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: 47


$\dfrac{9}{2} \pi$

Work Step by Step

The area is given as follows: $A=(2) \int_{0}^{\pi} (\dfrac{1}{2}) r^2 d\theta$ Then, $(2) \int_{0}^{\pi} (2-\cos \theta)^2= \int_{0}^{\pi} [4-4 \cos \theta +\cos^2 \theta ]d\theta$ After solving, we get $A=\dfrac{9}{2} \pi$
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.