Page 226 - Schaum's Outline of Theory and Problems of Advanced Calculus

CHAP. 9] MULTIPLE INTEGRALS 217

Let P be any point with coordinates ðx; yÞ or ðu; vÞ, where x ¼ f ðu; vÞ and y ¼ gðu; vÞ. Then the vector r
from O to P is given by r ¼ xi þ yj ¼ f ðu; vÞi þ gðu; vÞj. The tangent vectors to the coordinate curves u ¼ c 1
and v ¼ c 2 , where c 1 and c 2 are constants, are @r=@v and @r=@u, respectively. Then the area of region r of

@r @r
Fig. 9-11 is given approximately by u v.
@u @v
But

i j
k
@x @y
@x @y

@r @r @u @ðx; yÞ
¼ @u @u 0 ¼ @u k ¼ k

@u @v @x @y

@x @y @ðu; vÞ
@v @v

0
@v @v

@r @r @ðx; yÞ
so that u v ¼ u v
@u @v @ðu; vÞ
The double integral is the limit of the sum

X

@ðx; yÞ
u v
Ff f ðu; vÞ; gðu; vÞg

@ðu; vÞ
taken over the entire region r. Aninvestigation reveals that this limit is
ðð
@ðx; yÞ
du dv

Ff f ðu; vÞ; gðu; vÞg

@ðu; vÞ
r 0
where r is the region in the uv plane into which the region r is mapped under the transformation
0
x ¼ f ðu; vÞ; y ¼ gðu; vÞ.
Another method of justifying the above method of change of variables makes use of line integrals and
Green’s theorem in the plane (see Chapter 10, Problem 10.32).
2
2
9.7. If u ¼ x y and v ¼ 2xy, find @ðx; yÞ=@ðu; vÞ in terms of u and v.

2x 2y
@ðu; vÞ u x u y 2 2

¼ ¼ ¼ 4ðx þ y Þ
v x v y
2y 2x
@ðx; yÞ
2
2 2
2 2
2
2
From the identify ðx þ y Þ ¼ðx y Þ þð2xyÞ we have
p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2
2
2
2
2 2
2
ðx þ y Þ ¼ u þ v 2 and x þ y ¼ u þ v 2
Then by Problem 6.43, Chapter 6,
1 1 1
@ðx; yÞ
2 2
¼ ¼ ¼ p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2
@ðu; vÞ @ðu; vÞ=@ðx; yÞ 4 u þ v 2
4ðx þ y Þ
Another method: Solve the given equations for x and y in terms of u and v and find the Jacobian directly.
2
2
9.8. Find the polar moment of inertia of the region in the xy plane bounded by x y ¼ 1,
2
2
x y ¼ 9, xy ¼ 2; xy ¼ 4 assuming unit density.
2
2
Under the transformation x y ¼ u,2xy ¼ v the required region r in the xy plane [shaded in Fig.
9-12(a)] is mapped into region r of the uv plane [shaded in Fig. 9-12(b)]. Then:
0
ðð ðð
2 2 2 2 @ðx; yÞ

du dv
Required polar moment of inertia ¼ ðx þ y Þ dx dy ¼ ðx þ y Þ

@ðu; vÞ
r r 0
ðð ð 9 ð 8
p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi du dv 1
2
u þ v 2 du dv ¼ 8
¼ p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼
2
4 u þ v 2 4 u¼1 v¼4
r 0
where we have used the results of Problem 9.7.