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: 30



Work Step by Step

$$A=\int_{0}^{\pi/2}\frac{(1-cos\theta)^{2}}{2}d \theta=\int_{0}^{\pi/2}\frac{(1-2cos\theta+cos^{2}\theta)}{2}d \theta$$ Or, $$=\frac{1}{4}[3\theta-4sin\theta+\frac{1}{2}sin2\theta]_{0}^{\pi/2}$$ Hence, $$A=(\frac{3\pi}{2}-4)$$
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