Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 17 - Line and Surface Integrals - 17.4 Parametrized Surfaces and Surface Integrals - Exercises - Page 958: 28

Answer

The equation is symmetric in terms of $x$, $y$, and $z$. Therefore, we conclude that $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {x^2}{\rm{d}}S = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {y^2}{\rm{d}}S = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {z^2}{\rm{d}}S$ We show that $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {x^2}{\rm{d}}S = \frac{4}{3}\pi {R^4}$

Work Step by Step

The sphere $S$ of radius $R$ centered at the origin has equation: ${x^2} + {y^2} + {z^2} = {R^2}$ The equation is symmetric in terms of $x$, $y$, and $z$. Therefore, we conclude that $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {x^2}{\rm{d}}S = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {y^2}{\rm{d}}S = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {z^2}{\rm{d}}S$ Next, we evaluate $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} \left( {{x^2} + {y^2} + {z^2}} \right){\rm{d}}S$. In spherical coordinates the integral becomes $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} \left( {{x^2} + {y^2} + {z^2}} \right){\rm{d}}S = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {R^2}{\rm{d}}S = {R^2}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\rm{d}}S$ The surface area of the sphere is $4\pi {R^2}$. Therefore, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} \left( {{x^2} + {y^2} + {z^2}} \right){\rm{d}}S = 4\pi {R^4}$ $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {x^2}{\rm{d}}S + \mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {y^2}{\rm{d}}S + \mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {z^2}{\rm{d}}S = 4\pi {R^4}$ Since $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {x^2}{\rm{d}}S = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {y^2}{\rm{d}}S = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {z^2}{\rm{d}}S$, so $3\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {x^2}{\rm{d}}S = 4\pi {R^4}$ Hence, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {x^2}{\rm{d}}S = \frac{4}{3}\pi {R^4}$.
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