Calculus 8th Edition

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

Chapter 15 - Multiple Integrals - 15.1 Double Integrals over Rectangles - 15.1 Exercises - Page 1040: 30

Answer

$$\iint \limits_{R} \frac{\tan \theta}{\sqrt{1-t^{2}}} d A =\frac{\pi \ln 2}{6} $$

Work Step by Step

Given$$\iint \limits_{R} \frac{\tan \theta}{\sqrt{1-t^{2}}} d A, \quad R=\left\{(\theta, t) | 0 \leqslant \theta \leqslant \pi / 3,0 \leqslant t \leqslant \frac{1}{2}\right\}$$ So, we have \begin{aligned}I&= \iint \limits_{R} \frac{\tan \theta}{\sqrt{1-t^{2}}} dA \\ &=\int_{0}^{\pi / 3} \int_{0}^{1 / 2} \frac{\tan \theta}{\sqrt{1-t^{2}}} d t d \theta\\ &=\left[\int_{0}^{1 / 2} \frac{d t}{\sqrt{1-t^{2}}}\right]\left[\int_{0}^{\pi / 3} \tan \theta d \theta\right]\\ &=\left[\int_{0}^{1 / 2} \frac{d t}{\sqrt{1-t^{2}}}\right]\left[\int_{0}^{\pi / 3} \frac{\sin t}{\cos t} \theta d \theta\right]\\ &=[\arcsin t]_{0}^{1 / 2} \cdot[-\ln |\cos \theta|]_{0}^{\pi / 3}\\ &=\left[\arcsin \frac{1}{2}-\arcsin 0\right] \cdot\left[-\ln \left|\cos \frac{\pi}{3}\right|+\ln |\cos 0|\right]\\ &=\left[\frac{\pi}{6}-0\right] \cdot[-\ln \frac{1}{2}+0]\\ &=\left[\frac{\pi}{6} \right] \cdot[\ln 2 ]\\ &=\frac{\pi \ln 2}{6} \end{aligned}
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