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Thomas Calculus 13th [Solutions]

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Section 15.8 Substitutions in Multiple Integrals 1137<br />

(b) x 1 u 1, and x 2 u 2; y 1 uv 1 v 1 ,<br />

u<br />

and y 2 uv 2 v 2 ; thus,<br />

u<br />

2 2 2 2/ 2 2/ 2 2 2/ u<br />

y u<br />

uv<br />

u<br />

v<br />

2<br />

2 1<br />

1 1 x<br />

dy dx 1 1/ u u<br />

u dv du 1 1/ u uv dv du 1 u 2<br />

du 1 u du<br />

2 2<br />

1/ u<br />

u 2u<br />

3<br />

2<br />

3 2<br />

u 1 du ln u 3 ln 2;<br />

1 u<br />

1<br />

2 2 2 2<br />

2 2 y 2<br />

1 y 3<br />

2<br />

3 2<br />

dy dx dx dx ln x 3 ln 2<br />

1 1 x 1 x 2 2 1 x 2 1 2<br />

1<br />

( x, y) a cos ar sin<br />

2 2<br />

11. x ar cos and y ar sin J ( r, ) abr cos abr sin abr;<br />

( r, )<br />

b sin br cos<br />

2 2 2 1 2 2 2 2 2<br />

I0<br />

x y dA r a cos b sin J ( r, ) dr d<br />

0 0<br />

R<br />

2 1 3 2 2 2 2 2 2 2 2 2<br />

abr a cos b sin drd ab a cos b sin d<br />

0 0 4 0<br />

2 2 2 2 2<br />

2 2<br />

ab a b<br />

ab a a sin 2 b b sin 2<br />

4 2 4 2 4<br />

0<br />

4<br />

2 2<br />

12.<br />

( x, y)<br />

a 0<br />

J ( u, v) ab;<br />

1 1 2<br />

u<br />

A dy dx ab du dv 1 2<br />

( u, v)<br />

2 ab dv du 2ab 1 u du<br />

0 b<br />

1 1 u<br />

1<br />

R G<br />

1<br />

2 1 1 1 1<br />

2ab u 1 u sin u ab sin 1 sin ( 1) ab ab<br />

2 2 2 2<br />

1<br />

13. The region of integration R in the xy -plane is<br />

sketched in the figure at the right. The boundaries of<br />

the image G are obtained as follows, with G<br />

sketched at the right:<br />

xy -equations for<br />

Corresponding<br />

uv-equations<br />

Simplified<br />

the boundary of R for the boundary of G uv-equations<br />

x y 1 ( u 2 v) 1 ( u v )<br />

v 0<br />

3 3<br />

x 2 2y 1 ( u 2 v) 2 2 ( u v )<br />

u 2<br />

3 3<br />

y 0<br />

0 1 ( u v)<br />

v u<br />

3<br />

Copyright<br />

2014 Pearson Education, Inc.

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