University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 10 - Practice Exercises - Page 594: 51

Answer

$8$

Work Step by Step

Given: $r=-1+\cos \theta$ The length is given as follows: $L= \int_{0}^{2\pi} \sqrt{(-1+\cos \theta)^2+(-\sin \theta)^2} d\theta$ Then, $L=\int_{0}^{2\pi} (2) \sin (\dfrac{\theta}{2}) d\theta$ After solving, we get $A=-4(-1)-(-4)=8$
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