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 1078: 22


$20 \pi$

Work Step by Step

Let us consider that $V=\int_{-2}^2 \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} (5-z) dz dx $ and $\int_{-2}^2 (5z-z^2)_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}dx =\int_{-2}^{2} \sqrt{4-x^2} dx$ Consider $x=2 \sin \theta$ and $dx=2 \cos \theta d \theta$ Thus, we have $\int_{-2}^{2} \sqrt{4-(2 \sin \theta)^2} (2 \cos \theta d \theta)=(10) \int_{-\pi/2}^{\pi/2}(2 \cos \theta) (2\cos \theta) d \theta$ Hence, $V= 10\int_{-\pi/2}^{\pi/2}(2 \cos \theta)^2 d \theta= (20) \int_{-\pi/2}^{pi/2} (2\cos \theta+1)=20 \pi$
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