Calculus 8th Edition

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

Chapter 15 - Multiple Integrals - 15.1 Double Integrals over Rectangles - 15.1 Exercises - Page 1040: 29


$$\iint \limits_{R} \frac{x y^{2}}{x^{2}+1} d A=9 \ln 2$$

Work Step by Step

Given$$\iint \limits_{R} \frac{x y^{2}}{x^{2}+1} d A, \quad R=\{(x, y) | 0 \leqslant x \leqslant 1,-3 \leqslant y \leqslant 3\}$$ So, we have \begin{aligned}I&= \iint \limits_{R} \frac{x y^{2}}{x^{2}+1} dA \\ &= \int_{-3}^{3} \int_{0}^{1} \frac{x y^{2}}{x^{2}+1} d x \ dy\\ &=\left[\int_{-3}^{3} y^{2} d y\right]\left[\frac{1}{2}\int_{0}^{1} \frac{2x}{x^{2}+1} d x\right]\\ &=\left[ \frac{y^{3}}{3} \right]_{-3}^{3}\left[ \frac{1}{2} \ln [x^{2}+1] d x\right]_{0}^{1}\\ &=\left[ \frac{3^{3}}{3} -\frac{ (-3)^{3}}{3}\right] \left[ \frac{1}{2} \ln [1^{2}+1] - \frac{1}{2} \ln [0^{2}+1] \right] \\ &= \frac{2 (3^{3})}{3} \frac{1}{2} \ln 2\\ &= 9 \ln 2 \end{aligned}
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