Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 15: Multiple Integrals - Section 15.4 - Double Integrals in Polar Form - Exercises 15.4 - Page 894: 46

Answer

$\frac{\pi}{2}$

Work Step by Step

$ A=\int^{3\pi/4}_{\pi/4} \int^{2sin\theta}_{csc\theta}rdrd\theta $ =$\frac{1}{2}\int^{3\pi/4}_{\pi/4}(4sin^2\theta-csc^2\theta) d\theta $ =$\frac{1}{2}[2\theta-sin2\theta+cot\theta]^{3\\pi/4}_{\pi/4}$ =$\frac{\pi}{2}$
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