Page 260 - Schaum's Outline of Theory and Problems of Advanced Calculus

CHAP. 10] LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS 251

Face BCDE: n ¼ k; z ¼ 1. Then
1
1
ðð ð ð 1 ð ð 1
2
A n dS ¼ fð2x 1Þi þ x yj xkg k dx dy ¼ xdxdy 1=2
0 0 0 0
BCDE
Face AFGO: n ¼ k; z ¼ 0. Then
1
ðð ð ð 1
2
f2xi x yjg ð kÞ dx dy ¼ 0
A n dS ¼
0 0
AFGO
ðð
Adding, 3 1 1 1 11 : Since
2 2 3 2 6
A n dS ¼ þ þ þ 0 þ 0 ¼
S
1
1
ðð ð ð ð ð 1 11
2
r A dV ¼ ð2 þ x 2xzÞ dx dy dz ¼ 6
0 0 0
V
the divergence theorem is verified in this case.
ðð
10.24. Evaluate r n dS, where S is a closed surface.
S
By the divergence theorem,
ðð ðð ð
r r dV
r n dS ¼
S V
@ @ @
ðð ð
k ðxi þ yj þ zkÞ dV
@x @y @z
¼ i þ j þ
V
ðð ð ððð
@x @y @z
dV ¼ 3 dV ¼ 3V
¼ þ þ
@x @y @z
V V
where V is the volume enclosed by S.
ðð
2 2 3 2
10.25. Evaluate xz dy dz þðx y z Þ dz dx þð2xy þ y zÞ dx dy, where S is the entire surface of the
S p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2
2
2
a x y and z ¼ 0 (a)by the divergence theorem
hemispherical region bounded by z ¼
(Green’s theorem in space), (b) directly.
(a)Since dy dz ¼ dS cos ; dz dx ¼ dS cos ; dx dy ¼ dS cos
,the integral can be written
ðð ðð
2 2 3 2
fxz cos þðx y z Þ cos þð2xy þ y zÞ cos
g dS ¼ A n dS
S S
2
2
2
3
where A ¼ xz i þðx y z Þj þð2xy þ y zÞk and n ¼ cos i þ cos j þ cos
k,the outward drawn unit
normal.
Then by the divergence theorem the integral equals
ðð ð
@ 2 @ 2 3 @ 2 2 2 2
ðð ð ððð
r A dV ¼ ðxz Þþ ðx y z Þþ ð2xy þ y zÞ dV ¼ ðx þ y þ z Þ dV
@x @y @z
V V V
where V is the region bounded by the hemisphere and the xy plane.
By use of spherical coordinates, as in Problem 9.19, Chapter 9, this integral is equal to
ð =2 ð =2 ð 2 a 5
2
2
4 r r sin dr d d ¼
¼0 ¼0 r¼0 5