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: 37

Answer

$2\pi(2-\sqrt{e})$

Work Step by Step

$\int^{2\pi}_0 \int^{\sqrt{e}}_1 (\frac{lnr^2}{r})rdrd\theta $ =$\int^{2\pi}_0 \int^{\sqrt{e}}_1 2\ln rdrd\theta $ =$2\int^{2\pi}_0[rlnr-r]^{1/2}_1d\theta $ =$2\int^{2\pi}_0 \sqrt{r}[\frac{1}{2}-1+1]d\theta $ =$2\pi(2-\sqrt{e})$
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