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: 21

Answer

$\frac{2(1+\sqrt{2})}{3}$

Work Step by Step

$\int^1_0 \int^{\sqrt{2-x^2}}_{x} (x+2y)dydx $ =$\int^{\pi/2}_{\pi/4} \int^{\sqrt{2}}_0(rcos\theta+2rsin\theta) rdrd\theta $ =$\int^{\pi/2}_{\pi/4}[\frac{r^3}{3}cos\theta+\frac{2r^3}{3}sin\theta]^{\sqrt{2}}_0d\theta $ =$\int^{\pi/2}_{\pi/4}(\frac{2\sqrt{2}cos\theta}{3}+\frac{4\sqrt{2}}{3}sin\theta) d\theta $ =$[\frac{2\sqrt{2}}{3}sin\theta-\frac{4\sqrt{2}}{3}cos\theta]^{\pi/2}_{\pi/4}$ =$\frac{2(1+\sqrt{2})}{3}$
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