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\usepackage{epsfig}
\newcommand{\pa}{\partial}
\newcommand{\ds}{\displaystyle}
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\begin{center}
\textbf{Vector Fields}

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

\textbf{Question}

The vector field $\un{F}$ is given by
$$\un{F}(x,y,z) = \frac{2x}{z}\un{i} + \frac{2y}{z}\un{j} -
\frac{x^2+y^2}{z^2} \un{k}.$$
Show that $\un{F}$ is conservative, and find the potential. Describe
the equipotential surfaces and find the field lines of $\un{F}$.


\textbf{Answer}

\begin{eqnarray*}
F_1 &amp; = &amp; \frac{2x}{z}\\
F_2 &amp; = &amp; \frac{2y}{z}\\
F_3 &amp; = &amp; -\frac{x^2+y^2}{z^2}
\end{eqnarray*}
This gives
\begin{eqnarray*}
\frac{\pa F_1}{\pa y} &amp; = &amp; 0 = \frac{\pa F_2}{\pa x}\\
\frac{\pa F_1}{\pa z} &amp; = &amp; -\frac{2x}{z^2} = \frac{\pa F_3}{\pa x}\\
\frac{\pa F_2}{\pa z} &amp; = &amp; -\frac{2y}{z^2} = \frac{\pa F_3}{\pa y}.
\end{eqnarray*}
$\Rightarrow \un{F}$ can be conservative in $\Re^3$ except on the
plane $z=0$ where it is not defined. If $\un{F} = \nabla \phi$
\begin{eqnarray*}
\Rightarrow \frac{\pa \phi}{\pa x} &amp; = &amp;\frac{2x}{z}\\
\frac{\pa \phi}{\pa y} &amp; = &amp; \frac{2y}{z}\\
\frac{\pa \phi}{\pa z} &amp; = &amp; -\frac{x^2+y^2}{z^2}\\
\Rightarrow \phi(x,y,z) &amp; = &amp; \int \frac{2x}{z} \,dx\\
&amp; = &amp; \frac{x^2}{z} + C_1(y,z)\\
\frac{2y}{z} = \frac{\pa \phi}{\pa y} = \frac{\pa C_1}{\pa y}\\
\Rightarrow C_1(y,z) &amp; = &amp; \frac{y^2}{z} + C_2(z)\\
\phi(x,y,z) &amp; = &amp; \frac{x^2+y^2}{z} + C_2(z)\\
-\frac{x^2+y^2}{z^2} &amp; = &amp; \frac{\pa \phi}{\pa z} =
-\frac{x^2+y^2}{z^2} + C_2'(z)\\
\Rightarrow C_2(z) &amp; = &amp; 0
\end{eqnarray*}
So $\ds \phi(x,y,z) = \frac{x^2+y^2}{z}$ is a potential for $\un{F}$,
and $\un{F}$ is conservative on $\Re^3$, except where it is not
defined on ($z=0$).

The equipotential surfaces will have the equations $\ds
\frac{x^2+y^2}{z}=C$ or $Cz = x^2+y^2$. Therefore the surfaces are
circular paraboloids.

The field lines of $\un{F}$ satisfy
$$\frac{dx}{\frac{2x}{z}} = \frac{dy}{\frac{2y}{z}} =
\frac{dz}{-\frac{x^2+y^2}{z^2}}$$
So it can be seen that $\frac{dx}{x}=\frac{dy}{y}$, $\Rightarrow y=Ax$
for an arbitrary constant $A$.
\begin{eqnarray*}
\Rightarrow \frac{dx}{2x} &amp; = &amp; \frac{z \,dz}{-(x^2+y^2)}\\
&amp; = &amp; \frac{z \,dz}{-x^2(1+A^2)}\\
\Rightarrow -(1+a^2)x \,dx &amp; = &amp; 2z \,dz.
\end{eqnarray*}
And so 
$$\frac{1+A^2}{2}x^2 + z^2 = \frac{B}{2}$$
or
$$x^2+y^2+2z^2 = B$$
with $B$ being a second arbitrary constant. So the field lines of
$\un{F}$ are the ellipses in which the vertical planes containing the
$z$-axis intersects the ellipsoids $x^2+y^2+2z^2 = B$. These are
orthogonal to all the equipotential surfaces of $\un{F}$.


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