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: 9

Answer

$\frac{\pi}{2}$

Work Step by Step

$\int^1_{-1} \int ^{\sqrt{1-x^2}}_0 dydx $ =$\int ^{\pi}_0 \int ^{1}_0 r dr d\theta $ =$\frac{1}{2} \int ^{\pi}_0d\theta $ =$\frac{\pi}{2}$
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