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: 20



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, $$A=\frac{\pi}{5}$$
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