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 1051: 47

Answer

$\dfrac{136 \pi}{99}$

Work Step by Step

Apply the spherical coordinates system as: $x=\rho \sin \phi \cos \theta \\ y=\rho \sin \phi \sin \theta \\z=\rho \cos \phi$; So, $\rho=\sqrt {x^2+y^2+z^2} \implies \rho^2=x^2+y^2+z^2$ The equation of a bumpy sphere is $ \rho=1+\dfrac{1}{5} \sin 6 \theta \sin 5 \phi$ Therefore, $Volume= \iiint_{V} dV=\int_0^{\pi} \int_0^{2 \pi} \int_{0}^{1+\dfrac{1}{5} \sin 6 \theta \sin 5 \phi} \rho^2 \sin \phi d\rho \ d\theta d \phi $ Now, we will use a calculator. Thus, $Volume=\int_0^{\pi} \int_0^{2 \pi} \int_{0}^{1+\dfrac{1}{5} \sin 6 \theta \sin 5 \phi} \rho^2 \sin \phi d\rho \ d\theta d \phi \approx \dfrac{136 \pi}{99}$
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