Calculus (3rd Edition)

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

Chapter 17 - Line and Surface Integrals - 17.5 Surface Integrals of Vector Fields - Exercises - Page 968: 22

Answer

The flux is $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{F}}\cdot{\rm{d}}{\bf{S}} = - 4$

Work Step by Step

The ellipsoid ${\left( {\frac{x}{4}} \right)^2} + {\left( {\frac{y}{3}} \right)^2} + {\left( {\frac{z}{2}} \right)^2} = 1$ can be parametrized by $G\left( {\theta ,\phi } \right) = \left( {4\cos \theta \sin \phi ,3\sin \theta \sin \phi ,2\cos \phi } \right)$ So, ${{\bf{T}}_\theta } = \left( { - 4\sin \theta \sin \phi ,3\cos \theta \sin \phi ,0} \right)$ ${{\bf{T}}_\phi } = \left( {4\cos \theta \cos \phi ,3\sin \theta \cos \phi , - 2\sin \phi } \right)$ ${\bf{N}}\left( {\theta ,\phi } \right) = \left| {\begin{array}{*{20}{c}} {\bf{i}}&{\bf{j}}&{\bf{k}}\\ { - 4\sin \theta \sin \phi }&{3\cos \theta \sin \phi }&0\\ {4\cos \theta \cos \phi }&{3\sin \theta \cos \phi }&{ - 2\sin \phi } \end{array}} \right|$ $ = \left( { - 6\cos \theta {{\sin }^2}\phi } \right){\bf{i}} - \left( {8\sin \theta {{\sin }^2}\phi } \right){\bf{j}} + \left( { - 12\cos \phi \sin \phi } \right){\bf{k}}$ Since $x,y,z \le 0$, the portion of $S$ is in the range: $\pi \le \theta \le \frac{{3\pi }}{2}$ and $\frac{\pi }{2} \le \phi \le \pi $. In this range the $z$-component of ${\bf{N}}\left( {\theta ,\phi } \right)$ is always positive, thus it is upward-pointing normal. So, ${\bf{F}}\left( {G\left( {\theta ,\phi } \right)} \right) = 2\cos \phi {\bf{i}}$ ${\bf{F}}\left( {G\left( {\theta ,\phi } \right)} \right)\cdot{\bf{N}}\left( {\theta ,\phi } \right) = - 12\cos \theta \cos \phi {\sin ^2}\phi $ We evaluate the flux of ${\bf{F}} = z{\bf{i}}$ over the portion of $S$ using Eq. (3): $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{F}}\cdot{\rm{d}}{\bf{S}} = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{F}}\left( {G\left( {\theta ,\phi } \right)} \right)\cdot{\bf{N}}\left( {\theta ,\phi } \right){\rm{d}}\theta {\rm{d}}\phi $ $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{F}}\cdot{\rm{d}}{\bf{S}} = - 12\mathop \smallint \limits_{\theta = \pi }^{3\pi /2} \mathop \smallint \limits_{\phi = \pi /2}^\pi \cos \theta \cos \phi {\sin ^2}\phi {\rm{d}}\theta {\rm{d}}\phi $ $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{F}}\cdot{\rm{d}}{\bf{S}} = - 12\mathop \smallint \limits_{\theta = \pi }^{3\pi /2} \cos \theta {\rm{d}}\theta \mathop \smallint \limits_{\phi = \pi /2}^\pi \cos \phi {\sin ^2}\phi {\rm{d}}\phi $ Write $t = \sin \phi $. So, $dt = \cos \phi d\phi $. The integral becomes $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{F}}\cdot{\rm{d}}{\bf{S}} = - 12\left( {\sin \theta |_\pi ^{3\pi /2}} \right)\mathop \smallint \limits_{t = 1}^0 {t^2}{\rm{d}}t$ $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{F}}\cdot{\rm{d}}{\bf{S}} = - 12\left( { - 1} \right)\left( {\frac{1}{3}{t^3}} \right)|_1^0 = - 4$ So, the flux is $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{F}}\cdot{\rm{d}}{\bf{S}} = - 4$.
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