Page 261 - Schaum's Outline of Theory and Problems of Advanced Calculus

252 LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS [CHAP. 10

(b)If S 1 is the convex surface of the hemispherical region and S 2 is the base ðz ¼ 0Þ,then
p ffiffiffiffiffiffiffiffiffi p ffiffiffiffiffiffiffiffiffi
2
2
ðð ð a ð a y 2 q ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð a ð a y 2 q ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2 2 2 2 2 2 2 2 2
xz dy dz ¼ z a y z dz dy z a y z dz dy
y¼ a z¼0 y¼ a z¼0
S 1
ffiffiffiffiffiffiffiffiffiffi
2
p
ðð ð a ð a x 2 p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2 3 2 2 2 2 3
ðx y z Þ dz dx ¼ fx a x z z g dz dx
x¼ a x¼0
S 1
ffiffiffiffiffiffiffiffiffiffi
p
a a x p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ð ð 2 2
3
2
2
2
f x 2 a x z z g dz dx

x¼ a z¼0
p ffiffiffiffiffiffiffiffiffiffi
a a x q ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ðð ð ð 2 2
2 2 2 2 2
ð2xy þ y zÞ dx dy ¼ f2xy þ y a x y g dy dx
p ffiffiffiffiffiffiffiffiffiffi
2
x¼ a y¼ a x 2
S 1
ðð ðð
3
2
2
xz dy dz ¼ 0; ðx y z Þ dz dx ¼ 0;
S 2 S 2
p ffiffiffiffiffiffiffiffiffiffi
a a x
ðð ðð ð ð 2 2
2 2
ð2xy þ y zÞ dx dy ¼ f2xy þ y ð0Þg dx dy ¼ 2xy dy dx ¼ 0
p ffiffiffiffiffiffiffiffiffiffi
2
x¼ a y¼ a x 2
S 2 S 2
By addition of the above, we obtain
p ffiffiffiffiffiffiffiffiffiffi
2 p ffiffiffiffiffiffiffiffiffiffi
2
ð a ð a y 2 q ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð a ð a x 2
p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2
2
2
2
2
2
4 z 2 a y z dz dy þ 4 x 2 a x z dz dx
y¼0 x¼0 x¼0 z¼0
p ffiffiffiffiffiffiffiffiffiffi
2
ð a ð a x 2 q ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2
2
2
þ 4 y 2 a x y dy dx
x¼0 y¼0
Since by symmetry all these integrals are equal, the result is, on using polar coordinates,
p ffiffiffiffiffiffiffiffiffiffi
2
ð a ð a x 2 q ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð =2 ð a 5
p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 a
2
2
2
2
2
2
2
12 y 2 a x y dy dx ¼ 12 sin a d d ¼
x¼0 y¼0 ¼0 ¼0 5
STOKES’ THEOREM
10.26. Prove Stokes’ theorem.
Let S be a surface which is such that its projections on the xy, yz, and xz planes are regions bounded by
simple closed curves, as indicated in Fig. 10-20. Assume S to have representation z ¼ f ðx; yÞ or x ¼ gðy; zÞ
or y ¼ hðx; zÞ, where f ; g; h are single-valued, continuous, and differentiable functions. We must show that
ðð ðð
½r ðA 1 i þ A 2 j þ A 3 kފ n dS
ðr AÞ n dS ¼
S S
ð
A dr
¼
C
where C is the boundary of S.
ðð
Consider first ½r ðA 1 iފ n dS:
S

i j
k
@ @

@ @A 1 @A 1
k;
Since r ðA 1 iÞ¼ ¼ j
@x @y @z @z @y

0 0
A 1