( a sin ) k and r ( a sin sin ) i ( a sin cos ) j
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
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