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\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

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


\textbf{Answer}

\begin{eqnarray*}
\Rightarrow \frac{\pa F_1}{\pa y} = &  1 & = \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*}
So $\un{F}$ may be conservative.

If $\un{F} = \nabla \phi$
$$\Rightarrow \frac{\pa \phi}{\pa x}=y, \ \ \ \frac{\pa \phi}{\pa
y}=x, \ \ \ \frac{\pa \phi}{\pa z}=z^2.$$
\begin{eqnarray*}
\Rightarrow \phi (x,y,z) & = & \in y \,dx = xy+C_1(y,z)\\
x & = & \frac{\pa \phi}{\pa y} = x + \frac{\pa C_1}{\pa y}\\
\Rightarrow \frac{\pa C_1}{\pa y} & = & 0\\
C_1(y,z) & = & C_2(z), \ \ \ \phi(x,y,z) = xy+C_2(z)\\
z^2 & = & \frac{\pa \phi}{\pa z} = C_2'(z)\\
\Rightarrow C_2(z0 & = & \frac{z^3}{3}.
\end{eqnarray*}
So $\phi (x,y,z) = xy + \frac{z^3}{3}$ is a potential for $\un{F}$,
and $\un{F}$ is conservative on $\Re^3$.


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