Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 13 - Multiple Integration - 13.2 Double Integrals over General Regions - 13.2 Exercises - Page 983: 85

Answer

\[\ln \sqrt[4]{2}\]

Work Step by Step

\[\begin{align} & \text{Let the region }R=\left\{ \left( x,y \right):0\le y\le {{x}^{2}},\text{ }0\le x\le \sqrt{\pi }\text{/2 } \right\} \\ & \iint\limits_{R}{\left( x+y \right)}dA=\int_{0}^{\sqrt{\pi }/2}{\int_{0}^{{{x}^{2}}}{x{{\sec }^{2}}ydydx}} \\ & \text{Integrating} \\ & =\int_{0}^{\sqrt{\pi }/2}{\left[ x\tan y \right]_{0}^{{{x}^{2}}}}dx \\ & =\int_{0}^{\sqrt{\pi }/2}{x\tan {{x}^{2}}}dx \\ & =\frac{1}{2}\int_{0}^{\sqrt{\pi }/2}{2x\tan {{x}^{2}}}dx \\ & =\frac{1}{2}\left[ \ln \left| \sec {{x}^{2}} \right| \right]_{0}^{\sqrt{\pi }/2} \\ & =\frac{1}{2}\left[ \ln \left| \sec {{\left( \frac{\sqrt{\pi }}{2} \right)}^{2}} \right|-\ln \left| \sec {{\left( 0 \right)}^{2}} \right| \right] \\ & =\frac{1}{2}\left[ \ln \left| \sec \left( \frac{\pi }{4} \right) \right|-\ln \left| 1 \right| \right] \\ & =\frac{1}{2}\left[ \ln \left| \sqrt{2} \right| \right] \\ & =\ln \sqrt[4]{2} \\ \end{align}\]
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