Page 266 - Schaum's Outline of Theory and Problems of Advanced Calculus

CHAP. 10] LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS 257

Alternatively, we may notice by inspection that
3 2 2 2
F dr ¼ð2xz dx þ 3x z dzÞþð6ydx þ 6xdyÞ ð2yz dy þ y dzÞ
2 3 2 2 3 2
¼ dðx z Þþ dð6xyÞ dðy zÞ¼ dðx z þ 6xy y z þ cÞ
from which is determined.
Method 2: Since the integral is independent of the path, we can choose any path to evaluate it; in
particular, we can choose the path consisting of straight lines from ð1; 1; 1Þ to ð2; 1; 1Þ,thento
ð2; 1; 1Þ and then to ð2; 1; 1Þ. The result is

ð 2 ð 1 ð 1
2
ð12z 1Þ dz ¼ 15
ð2x 6Þ dx þ ð12 2yÞ dy þ
x¼1 y¼ 1 z¼1
where the first integral is obtained from the line integral by placing y ¼ 1; z ¼ 1; dy ¼ 0; dz ¼ 0;
the second integral by placing x ¼ 2; z ¼ 1; dx ¼ 0; dz ¼ 0; and the third integral by placing
x ¼ 2; y ¼ 1; dx ¼ 0; dy ¼ 0.
ð
(c) Physically F dr represents the work done in moving an object from ð1; 1; 1Þ to ð2; 1; 1Þ along C.
C
In a conservative force field the work done is independent of the path C joining these points.

MISCELLANEOUS PROBLEMS

10.32. (a)If x ¼ f ðu; vÞ; y ¼ gðu; vÞ defines a transformation which maps a region r of the xy plane into
a region r of the uv plane, prove that
0
ðð
ðð
@ðx; yÞ

du dv
dx dy ¼

@ðu; vÞ
r r 0
(b) Interpret geometrically the result in (a).
(a)If C (assumed to be a simple closed curve) is the boundary of r,then byProblem 10.8,
ðð þ
1
xdy ydx
dx dy ¼ ð1Þ
2 C
r
Under the given transformation the integral on the right of (1)becomes
ð
1 @y @y @x @x 1 @y @x @y @x
þ
x du þ dv y du þ dv ¼ x y du þ x y dv ð2Þ
2 C 0 @u @v @u @v 2 C 0 @u @u @v @v
where C is the mapping of C in the uv plane (we suppose the mapping to be such that C is a simple
0
0
closed curve also).
By Green’s theorem if r is the region in the uv plane bounded by C ,the right side of (2)equals
0
0
ðð ðð
1 @ @y @x @ @y @x @x @y @x @y
x y x y du dv
du dv ¼
2 @u @v @v @v @u @u @u @v @v @u
r 0 r 0
ðð
@ðx; yÞ
du dv
¼

@ðu; vÞ
r 0
ðð
where we have inserted absolute value signs so as to ensure that the result is non-negative as is dx dy
In general, we can show (see Problem 10.83) that r

ðð ðð

@ðx; yÞ
du dv
Fðx; yÞ dx dy ¼ Ff f ðu; vÞ; gðu; vÞg ð3Þ

@ðu; vÞ
r r 0