Answer
${\bf{F}} = \nabla f = \left( {y{{\rm{e}}^z},x{{\rm{e}}^z},xy{{\rm{e}}^z}} \right)$
(a) $\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = 2$
(b) $\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = 0$
Work Step by Step
We have $f\left( {x,y,z} \right) = xy{{\rm{e}}^z}$. So,
${\bf{F}} = \nabla f = \left( {\frac{{\partial f}}{{\partial x}},\frac{{\partial f}}{{\partial y}},\frac{{\partial f}}{{\partial z}}} \right) = \left( {y{{\rm{e}}^z},x{{\rm{e}}^z},xy{{\rm{e}}^z}} \right)$
(a) Since there is a function $f\left( {x,y,z} \right) = xy{{\rm{e}}^z}$ such that ${\bf{F}} = \nabla f = \left( {y{{\rm{e}}^z},x{{\rm{e}}^z},xy{{\rm{e}}^z}} \right)$, by definition, ${\bf{F}}$ is conservative. Thus, the line integral of ${\bf{F}}$ along any curve from $\left( {1,1,0} \right)$ to $\left( {3,{\rm{e}}, - 1} \right)$ depends only on the endpoints, that is
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = f\left( {3,{\rm{e}}, - 1} \right) - f\left( {1,1,0} \right) = 3 - 1 = 2$
So, $\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = 2$.
(b) From part (a) we know that ${\bf{F}}$ is conservative. Since the circulation is around a closed curve, by Theorem 1 of Section 17.3:
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = 0$
