\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

$\un{F}(x,y,z) = x\un{i} -2y\un{j}+3z\un{k}$


\textbf{Answer}

$F_1=x$, $F_2=-2y$, $F_3=3z$.
\begin{eqnarray*}
\frac{\pa F_1}{\pa y} = & 0 & = \frac{\pa F_2}{\pa x}\\
\frac{\pa F_1}{\pa z} = & 0 & = \frac{\pa F_3}{\pa x}\\
\frac{\pa F_2}{\pa_z} = & 0 & = \frac{\pa F_3}{\pa y}
\end{eqnarray*}
$\Rightarrow \un{F}$ may be conservative.

If $\un{F}=\nabla \phi$
$$\Rightarrow \frac{\pa \phi}{\pa x} = x, \ \ \ \frac{\pa \phi}{\pa
y}=-2y, \ \ \ \frac{\pa \phi}{\pa z} =3z.$$
So $\phi(x,y,z) = \frac{x^2}{2} - y^2 + \frac{3z^2}{2}$ is a potential
for $\un{F}$, and so $\un{F}$ is conservative on $\Re^3$.

\end{document}