Page 267 - Schaum's Outline of Theory and Problems of Advanced Calculus

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

ðð
ðð
@ðx; yÞ
(b) dx dy and du dv represent the area of region r,the first expressed in rectangular

@ðu; vÞ
r r 0
coordinates, the second in curvilinear coordinates. See Page 212, and the introduction of the differ-
ential element of surface area for an alternative to Part (a).
yi þ xj þ
. (a) Calculate r F: ðbÞ Evaluate F dr around any closed path and
x þ y
10.33. Let F ¼ 2 2
explain the results.

i j
k
@ @

@

@x @y ¼ 0 in any region excluding ð0; 0Þ:
ðaÞ r F ¼
@z
y x

x þ y x þ y
2 2 2 2 0
ydx þ xdy
þ þ
(b) F dr ¼ . Let x ¼ cos ; y ¼ sin , where ð ; Þ are polar coordinates. Then
2
x þ y 2
dx ¼ sin d þ d cos ; dy ¼ cos d þ d sin
ydx þ xdy y
and so ¼ d ¼ d arc tan
2
x þ y 2 x
For a closed curve ABCDA [see Fig. 10-22(a)below] surrounding the origin, ¼ 0at A and ¼ 2
ð 2
after a complete circuit back to A. Inthis case the line integral equals d ¼ 2 .
0
y y
Q
B
P
C
φ A
x
O φ
S
R
φ 0
D x
O
(a) (b)
Fig. 10-22
For a closed curve PQRSP [see Fig. 10-22(b) above] not surrounding the origin, ¼ 0 at P and
ð
0
¼ 0 after a complete circuit back to P. Inthis case the line integral equals d ¼ 0.
0
Since F ¼ Pi þ Qj; r F ¼ 0 is equivalent to @P=@y ¼ @Q=@x and the results would seem to con-
y x
tradict those of Problem 10.11. However, no contradiction exists since P ¼ 2 2 and Q ¼ 2 2
x þ y x þ y
do not have continuous derivatives throughout any region including ð0; 0Þ, and this was assumed in
Problem 10.11.
10.34. If div A denotes the divergence of a vector field A at a point P, show that
ðð
A n dS
s
div A ¼ lim
V!0 V