Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 15 - Section 15.8 - Triple Integrals in Cylindrical Coordinates - 15.8 Exercise - Page 1050: 24



Work Step by Step

In the spherical coordinates system, we have $x=\rho \sin \phi \cos \theta; y=\rho \sin \phi \sin \theta;z=\rho \cos \phi$ and $\rho=\sqrt {x^2+y^2+z^2} \implies \rho^2=x^2+y^2+z^2$ Now, consider $I=\int_0^{\pi} \int_{0}^{\pi}\int_{0}^{3} (\rho^2 \sin^2 \phi \sin^2 \theta) \times (\rho^2 \sin \phi) d\rho d\theta d\phi=\int_0^{\pi} \int_{0}^{\pi} [\dfrac{\rho^5}{5}\sin^3 \phi \sin^2 \theta]_0^3 d\theta d\phi$ $= \dfrac{243}{5} \int_0^{\pi} \int_0^{\pi}( \sin^3 \phi) (\sin^2 \theta) d\theta d\phi$ $=\int_0^{\pi} \int_0^{\pi} (\dfrac{243}{5}) ( \sin^3 \phi) \times (\dfrac{1}{2}-\dfrac{1}{2} \cos 2 \theta) d\theta d\phi$ $=(\dfrac{243 \pi}{10} )\int_0^{\pi} \sin \phi -\sin \phi\cos^2 \phi$ $=\dfrac{162\pi}{5}$
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