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: 12

Answer

$\pi a^2$

Work Step by Step

$\int ^a_{-a} \int ^{\sqrt{a^2-x^2}}_\sqrt{a^2-x^2}dydx $ =$\int ^{2\pi}_0 \int^a_0 rdr d\theta $ =$\frac{a^2}{2} \int^{2\pi}_0 d\theta $ =$\pi a^2$
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