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: 31

Answer

$\frac{3\pi}{8}+1$

Work Step by Step

We calculate the area as follows: $ A=\int^{\pi/2}_0 \int^{1+sin\theta}_0 rdrd\theta $ $\frac{1}{2} \int^{\pi/2}_0 (\frac{3}{2}+2sin\theta -\frac{cos2\theta}{2})d\theta $ =$\frac{3\pi}{8}+1$
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