Multivariable Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 0-53849-787-4
ISBN 13: 978-0-53849-787-9

Chapter 15 - Multiple Integrals - 15.9 Exercises - Page 1062: 27

Answer

$ \dfrac{(\sqrt 3 -1)}{3} \pi a^3$

Work Step by Step

Consider the spherical coordinates system as: $x=\rho \sin \phi \cos \theta \\ y=\rho \sin \phi \sin \theta \\z=\rho \cos \phi$; So, $\rho=\sqrt {x^2+y^2+z^2}$ or, $ \rho^2=x^2+y^2+z^2$ Write the jacobian for spherical coordinates: $\rho^2 \sin \rho$ Therefore, $\iiint{E} \sqrt {x^2+y^2+z^2} dV =\iiint_{E} \rho^2 \sin \rho d\rho d\theta d\phi$ Therefore, $I=\int_{\pi/6}^{\pi/3} \int_{0}^{2\pi}\int_{0}^{a} \rho^2 \sin \phi d\rho d\theta d\phi \\=\int_{\pi/6}^{\pi/3} \int_{0}^{2\pi}[\dfrac{1}{3} \rho^3 \sin \phi ]_0^a d\theta d\phi \\ = \dfrac{1}{3}\int_{\pi/6}^{\pi/3} \int_{0}^{2\pi} a^3 \sin \phi d\theta d \phi \\= [\dfrac{-2 \cos\phi \pi a^3 }{3} ]_{\pi/6}^{\pi/3}$ Hence, $I=\dfrac{1}{3} \sqrt 3 \pi a^3-\dfrac{1}{3} \pi a^3 =\dfrac{(\sqrt 3 -1)}{3} \pi a^3$
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