Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 15 - Multiple Integrals - 15.9 Change of Variables in Multiple Integrals - 15.9 Exercises - Page 1100: 7

Answer

The parallelogram with vertices $(0,0), (6,3), (12,1), (6,-2)$

Work Step by Step

Let us consider $u=\dfrac{x+3y}{5}; y=\dfrac{x-2y}{5}$ This gives: $0 \leq u \leq 3$ and $0 \leq v \leq 2$ This can be written as: $0 \leq \dfrac{x+3y}{5} \leq 3$ and $0 \leq \dfrac{x-2y}{5} \leq 2$ This implies that $0 \leq x+3y \leq 15$ and $0 \leq x-2y \leq 10$ Hence, the parallelogram with vertices $(0,0), (6,3), (12,1), (6,-2)$
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