University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 10 - Section 10.5 - Areas and Lengths in Polar Coordinates - Exercises - Page 584: 2



Work Step by Step

The area of a shaded region can be found as: $A=\dfrac{1}{2}\int_p^q r^2 d \theta$ Here, we have $A=\dfrac{1}{2}\int_{\pi/4}^{\pi/2} (2 \sin \theta)^2 d \theta$ $\dfrac{4}{2}\int_{\pi/4}^{\pi/2}[\dfrac{1-\cos 2 \theta}{2}] d \theta=[\theta-\dfrac{1-\sin 2 \theta}{2}]_{\pi/4}^{\pi/2}$ Thus, $A=(\dfrac{\pi}{2}-0)-(\dfrac{\pi}{4}-\dfrac{1}{2})=\dfrac{\pi+2}{4}$
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.