Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 15 - Section 15.1 - Double Integrals over Rectangles - 15.1 Exercise - Page 1000: 30

Answer

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

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}
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.