Answer
$=\frac{\pi}{16}$
Work Step by Step
$r=sin4\theta$
Area enclosed by one loop is
$A=\int_{0}^{\pi/4}\frac{( {sin4\theta})^{2}}{2}d \theta$
$=\int_{0}^{\pi/4}\frac{( {1-cos8\theta})}{4}d \theta$
$=\frac{\pi}{16}$
Calculus: Early Transcendentals 8th Edition
by Stewart, James
Published by
Cengage Learning
ISBN 10:
1285741552
ISBN 13:
978-1-28574-155-0
Chapter 10 - Section 10.2 - Areas and Lengths in Polar Coordinates - 10.4 Exercises - Page 673: 19
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