Answer
$8$
Work Step by Step
Consider $\iint_{R} f(u,v) dA= \int_{-2}^{0} \int_v^{-v} 2 dp dv= \int_{-2}^{0} [2p]_{v}^{-v} dv$
or, $= \int_{-2}^{0} -2v-2v dv$
or, $= [-2v^2]_{-2}^{0}$
Thus, we have $I=2\times (-2)^2 =8$
Thomas' Calculus 13th Edition
by Thomas Jr., George B.
Published by
Pearson
ISBN 10:
0-32187-896-5
ISBN 13:
978-0-32187-896-0
Chapter 15: Multiple Integrals - Section 15.2 - Double Integrals over General Regions - Exercises 15.2 - Page 882: 29
Update this answer!
You can help us out by revising, improving and updating this answer.
Update this answerAfter 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.
