James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

1126 Chapter 16 Vector Calculus
Therefore
Z
i
r 3 r z −
j k
2sin cos 0
0 0 1
Z − cos i 1 sin j
and | r 3 r z | − scos 2 1 sin 2 − 1
Thus the surface integral over S 1 is
yy z dS − yy z | r 3 r z | dA
S 1 D
− y 2
− 1 2 y 2
y 11cos
0 0
Since S 2 lies in the plane z − 0, we have
z dz d − y 2
0
1
2 s1 1 cos d2 d
0
f1 1 2 cos 1 1 2 s1 1 cos 2d g d
− 1 2 f 3 2 1 2 sin 1 1 4 sin 2g 2 3
0 −
2
yy z dS − yy 0 dS − 0
S 2 S 2
The top surface S 3 lies above the unit disk D and is part of the plane z − 1 1 x. So,
taking tsx, yd − 1 1 x in Formula 4 and converting to polar coordinates, we have
yy z dS − y
S 3 D
y s1 1 xdÎ1 1S −z
− y 2
0
y 1
0
− s2 y 2
0
y 1
2
−xD
1S
−yD
−z
2
dA
s1 1 r cos ds1 1 1 1 0 r dr d
0
sr 1 r 2 cos d dr d
− s2 y 2
( 1 2 1 1 3 cos d d
0
− s2 F 2 1 sin
2
− s2
3
G0
Therefore
yy z dS − yy z dS 1 yy z dS 1 yy z dS
S
S 1 S 2 S 3
− 3 2 1 0 1 s2 − ( 3 2 1 s2 ) ■
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