University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 14 - Section 14.3 - Area by Double Integration - Exercises - Page 772: 9



Work Step by Step

Our aim is to integrate the integral as follows: $ Area =\iint_D dA $ $\int^{2}_{0} \int^{3y}_{y} (1) \space dx \space dy = \int^{2}_{0}[x]^{3y}_y \space dy $ or, $=\int^{2}_{0}(2y)dy $ or, $=[y^2]^2_0$ So, $ Area=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.