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: 27

Answer

$2(\pi -1)$

Work Step by Step

$\int^{\pi/2}_0 \int^{2\sqrt{2-\sin2\theta}rdrd\theta}$ =$2\int^{\pi/2}_0(2-sin2\theta) d\theta $ =$2(\pi -1)$
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