Calculus 8th Edition

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

Chapter 15 - Multiple Integrals - 15.6 Triple Integrals - 15.6 Exercises - Page 1077: 7



Work Step by Step

$ \int_{0}^{\pi} \int_{0}^{1} \int_{0}^{\sqrt{1-z^2}} z \sin x dy dz dx=(2) \int_{0}^{1} z\sqrt{1-z^2} dz$ Consider $p=1-z^2; dz=(-\dfrac{1}{2z}) dp$ Then, we have $(2) \int_{0}^{1} z\sqrt{1-z^2} dz= (2)(-\dfrac{1}{2})\int_{-1}^0 p^{1/2} dp \\= [\dfrac{2p^{3/2}}{3}]_0^1 \\Hence, \int_{0}^{\pi} \int_{0}^{1} \int_{0}^{\sqrt{1-z^2}} z \sin x dy dz dx=\dfrac{2}{3}$
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