Chapter 17 - Line and Surface Integrals - Chapter Review Exercises - Page 971: 25

The vector field is conservative. The potential function is $f\left( {x,y,z} \right) = y\ln \left( {{x^2} + z} \right) + C$, where $C$ is a constant.

We have ${\bf{F}}\left( {x,y,z} \right) = \left( {\frac{{2xy}}{{{x^2} + z}},\ln \left( {{x^2} + z} \right),\frac{y}{{{x^2} + z}}} \right)$. 1. We check if ${\bf{F}}$ satisfies the cross-partials condition: $\frac{{\partial {F_1}}}{{\partial y}} = \frac{{\partial {F_2}}}{{\partial x}}$, ${\ \ \ \ }$ $\frac{{\partial {F_2}}}{{\partial z}} = \frac{{\partial {F_3}}}{{\partial y}}$, ${\ \ \ \ }$ $\frac{{\partial {F_3}}}{{\partial x}} = \frac{{\partial {F_1}}}{{\partial z}}$ We get $\frac{{2x}}{{{x^2} + z}} = \frac{{2x}}{{{x^2} + z}}$, ${\ \ \ \ }$ $\frac{1}{{{x^2} + z}} = \frac{1}{{{x^2} + z}}$, ${\ \ \ \ }$ $ - \frac{{2xy}}{{{{\left( {{x^2} + z} \right)}^2}}} = - \frac{{2xy}}{{{{\left( {{x^2} + z} \right)}^2}}}$ From these results, we conclude that ${\bf{F}} = \left( {\frac{{2xy}}{{{x^2} + z}},\ln \left( {{x^2} + z} \right),\frac{y}{{{x^2} + z}}} \right)$ satisfies the cross-partials condition. Therefore, by Theorem 4 in Section 17.3, ${\bf{F}}$ is conservative. Thus, 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 ${\bf{F}} = \nabla f = \left( {\frac{{\partial f}}{{\partial x}},\frac{{\partial f}}{{\partial y}},\frac{{\partial f}}{{\partial z}}} \right)$. So, a. taking the integral of $\frac{{\partial f}}{{\partial x}}$ with respect to $x$ gives $f\left( {x,y,z} \right) = \smallint \frac{{2xy}}{{{x^2} + z}}{\rm{d}}x = y\ln \left( {{x^2} + z} \right) + m\left( {y,z} \right)$ b. taking the integral of $\frac{{\partial f}}{{\partial y}}$ with respect to $y$ gives $f\left( {x,y,z} \right) = \smallint \ln \left( {{x^2} + z} \right){\rm{d}}y = y\ln \left( {{x^2} + z} \right) + n\left( {x,z} \right)$ c. taking the integral of $\frac{{\partial f}}{{\partial z}}$ with respect to $z$ gives $f\left( {x,y,z} \right) = \smallint \frac{y}{{{x^2} + z}}{\rm{d}}z = y\ln \left( {{x^2} + z} \right) + o\left( {x,y} \right)$ Since the three ways of expressing $f\left( {x,y,z} \right)$ must be equal, we get $y\ln \left( {{x^2} + z} \right) + m\left( {y,z} \right) = y\ln \left( {{x^2} + z} \right) + n\left( {x,z} \right) = y\ln \left( {{x^2} + z} \right) + o\left( {x,y} \right)$ From here, we conclude that the potential function is $f\left( {x,y,z} \right) = y\ln \left( {{x^2} + z} \right) + C$, where $C$ is a constant.