\documentclass[a4paper,12pt]{article}
\usepackage{epsfig}
\newcommand{\pa}{\partial}
\newcommand{\ds}{\displaystyle}
\newcommand{\un}{\underline}
\begin{document}
\parindent=0pt

\begin{center}
\textbf{Vector Fields}

\textit{\textbf{Conservative Fields}}
\end{center}

\textbf{Question}

For the following vector field, find whether it is conservative. If
so, find a corresponding potential

$\un{F}(x,y,z) = (2xy-z^2)\un{i} +(2yz+x^2)\un{j} - (2zx - y^2)\un{k}$


\textbf{Answer}

\begin{eqnarray*}
\Rightarrow F_1 & = & 2xy-z^2\\
F_2 & = & 2yz + x^2\\
F_3 & = & y^2-2zx\\
\Rightarrow \frac{\pa F_1}{\pa y} & = & 2x = \frac{\pa F_2}{\pa x}\\
\frac{\pa F_1}{\pa z} & = & -2z \frac{\pa F_3}{\pa x}\\
\frac{\pa F_2}{\pa z} & = & 2y = \frac{\pa F_3}{\pa y}
\end{eqnarray*}
$\Rightarrow \un{F}$ can be conservative

If $\un{F} = \nabla \phi$
\begin{eqnarray*}
\Rightarrow \frac{\pa \phi}{\pa x} & = & 2xy-z^2\\
\frac{\pa \phi}{\pa y} & = & 2yz + x^2\\
\frac{\pa \phi}{\pa z} & = & y^2-2zx\\
\Rightarrow \phi(x,y,z) & = & \int (2xy - z^2) \,dx\\
& = & x^2y - xz^2 + C_1(y,z)\\
2yx+x^2 & = & \frac{\pa \phi}{\pa y} = x^2 + \frac{\pa C_1}{\pa y}\\
\Rightarrow \frac{\pa C_1}{\pa y} & = & 2yz\\
\Rightarrow C_1(y,z) & = & y^2z+ C_2(z)\\
& & \\
\phi (x,y,z) & = & x^y-xz^2+y^2z+C_2(z)\\
y^2-2zx & = & \frac{\pa \phi}{\pa z} = -2xz + y^2 +C_2'(z)\\
\Rightarrow C_2'(z) & = & 0.
\end{eqnarray*}
So $\phi (x,y,z) = x^2y-xz^2+y^2z$ is a scalar potential for $\un{F}$,
and $\un{F}$ is conservative on $\Re^3$.

\end{document}