Calculus 8th Edition

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

Chapter 15 - Multiple Integrals - Review - Exercises - Page 1102: 28

Answer

$\frac{2\sqrt 2-1}{3}$

Work Step by Step

$\int\int_{D}xdA=\int_{0}^{\pi/2}\int_{1} ^{\sqrt 2} (rcos\theta)rdrd\theta$ $=\int_{0}^{\pi/2}(cos\theta)d\theta\int_{1} ^{\sqrt 2} r^{2}dr$ $=(sin\theta)_{0}^{\pi/2}[\frac{r^{3}}{3}]_{1} ^{\sqrt 2}$ $=(1)\frac{2\sqrt 2-1}{3}$ $=\frac{2\sqrt 2-1}{3}$
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