Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 14 - Multiple Integrals - 14.2 Double Integrals Over Nonrectangular Regions - Exercises Set 14.2 - Page 1017: 42

Answer

$$\frac{8}{15}$$

Work Step by Step

Volume under the function $ f (x, y)=z $ and over the region $ R $ is given by \[ \begin{array}{c} \iint_{R} -x +1d A=V \\ =\int_{-1}^{1} \int_{y^{2}}^{1} 1-x d x d y \\ =\int_{-1}^{1}\left[x-\frac{x^{2}}{2}\right]_{y^{2}}^{1} d y \\ =\int_{-1}^{1}\left[1-\frac{1^{2}}{2}\right]-\left[y^{2}-\frac{\left(y^{2}\right)^{2}}{2}\right] d y \\ =\int_{-1}^{1} \frac{1}{2}-y^{2}+\frac{y^{4}}{2} d y \\ =\left[\frac{y}{2}-\frac{y^{3}}{3}+\frac{y^{5}}{10}\right]_{-1}^{1} \\ \frac{8}{15}=\left[-\frac{1}{3}+\frac{1}{10}+\frac{1}{2}\right]2 \end{array} \]
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