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



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 $A=-4(-1)-(-4)=8$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.