\documentclass[a4paper,12pt]{article}
\usepackage{epsfig}
\newcommand{\pa}{\partial}
\newcommand{\dci}{\rlap{$\displaystyle\int\!\!\!\int_{B_{a,b,c}}$}
{\hspace{2.2pt}\bigcirc} \ \ }
\newcommand{\un}{\underline}
\begin{document}
\parindent=0pt


\begin{center}
\textbf{Vector Calculus}

\textit{\textbf{Grad, Div and Curl Identities}}
\end{center}

\textbf{Question}

If $\un{F}$ is given by
$$\un{F} = xe^{2z} \un{i} + ye^{2z} \un{j} - e^{2z} \un{k},$$
show that $\un{F}$ is a solenoidal vector field. Find a vecotr field
for $\un{F}$.


\textbf{Answer}

Given $\un{F}$
$$\Rightarrow \textrm{div}\un{F} = e^{2z} + e{2z}-2 e^{2z} =0$$
so $\un{F}$ is solenoidal.

If $\un{F} = \nabla \times \un{G}$
\begin{eqnarray*}
\Rightarrow
\frac{\pa G_3}{\pa y} - \frac{\pa G_2}{\pa z} & = & xe^{2z}\\
\frac{\pa G_1}{\pa z} - \frac{\pa G_3}{\pa x} & = & ye^{2z}\\
\frac{\pa G_2}{\pa x} - \frac{\pa G_1}{\pa y} & = & -e^{2z}
\end{eqnarray*}
Find a solution with $G_2=0$.
$$\Rightarrow G_3 = \int x e^{2z} \,dy = xye^{2z} + M(x,z).$$
Try setting $M(x,z)=0$, $\Rightarrow G_3=xye^{2z}$. So now
\begin{eqnarray*}
\frac{\pa G_1}{\pa z} & = & ye^{2z} + \frac{\pa G_3}{\pa x} =
2ye^{2z}\\
\Rightarrow G_1 & =& \in 2ye^{2z} \,dz = ye^{2z} + N(x,y)\\
\textrm{As } -e^{2z} & = & =\frac{\pa G_1}{\pa y} = -e^{2z} -
\frac{\pa N}{\pa y},\\
\textrm{take } N(x,y) & = & 0
\end{eqnarray*}
So a vector potetnial for $\un{F}$ is given by
$$\un{G} = ye^{2z} \un{i} + xy e^{2z} \un{k}.$$

\end{document}