Calculus: Early Transcendentals 8th Edition

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

Chapter 15 - Review - Exercises - Page 1062: 25

Answer

$8$

Work Step by Step

$\int\int_{D}ydA=\int_{0}^{2}\int_{y^{2}} ^{8-y^{2}} ydxdy$ $=\int_{0}^{2} [y(8-y^{2}-y^{2})]dy$ $=\int_{0}^{2} [8y-y^{3}]dy$ $=[\frac{8y^{2}}{2}-\frac{2y^{4}}{4}]_{0}^{2}$ $=16-8$ $=8$
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