Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 15 - Multiple Integrals - Review - Exercises - Page 1102: 26

Answer

$\frac{4}{3}$

Work Step by Step

$\int\int_{D}ydA=\int_{1}^{2}\int_{1/y} ^{y} ydxdy$ $=\int_{1}^{2} [yx]|_{1/y} ^{y}dy$ $=\int_{1}^{2} y^{2}-1dy$ $=[\frac{y^{3}}{3}-y]_{1}^{2}$ $=[\frac{8}{3}-2]-[\frac{1}{3}-1]$ $=\frac{4}{3}$
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