Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 10 - Section 10.2 - Areas and Lengths in Polar Coordinates - 10.4 Exercises - Page 673: 12

Answer

The area that the curve encloses: $$ r=2- \cos \theta \quad \text {for } \theta =0 \quad \text {to} \quad \theta=2 \pi $$ is equal to $$ A =\int_{0}^{2\pi} \frac{1}{2} r^{2} d \theta =\frac{9}{2} \pi $$

Work Step by Step

The area that the curve encloses: $$ r=2- \cos \theta \quad \text {for } \theta =0 \quad \text {to} \quad \theta=2 \pi $$ is equal to $$ \begin{aligned} A &=\int_{0}^{2\pi} \frac{1}{2} r^{2} d \theta \\ &=\int_{0}^{2 \pi} \frac{1}{2}(2- \cos \theta)^{2} d \theta\\ &=\frac{1}{2} \int_{0}^{2 \pi}\left(4-4\cos \theta+ \cos ^{2} \theta\right) d \theta \\ &=\frac{1}{2} \int_{0}^{2 \pi}\left[4-4 \cos \theta+ \frac{1}{2}(1+\cos 2 \theta)\right] d \theta \\ &=\frac{1}{2} \int_{0}^{2 \pi}(\frac{9}{2}-4 \cos \theta+\frac{1}{2} \cos 2 \theta) d \theta \\ &= \frac{1}{2}[\frac{9}{2} \theta-4 \sin \theta+\frac{1}{4}\sin 2 \theta]_{0}^{2 \pi} \\ &=\frac{1}{2}(9\pi)\\ &=\frac{9}{2} \pi \end{aligned} $$
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.