Answer
$$A=(\pi-\frac{3\sqrt 3}{2})$$
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, $$A=(\pi-\frac{3\sqrt 3}{2})$$
Calculus 8th Edition
by Stewart, James
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
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