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: 21


$$A=(\pi-\frac{3\sqrt 3}{2})$$

Work Step by Step

$$A=\int_{7\pi/6}^{11\pi/6}\frac{1}{2}(1+2sin\theta)^{2}d \theta=\int_{7\pi/6}^{11\pi/6}\frac{1}{2}(1+4sin^{2}\theta+4sin\theta)d \theta$$ Hence, $$A=(\pi-\frac{3\sqrt 3}{2})$$
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