Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 15: Multiple Integrals - Section 15.7 - Triple Integrals in Cylindrical and Spherical Coordinates - Exercises 15.7 - Page 921: 69


$$ \bar{z}=\frac{M_{x y}}{M}=(\pi)\left(\frac{3}{8 \pi}\right)=\frac{3}{8},\ \bar{x}=\bar{y}=0 $$

Work Step by Step

Since \begin{align*} M&=\frac{8 \pi}{3} \\ M_{x y}&=\int_{0}^{2 \pi} \int_{\pi / 3}^{\pi / 2} \int_{0}^{2} z \rho^{2} \sin \phi d \rho d \phi d \theta\\ &=\int_{0}^{2 \pi} \int_{\pi / 3}^{\pi / 2} \int_{0}^{2} \rho^{3} \cos \phi \sin \phi d \rho d \phi d \theta\\ &=4 \int_{0}^{2 \pi} \int_{\pi / 3}^{\pi / 2} \cos \phi \sin \phi d \phi d \theta\\ &=4 \int_{0}^{2 \pi}\left[\frac{\sin ^{2} \phi}{2}\right]_{\pi / 3}^{\pi / 2} d \theta\\ &=4 \int_{0}^{2 \pi}\left(\frac{1}{2}-\frac{3}{8}\right) d \theta\\ &=\frac{1}{2} \int_{0}^{2 \pi} d \theta=\pi \end{align*} Then by symmetry $$ \bar{z}=\frac{M_{x y}}{M}=(\pi)\left(\frac{3}{8 \pi}\right)=\frac{3}{8},\ \bar{x}=\bar{y}=0 $$
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