Answer
$=\frac{\pi}{5}$
Work Step by Step
$r=2sin5\theta$
Area enclosed by one loop is
$A=\frac{1}{2}\int_{0}^{\pi/5}4sin^{2}(5\theta)d \theta$
$=2\int_{0}^{\pi/5}( {\frac{1}{2}-\frac{1}{2}cos(10\theta)}) d \theta$
$=\frac{\pi}{5}$
Multivariable Calculus, 7th Edition
by Stewart, James
Published by
Brooks Cole
ISBN 10:
0-53849-787-4
ISBN 13:
978-0-53849-787-9
Chapter 10 - Parametric Equations and Polar Coordinates - 10.4 Exercises - Page 693: 20
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