Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 15: Multiple Integrals - Section 15.8 - Substitutions in Multiple Integrals - Exercises 15.8 - Page 931: 8

Answer

$-3$

Work Step by Step

$\int \int_R 2(x-y)dx dy $ =$\int \int_G (-2v)|\frac{\sigma(x,y)}{\sigma(u,v)}|$ du dv =$\int \int_G(-2v) du dv $ =$\int \int_G(-2v)$ du dv =$\int^1_0 \int^{3-3v}_{-3v}(-2v)$ du dv =$\int^1_0 (-2v)(3-3v+3v)dv $ =$\int^1_0 (-6v)dv $ =$[-3v^2]^1_0$ =-3
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