Page 256 - Schaum's Outline of Theory and Problems of Advanced Calculus

CHAP. 10] LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS 247

(c) If Uðx; y; zÞ¼ 3z,(2)becomes
ðð ðð
q ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi q ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2 2 2 2 2 2
3z 1 þ 4x þ 4y dx dy ¼ 3f2 ðx þ y Þg 1 þ 4x þ 4y dx dy
r r
or in polar coordinates,
2 p ffiffi
ð ð 2
2 p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 111
3 ð2 Þ 1 þ 4 d d ¼
¼0 ¼0 10
Physically this could represent the mass of S assuming a density ¼ 3z,or three times the first
moment of S about the xy plane.

10.18. Find the surface area of a hemisphere of radius a cut off
by a cylinder having this radius as diameter.

Equations for the hemisphere and cylinder (see Fig. 10-15)
2
2
2
are given respectively by x þ y þ z ¼ a 2 (or z ¼
p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 2 2 2
2
2
2
a x y Þ and ðx a=2Þ þ y ¼ a =4(or x þ y ¼ ax).
Since
x y
and
z x ¼ p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi z y ¼ p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2
2
2
2
a x y 2 a x y 2
we have
Fig. 10-15
ðð
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a
ðð q
2
2
Required surface area ¼ 2 1 þ z x þ z y dx dy ¼ 2 p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi dx dy
2
2
a x y 2
r r
Two methods of evaluation are possible.
Method 1: Using polar coordinates.
2
2
Since x þ y ¼ ax in polar coordinates is ¼ a cos ,the integral becomes
ð =2 ð a cos a ð =2 p a cos
2
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2 p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi d d ¼ 2a a 2 d
2
¼0 ¼0 a 2 ¼0 ¼0
ð =2
¼ 2a 2 ð1 sin Þ d ¼ð 2Þa 2
0
Method 2: The integral is equal to
p ffiffiffiffiffiffiffiffiffiffi
p ffiffiffiffiffiffiffiffiffiffi
ð a ð ax x 2 a ð a y ax x 2
2 p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi dy dx ¼ 2a sin 1 p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi dx
2
2
x¼0 y¼0 a x y 2 x¼0 ax x 2 y¼0
x
ð a r ffiffiffiffiffiffiffiffiffiffiffi
¼ 2a sin 1 dx
0 a þ x
2
Letting x ¼ a tan ,this integral becomes
ð =4 ð =4
2
2
2
4a 2 tan sec d ¼ 4a 2 1 2 tan j =4 1 2 tan d
0
0 0
=4
ð
2
2
¼ 2a 2 tan j =4 ðsec 1Þ d
0
0
n o
2 =4 2
¼ 2a =4 ðtan Þj 0 ¼ð 2Þa
Note that the above integrals are actually improper and should be treated by appropriate limiting
procedures (see Problem 5.74, Chapter 5, and also Chapter 12).