Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 10 - Parametric Equations and Polar Coordinates - 10.4 Areas and Lengths in Polar Coordinates - 10.4 Exercises - Page 713: 32


$$A=(11\pi-2\sqrt 2)$$ or $$A=17.6$$

Work Step by Step

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