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\newcommand{\ds}{\displaystyle}
\newcommand{\pl}{\partial}
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{\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} & = & \left|
\begin{array}{ccc} {\bf i} & {\bf j} & {\bf k} \\ \frac{\pl}{\pl
x} & \frac{\pl}{\pl y} & \frac{\pl}{\pl z} \\ y^2 -2xyz^3 & 3+2xy
- x^2z^3 & 6z^3 - 3x^2yz^2 \end{array} \right| \\ & = & {\bf
i}(-3x^2z^2 - 3x^2z^2) - {\bf j} (6xyz^2 - 6xyz^2) \\ & & + {\bf
k}(2y - 2x^2z^3 - 2y +2xz^3) \\ & = & 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:

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

Comparing:

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

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

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

\end{description}



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