Answer
We verify that ${\bf{F}} = \nabla f$.
$\mathop \smallint \limits_{\cal C} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = 32{\rm{e}} - 1$
Work Step by Step
1. We have ${\bf{F}}\left( {x,y,z} \right) = y{{\rm{e}}^z}{\bf{i}} + x{{\rm{e}}^z}{\bf{j}} + xy{{\rm{e}}^z}{\bf{k}}$ and $f\left( {x,y,z} \right) = xy{{\rm{e}}^z}$.
We evaluate
$\nabla f = \left( {\frac{{\partial f}}{{\partial x}},\frac{{\partial f}}{{\partial y}},\frac{{\partial f}}{{\partial z}}} \right) = \left( {y{{\rm{e}}^z},x{{\rm{e}}^z},xy{{\rm{e}}^z}} \right)$
Since ${\bf{F}}\left( {x,y,z} \right) = y{{\rm{e}}^z}{\bf{i}} + x{{\rm{e}}^z}{\bf{j}} + xy{{\rm{e}}^z}{\bf{k}}$, so ${\bf{F}} = \nabla f$.
2. Evaluate the line integral of ${\bf{F}}$ over the path ${\bf{r}}\left( t \right) = \left( {{t^2},{t^3},t - 1} \right)$ for $1 \le t \le 2$.
For $t=1$ and $t=2$, we get the points $P = \left( {1,1,0} \right)$ and $Q = \left( {4,8,1} \right)$, respectively. Since ${\bf{F}} = \nabla f$, so ${\bf{F}}$ is conservative. By Theorem 1,
$\mathop \smallint \limits_{\cal C} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = f\left( Q \right) - f\left( P \right) = f\left( {4,8,1} \right) - f\left( {1,1,0} \right) = 32{\rm{e}} - 1$
