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: 36

Answer

$\frac{3}{2}$

Work Step by Step

We calculate the average as follows: average=$\frac{1}{\pi} \int \int _R [(1-x)^2+y^2]dydx $ =$\frac{1}{\pi}\int ^{2\pi}_0 \int^{1}_0[(1-rcos\theta)^2+r^2sin^2\theta]rdrd\theta $ =$\frac{1}{\pi} \int^{2\pi}_0 \int^1_0 (r^3-2r^2cos\theta+r)drd\theta $ =$\frac{1}{\pi} \int^{2\pi}_0(\frac{3}{4}-\frac{2cos\theta}{3})d\theta $ =$\frac{1}{\pi}[\frac{3}{4}\theta-\frac{2sin\theta}{3}]^{2\pi}_0$ =$\frac{3}{2}$
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