Answer
$$A=\frac{\pi}{5}$$
Work Step by Step
Given: $r=2sin5\theta$
$$A=\frac{1}{2}\int_{0}^{\pi/5}4sin^{2}(5\theta)d \theta=2\int_{0}^{\pi/5}( {\frac{1}{2}-\frac{1}{2}cos(10\theta)}) d \theta$$
Hence, $$A=\frac{\pi}{5}$$
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: 20
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