Answer
$\frac{4}{3}$
Work Step by Step
$\int\int_{D}ydA=\int_{1}^{2}\int_{1/y} ^{y} ydxdy$
$=\int_{1}^{2} [yx]|_{1/y} ^{y}dy$
$=\int_{1}^{2} y^{2}-1dy$
$=[\frac{y^{3}}{3}-y]_{1}^{2}$
$=[\frac{8}{3}-2]-[\frac{1}{3}-1]$
$=\frac{4}{3}$
Calculus 8th Edition
by Stewart, James
Published by
Cengage
ISBN 10:
1285740629
ISBN 13:
978-1-28574-062-1
Chapter 15 - Multiple Integrals - Review - Exercises - Page 1102: 26
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