Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 14 - Multiple Integrals - 14.3 Double Integrals In Polar Coordinates - Exercises Set 14.3 - Page 1024: 9

Answer

$\frac{\pi}{16}$

Work Step by Step

$\sin 2 \theta=r, 1=r$ $\frac{\pi}{4} \leqslant \theta \leqslant \frac{\pi}{2}$ so $\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \int_{\sin 2 \theta}^{1} r d r d \theta=A$ integrate with respect to $r$ $A=\int_{\frac{\pi}{2}}^{\frac{\pi}{4}}\left[\frac{r^{2}}{2}\right]_{\sin \theta}^{1} d \theta$ $A=\frac{1}{2} \int_{\frac{\pi}{2}}^{\frac{\pi}{4}}\left(-\sin ^{2} 2 \theta+1\right) d \theta$ and then $A=\frac{1}{2} \int_{\frac{\pi}{2}}^{\frac{\pi}{4}}\left(1-\frac{1}{2}+\frac{\cos 4 \theta}{2}\right) d \theta$ $A=\frac{1}{2} \int_{\frac{\pi}{2}}^{\frac{\pi}{4}}\left(\frac{1}{2}+\frac{\cos 4 \theta}{2}\right) d \theta$ integrate $A=\frac{1}{2}\left[\frac{\theta}{2}-\frac{\sin 4 \theta}{8}\right]_{\frac{\pi}{2}}^{\frac{\pi}{2}}$ evaluate $-\frac{1}{2}\left[\frac{\pi}{4}-\frac{\sin 2 \pi}{8}\right]+\frac{1}{2}\left[\frac{\pi}{8}-\frac{\sin \pi}{8}\right]=A$ $A=-\frac{\pi}{8}+\frac{\pi}{16}=\frac{\pi}{16}$
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