University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 10 - Section 10.5 - Areas and Lengths in Polar Coordinates - Exercises - Page 585: 4


$\dfrac{3\pi a^2}{2}$

Work Step by Step

The area of a shaded region can be found as: $A=\dfrac{1}{2}\int_p^q r^2 d \theta$ Here, we have $A=\int_{0}^{2\pi} \dfrac{a(1+ \cos \theta)^2}{2} d \theta$ $\dfrac{a^2}{2}\int_{0}^{2\pi} 1+\dfrac{1}{2}+\dfrac{\cos2 \theta}{2}+2\cos \theta d \theta=\dfrac{a^2}{2}[ 3\theta+\dfrac{\sin 2 \theta}{4}+2 \sin \theta]_{0}^{2\pi} $ Thus, $A=(\dfrac{a^2}{2}) [\dfrac{6\pi}{2}]=\dfrac{3\pi a^2}{2}$
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