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: 22

Answer

$\frac {41}{24}$

Work Step by Step

Since, $D=(x,y) | 0\leq y\leq 1, y^{2}\leq x \leq y+2$ $\int\int_{D} xydA=\int_{0}^{1}\int_{y^{2}} ^{y+2}xydxdy$ $=\int_{0}^{1}[\frac{x^{2}}{y}]_{y^{2}} ^{y+2}dxdy$ $=(\frac {y^{4}}{8}+\frac{2}{3}y^{3}+y^{2}-\frac{y^{6}}{12}]_{0} ^{1}$ $=\frac{1}{8}+\frac{2}{3}+1-\frac{1}{12}$ $=\frac {41}{24}$
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