Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 10 - Section 10.2 - Areas and Lengths in Polar Coordinates - 10.4 Exercises - Page 673: 32


$11\pi-12\sqrt 2$ or $17.6$

Work Step by Step

$A=2.\frac{1}{2}\int_{\pi/4}^{5\pi/4}(3+2cos\theta)^{2}d \theta$ $=\int_{\pi/4}^{5\pi/4}(11+12cos\theta+2cos2\theta)d \theta$ $=[11\theta+12sin\theta+sin2\theta]_{\pi/4}^{5\pi/4}$ $=11\pi-12\sqrt 2$ or $=17.6$
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