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: 23

Answer

The flow rate is $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{v}}\cdot{\rm{d}}{\bf{S}} = \frac{2}{3}\pi $ ${m^3}/s$

Work Step by Step

In spherical coordinates: $x = r\cos \theta \sin \phi $, ${\ \ \ }$ $y = r\sin \theta \sin \phi $, ${\ \ \ }$ $z = r\cos \phi $ So, ${{\bf{e}}_r} = \left( {\frac{x}{r},\frac{y}{r},\frac{z}{r}} \right) = \left( {\cos \theta \sin \phi ,\sin \theta \sin \phi ,\cos \phi } \right)$. The upper hemisphere of ${x^2} + {y^2} + {z^2} = 1$ can be parametrized by $G\left( {\theta ,\phi } \right) = \left( {\cos \theta \sin \phi ,\sin \theta \sin \phi ,\cos \phi } \right)$ for $0 \le \theta \le 2\pi $, $0 \le \phi \le \frac{\pi }{2}$. So, ${{\bf{T}}_\theta } = \frac{{\partial G}}{{\partial \theta }} = \left( { - \sin \theta \sin \phi ,\cos \theta \sin \phi ,0} \right)$ ${{\bf{T}}_\phi } = \frac{{\partial G}}{{\partial \phi }} = \left( {\cos \theta \cos \phi ,\sin \theta \cos \phi , - \sin \phi } \right)$ By Eq. (2) in Section 17.4, the outward-pointing normal vector is ${\bf{N}}\left( {\theta ,\phi } \right) = \sin \phi {{\bf{e}}_r}$. Since $0 \le \phi \le \frac{\pi }{2}$, the normal vector ${\bf{N}}\left( {\theta ,\phi } \right)$ is positive value, thus it is outward-pointing. Evaluate: ${\bf{v}}\left( {G\left( {\theta ,\phi } \right)} \right) = \cos \phi {\bf{k}}$ ${\bf{v}}\left( {G\left( {\theta ,\phi } \right)} \right)\cdot{\bf{N}}\left( {\theta ,\phi } \right) = \left( {\cos \phi {\bf{k}}} \right)\cdot\left( {\sin \phi {{\bf{e}}_r}} \right) = \cos \phi \sin \phi {\bf{k}}\cdot{{\bf{e}}_r}$ Since ${{\bf{e}}_r} = \left( {\cos \theta \sin \phi ,\sin \theta \sin \phi ,\cos \phi } \right)$, so ${\bf{v}}\left( {G\left( {\theta ,\phi } \right)} \right)\cdot{\bf{N}}\left( {\theta ,\phi } \right) = {\cos ^2}\phi \sin \phi $ We calculate the flow rate through the upper hemisphere by Eq. (3): $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{v}}\cdot{\rm{d}}{\bf{S}} = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{v}}\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{v}}\cdot{\rm{d}}{\bf{S}} = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\phi = 0}^{\pi /2} {\cos ^2}\phi \sin \phi {\rm{d}}\theta {\rm{d}}\phi $ $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{v}}\cdot{\rm{d}}{\bf{S}} = \mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta \mathop \smallint \limits_{\phi = 0}^{\pi /2} {\cos ^2}\phi \sin \phi {\rm{d}}\phi $ Write $t = \cos \phi $. So, $dt = - \sin \phi d\phi $. The integral becomes $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{v}}\cdot{\rm{d}}{\bf{S}} = - 2\pi \mathop \smallint \limits_{t = 1}^0 {t^2}{\rm{d}}t$ $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{v}}\cdot{\rm{d}}{\bf{S}} = - 2\pi \left( {\frac{1}{3}{t^3}} \right)|_1^0 = \frac{2}{3}\pi $ So, the flow rate is $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{v}}\cdot{\rm{d}}{\bf{S}} = \frac{2}{3}\pi $ ${m^3}/s$.
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