## Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

# Chapter 10 - Section 10.2 - Areas and Lengths in Polar Coordinates - 10.4 Exercises - Page 673: 19

#### Answer

$=\frac{\pi}{16}$

#### Work Step by Step

$r=sin4\theta$ Area enclosed by one loop is $A=\int_{0}^{\pi/4}\frac{( {sin4\theta})^{2}}{2}d \theta$ $=\int_{0}^{\pi/4}\frac{( {1-cos8\theta})}{4}d \theta$ $=\frac{\pi}{16}$

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