Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 15 - Section 15.8 - Triple Integrals in Cylindrical Coordinates - 15.8 Exercise - Page 1107: 46

Answer

$2.4 \times 10^{18} \ kilograms $

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, $\rho=\sqrt {x^2+y^2+z^2} \implies \rho^2=x^2+y^2+z^2$ Therefore, $Mass= \int_0^{\pi} \int_0^{2 \pi} \int_{6.370 \times 10^{6}}^{6.375 \times 10^{6}} (619.09 -0.00007 \rho) \rho^2 \sin \phi d\rho \ d\theta d \phi \\=\int_0^{\pi} \sin \phi d \phi \int_0^{2 \pi} d\theta \int_{6.370 \times 10^{6}}^{6.375 \times 10^{6}} (619.09 \rho^2 -0.00007 \rho^3 \ d\rho \\= 4 \pi [206.36 \rho^3 -0.00002425 \rho^4 ]_{6.370 \times 10^{6}}^{6.375 \times 10^{6}} $ Now, we will use a calculator. Thus, $Mass =4 \pi [206.36 \rho^3 -0.00002425 \rho^4 ]_{6.370 \times 10^{6}}^{6.375 \times 10^{6}} \approx 2.4 \times 10^{18} \ kilograms $
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