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: 10

Answer

$\frac{\pi}{8}$

Work Step by Step

$\int _0^1 \int ^{\sqrt{1-y^2}}_0 (x^2+y^2)dxdy $ =$\int^{\pi/2}_0 \int ^1_0 r^3 drd\theta $ =$\frac{1}{4}\int ^{\pi/2}_0 d\theta $ =$\frac{\pi}{8}$
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