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 893: 16

Answer

=$2-\frac{\pi}{2}$

Work Step by Step

$\int^2_{\sqrt{2}} \int^y_{\sqrt{4-y^2}}dydx $ =$\int^{\pi/2}_{\pi/4} \int^{2csc\theta}_{2}rdrd\theta $ =$\int^{\pi/4}_{\pi/6}(2csc^2\theta-2)d\theta $ =$[-2cot\theta-\frac{1}{2}\theta]^{\pi/2}_{\pi/6}$ =$2-\frac{\pi}{2}$
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