Calculus (3rd Edition)

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

Chapter 17 - Line and Surface Integrals - 17.3 Conservative Vector Fields - Exercises - Page 945: 16

Answer

The potential function is $f\left( {x,y,z} \right) = z{{\rm{e}}^{xy}} - yz + C$, where $C$ is a constant.

Work Step by Step

1. Assuming that ${\bf{F}}$ is a vector field on a simply connected domain ${\cal D}$, we check if ${\bf{F}} = \left( {{F_1},{F_2},{F_3}} \right) = \left( {yz{{\rm{e}}^{xy}},xz{{\rm{e}}^{xy}} - z,{{\rm{e}}^{xy}} - y} \right)$ satisfies the cross-partials condition: $\dfrac{{\partial {F_1}}}{{\partial y}} = \dfrac{{\partial {F_2}}}{{\partial x}}$, ${\ \ \ }$ $\dfrac{{\partial {F_2}}}{{\partial z}} = \dfrac{{\partial {F_3}}}{{\partial y}}$, ${\ \ \ }$ $\dfrac{{\partial {F_3}}}{{\partial x}} = \dfrac{{\partial {F_1}}}{{\partial z}}$ We get $z{{\rm{e}}^{xy}} + xyz{{\rm{e}}^{xy}} = z{{\rm{e}}^{xy}} + xyz{{\rm{e}}^{xy}}$, $x{{\rm{e}}^{xy}} - 1 = x{{\rm{e}}^{xy}} - 1$, $y{{\rm{e}}^{xy}} = y{{\rm{e}}^{xy}}$ From these results, we conclude that ${\bf{F}} = \left( {{F_1},{F_2},{F_3}} \right) = \left( {yz{{\rm{e}}^{xy}},xz{{\rm{e}}^{xy}} - z,{{\rm{e}}^{xy}} - y} \right)$ satisfies the cross-partials condition. Therefore, by Theorem 4, there is a potential function for ${\bf{F}}$. 2. Find a potential function for ${\bf{F}}$. Let the potential function for ${\bf{F}}$ be $f\left( {x,y,z} \right)$ such that ${\bf{F}} = \nabla f = \left( {\dfrac{{\partial f}}{{\partial x}},\dfrac{{\partial f}}{{\partial y}},\dfrac{{\partial f}}{{\partial z}}} \right)$. So, a. taking the integral of $\dfrac{{\partial f}}{{\partial x}}$ with respect to $x$ gives $f\left( {x,y,z} \right) = \smallint yz{{\rm{e}}^{xy}}{\rm{d}}x = z{{\rm{e}}^{xy}} + j\left( {y,z} \right)$ b. taking the integral of $\dfrac{{\partial f}}{{\partial y}}$ with respect to $y$ gives $f\left( {x,y,z} \right) = \smallint \left( {xz{{\rm{e}}^{xy}} - z} \right){\rm{d}}y = z{{\rm{e}}^{xy}} - yz + g\left( {x,z} \right)$ c. taking the integral of $\dfrac{{\partial f}}{{\partial z}}$ with respect to $z$ gives $f\left( {x,y,z} \right) = \smallint \left( {{{\rm{e}}^{xy}} - y} \right){\rm{d}}z = z{{\rm{e}}^{xy}} - yz + h\left( {x,y} \right)$ Since the three ways of expressing $f\left( {x,y,z} \right)$ must be equal, we get $z{{\rm{e}}^{xy}} + j\left( {y,z} \right) = z{{\rm{e}}^{xy}} - yz + g\left( {x,z} \right) = z{{\rm{e}}^{xy}} - yz + h\left( {x,y} \right)$ From here, we conclude that the potential function is $f\left( {x,y,z} \right) = z{{\rm{e}}^{xy}} - yz + C$, where $C$ is a constant.
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