Calculus: Early Transcendentals 8th Edition

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

Chapter 10 - Review - Exercises - Page 690: 35

Answer

$=\frac{\pi-1}{2}$

Work Step by Step

$r=2 sin \theta$ and $r=sin \theta+cos \theta$ $Area,A= \frac{1}{2}\int_{0}^{\pi/4}2sin\theta)^2d\theta+\frac{1}{2}\int_{\pi/4}^{3\pi/4}(\sqrt 2sin( \theta+ \pi/4))^2d\theta$ $=[\theta-\frac{1}{2}sin 2\theta]_{0}^{\pi/4}+[\frac{1}{2} \theta -\frac{1}{4} sin (2 \theta+ \pi/2)]_{\pi/4}^{3\pi/4}$ $=\frac{\pi-1}{2}$
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.