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: 7

Answer

$\frac{2}{3}$

Work Step by Step

$\int_{0}^{\pi}\int_{0}^{1}\int_{0}^{\sqrt {1-y^{2}}}$$y$$sinx$$dz$$dy$$dx$$=\int_{0}^{\pi}\int_{0}^{1}$$y$$sinx$$z|_{0}^\sqrt {1-y^{2}}dydx$ $=\int_{0}^{\pi}\int_{0}^{1}y\sqrt {1-y^{2}}sinxdydx$ Suppose $u=1-y^{2}$, $du=-2ydy$ and $ydy=\frac{-du}{2}$ when $y=0$ , $u=1$ and when $y=1$ $u=0$ $=\int_{0}^{\pi}\int_{0}^{1}\sqrt {u}(\frac{-du}{2})sinxdudx$ $=\frac{1}{2}\int_{0}^{\pi}\frac{2}{3}u^{3/2}|_{0}^{1}sinxdx$ $=\frac{1}{3}\int_{0}^{\pi}sinxdx$ $=\frac{1}{3}[-cosx]|_{0}^{\pi}$ $=\frac{1}{3}(1-(-1))$ $=\frac{2}{3}$
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