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: 26

Answer

$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} x{\rm{d}}S = \frac{1}{2}\sqrt 2 \pi $

Work Step by Step

The surface is a unit sphere centered at the origin, so we can parametrize it by $G\left( {\theta ,\phi } \right) = \left( {\cos \theta \sin \phi ,\sin \theta \sin \phi ,\cos \phi } \right)$. ${{\bf{T}}_\theta } = \frac{{\partial G}}{{\partial \theta }} = \left( {\sin \theta \sin \phi ,\cos \theta \sin \phi ,0} \right)$ ${{\bf{T}}_\phi } = \frac{{\partial G}}{{\partial \phi }} = \left( {\cos \theta \cos \phi ,\sin \theta \cos \phi , - \sin \phi } \right)$ By Eq. (2), the normal vector is ${\bf{N}}\left( {\theta ,\phi } \right) = \sin \phi {{\bf{e}}_r}$. So, $||{\bf{N}}\left( {\theta ,\phi } \right)|| = \sin \phi $. From $\left| y \right| \le x$, we obtain $ - x \le y \le x$. In spherical coordinates we get $ - \cos \theta \sin \phi \le \sin \theta \sin \phi \le \cos \theta \sin \phi $ Dividing the inequality above by $\cos \theta \sin \phi $ gives $ - 1 \le \tan \theta \le 1$ $ - \frac{\pi }{4} \le \theta \le \frac{\pi }{4}$ Thus, the domain description is ${\cal D} = \left\{ {\left( {\theta ,\phi } \right): - \frac{\pi }{4} \le \theta \le \frac{\pi }{4},0 \le \phi \le \pi } \right\}$. Since $f\left( {x,y,z} \right) = x$, so $f\left( {G\left( {\theta ,\phi } \right)} \right) = \cos \theta \sin \phi $. By Eq. (7) of Theorem 1, we have $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} f\left( {x,y,z} \right){\rm{d}}S = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {G\left( {\theta ,\phi } \right)} \right)||{\bf{N}}\left( {\theta ,\phi } \right)||{\rm{d}}\theta {\rm{d}}\phi $ $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} f\left( {x,y,z} \right){\rm{d}}S = \mathop \smallint \limits_{\theta = - \pi /4}^{\pi /4} \mathop \smallint \limits_{\phi = 0}^\pi \cos \theta {\sin ^2}\phi {\rm{d}}\theta {\rm{d}}\phi $ Using the identity $\cos 2\phi = 1 - 2{\sin ^2}\phi $, we get $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} f\left( {x,y,z} \right){\rm{d}}S = \left( {\mathop \smallint \limits_{\theta = - \pi /4}^{\pi /4} \cos \theta {\rm{d}}\theta } \right)\left[ {\frac{1}{2}\mathop \smallint \limits_{\phi = 0}^\pi \left( {1 - \cos 2\phi } \right){\rm{d}}\phi } \right]$ $ = \frac{1}{2}\left( {\sin \theta |_{ - \pi /4}^{\pi /4}} \right)\left( {\phi - \frac{1}{2}\sin \phi } \right)|_0^\pi = \frac{1}{2}\sqrt 2 \pi $ So, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} x{\rm{d}}S = \frac{1}{2}\sqrt 2 \pi $.
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