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: 20

Answer

$16 \pi$

Work Step by Step

Here, we have $V=\int_{-2}^2 \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} \int_{x^2+z^2}^{8-x^2-z^2} dy dz dx $ This implies that $V=\int_{-2}^2 \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} [y]_{x^2+z^2}^{8-x^2-z^2} dz dx $ In polar coordinate system, we have $x=r \cos \theta \\ y=r \sin \theta \\z=z$ and $r^2=x^2+y^2 \\ \tan \theta=\dfrac{y}{x} \\z=z$ and $\int_{-2}^2 \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} [8-2(x^2+z^2)] dz dx =\int_{0}^{2\pi} \int_0^2 (8-2r^2) r dr d\theta$ Now, we have $\int_0^2 (8r-2r^3) dr \int_{0}^{2\pi} d \theta= (2 \pi) \times [4r^2-(\dfrac{1}{2}) r^4]_0^2$ or, $V= (2 \pi) [4(2-0)^2-(\dfrac{1}{2}) (2-0)^4]= (2 \pi) (16-8)= 16 \pi$
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