<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\pl}{\partial}
\begin{document}
\parindent=0pt


{\bf Question}

\begin{description}
\item[(a)] Prove that the force field

$$ {\bf F}  = (y^2 -2xyz^3) {\bf i} + (3+2xy - x^2z^3){\bf j}
 + (6z^3 - 3x^2yz^2){\bf k} $$ is conservative.
\item[(b)] Find the potential $U(x,y,z)$ associated with the force
field.
\item[(c)] Find the work done by the field in moving a particle
from the point (2,-1,2) to (-1,3,-2)
\end{description}

\vspace{.25in}

{\bf Answer}


\begin{description}
\item[(a)]
\begin{eqnarray*} \nabla \times {\bf F} &amp; = &amp; \left|
\begin{array}{ccc} {\bf i} &amp; {\bf j} &amp; {\bf k} \\ \frac{\pl}{\pl
x} &amp; \frac{\pl}{\pl y} &amp; \frac{\pl}{\pl z} \\ y^2 -2xyz^3 &amp; 3+2xy
- x^2z^3 &amp; 6z^3 - 3x^2yz^2 \end{array} \right| \\ &amp; = &amp; {\bf
i}(-3x^2z^2 - 3x^2z^2) - {\bf j} (6xyz^2 - 6xyz^2) \\ &amp; &amp; + {\bf
k}(2y - 2x^2z^3 - 2y +2xz^3) \\ &amp; = &amp; 0 \end{eqnarray*}

Therefore ${\bf F}$ is conservative.

\item[(b)]
$$ {\bf F} = -\nabla U = -{\bf i} \frac{\pl U}{\pl x} - {\bf
j}\frac{\pl U}{\pl y} - {\bf k}\frac{\pl U}{\pl z} $$

Equating components:

$\begin{array}{rclr} \ds-\frac{\pl U}{\pl x} &amp; = &amp; y^2x - 2xyz^3 &amp;
 \\ \Rightarrow  U &amp; = &amp; -y^2 + x^2yz^3 +f_1(y,z) &amp;(1) \\ \\ \ds -\frac{\pl
U}{\pl y} &amp; = &amp;  3 + 2xy - x^2z^3 \\ \Rightarrow  U &amp; = &amp; -3y +
x^2y^2 - yx^2z^2 +f_2(x,z) &amp; (2) \\ \\ \ds-\frac{\pl U}{\pl z} &amp; =
&amp; 6z^3 - 3x^2yz^2 \\ \Rightarrow  U &amp; = &amp; -\frac{3}{2}z^4 +
x^2yz^3 +f_3(x,z) &amp; (3) \end{array}$

Comparing:

$\begin{array}{lcrclrcl}  (1) {\rm \ and\ } (2) &amp; \Rightarrow &amp;
f_1(y,z) &amp; = &amp; -3y + g(z), &amp; f_2 &amp; = &amp; g(z) \\  (2) {\rm \ and\ }
(3) &amp; \Rightarrow &amp; g(z) &amp; = &amp; -\frac{3}{2}z^4, &amp; f_3 &amp; = &amp; -3y -
xy^2 \end{array}$

Therefore $U = -y^2x + x^2yz^3 - 3y -\frac{3}{2}z^4 (+ {\rm
constant})$

\item[(c)]
Work done is the difference in $-U$

\begin{eqnarray*} U(2,-1,2) &amp; = &amp; (-1)^2  2 + 2^2 (-1)
 2^2 - 3 (-1) \frac{3}{2}  2^4 \\ &amp; = &amp; -2 -32
+ 3 - 42 \\ &amp; = &amp; -55 \\ U(-1,3,-2) &amp; = &amp; (-9)  (-1) + (-1)^2
 3  (-2)^3  - 3 3 \frac{3}{2}  (-2)^4 \\
&amp; = &amp; -9 -24 - 9 - 24 \\ &amp; = &amp; -48 \end{eqnarray*}

Work done is $-(-48 - -55) = -7$

\end{description}



\end{document}
</pre></body></html>