Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 11: Parametric Equations and Polar Coordinates - Section 11.5 - Areas and Lengths in Polar Coordinates - Exercises 11.5 - Page 670: 11


$5\pi-8 $

Work Step by Step

The area of between the polar curve is given by as follows: $A=(\dfrac{1}{2})\int_p^q [(r_2(\theta))^2-(r_1(\theta))^2] d \theta$ $\implies A=(\dfrac{1}{2})\int_{(-\pi/2)}^{(\pi/2)} [2^2-(2-2\cos \theta)^2] d\theta $ $\implies A=(\dfrac{1}{2})\int_{-\pi/2}^{\pi/2} [8 \cos \theta-4 \cos^2 \theta] d\theta$ This implies that $A=\dfrac{1}{2}[8 \sin \theta-4 (1/2)\sin 2 \theta/2+\theta]_{-\pi/2}^{\pi/2}$ Thus, $A=\pi(2)^2-(8-\pi)=5\pi-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.