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 970: 34

Answer

The rate of heat flow: $ - K\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} \nabla w\cdot{\rm{d}}{\bf{S}} = 16000\pi $

Work Step by Step

We are given the temperature function: $w\left( {x,y,z} \right) = 20 - 5\left( {{x^2} + {y^2} + {z^2}} \right)$ So, $\nabla w = \left( {\frac{{\partial w}}{{\partial x}},\frac{{\partial w}}{{\partial y}},\frac{{\partial w}}{{\partial z}}} \right) = \left( { - 10x, - 10y, - 10z} \right) = - 10\left( {x,y,z} \right)$ Since $K=400$, the rate of heat flow out of a sphere of radius $1$ is $ - K\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} \nabla w\cdot{\rm{d}}{\bf{S}} = 4000\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} \left( {x,y,z} \right)\cdot{\rm{d}}{\bf{S}}$ To evaluate this integral we use the spherical coordinates: $x = r\cos \theta \sin \phi $, ${\ \ \ }$ $y = r\sin \theta \sin \phi $, ${\ \ \ }$ $z = r\cos \phi $ By Eq. (2) in Section 17.4, the outward-pointing normal vector is ${\bf{N}}\left( {\theta ,\phi } \right) = \sin \phi {{\bf{e}}_r}$. The vector $\left( {x,y,z} \right)$ in spherical coordinates is ${\bf{r}} = r{{\bf{e}}_r}$. And since the radius is $1$, we have ${\bf{r}} = {{\bf{e}}_r}$. Thus, by Eq. (3) the rate of heat flow is $ - K\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} \nabla w\cdot{\rm{d}}{\bf{S}} = 4000\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} \left( {x,y,z} \right)\cdot{\rm{d}}{\bf{S}}$ $ = 4000\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\phi = 0}^\pi {{\bf{e}}_r}\cdot{\bf{N}}\left( {\theta ,\phi } \right){\rm{d}}\theta {\rm{d}}\phi $ $ = 4000\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\phi = 0}^\pi \sin \phi {\rm{d}}\theta {\rm{d}}\phi $ $ = 4000\mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta \mathop \smallint \limits_{\phi = 0}^\pi \sin \phi {\rm{d}}\phi $ $ = 8000\pi \left( { - \cos \phi } \right)|_0^\pi = 16000\pi $ So, the rate of heat flow is $ - K\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} \nabla w\cdot{\rm{d}}{\bf{S}} = 16000\pi $.
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