Answer
(a) we verify that ${\bf{F}} = \nabla f$
(b) $\mathop \smallint \limits_C {\bf{F}}\cdot{\rm{d}}{\bf{r}} = 78$
(c) $\mathop \smallint \limits_C {\bf{F}}\cdot{\rm{d}}{\bf{r}} = 4\ln 2$
(d) because ${\bf{F}}$ and $f$ contain the logarithmic function $\ln \left( {xy} \right)$, they are not defined for $x \le 0$ and $y \le 0$. Therefore, it is necessary to specify that the path lies in the region where $x$ and $y$ are positive.
Work Step by Step
(a) Using $f\left( {x,y,z} \right) = z\ln \left( {xy} \right)$, we evaluate
$\nabla f = \left( {\dfrac{{\partial f}}{{\partial x}},\dfrac{{\partial f}}{{\partial y}}} \right) = \left( {\dfrac{z}{{xy}}\cdot y,\dfrac{z}{{xy}}\cdot x,\ln \left( {xy} \right)} \right) = \left( {{x^{ - 1}}z,{y^{ - 1}}z,\ln \left( {xy} \right)} \right)$
Since ${\bf{F}}\left( {x,y,z} \right) = \left( {{x^{ - 1}}z,{y^{ - 1}}z,\ln \left( {xy} \right)} \right)$, so ${\bf{F}} = \nabla f$.
(b) In part (a) we verified that ${\bf{F}} = \nabla f$. By definition ${\bf{F}}$ is conservative.
Using ${\bf{r}}\left( t \right) = \left( {{{\rm{e}}^t},{{\rm{e}}^{2t}},{t^2}} \right)$, the values $t=1$ and $t=3$ correspond to the points $P = \left( {{\rm{e}},{{\rm{e}}^2},1} \right)$ and $Q = \left( {{{\rm{e}}^3},{{\rm{e}}^6},9} \right)$, respectively.
Thus, by Theorem 1, the line integral over any path for $1 \le t \le 3$ has the value
$\mathop \smallint \limits_C {\bf{F}}\cdot{\rm{d}}{\bf{r}} = f\left( {{{\rm{e}}^3},{{\rm{e}}^6},9} \right) - f\left( {{\rm{e}},{{\rm{e}}^2},1} \right)$
Since $f\left( {x,y,z} \right) = z\ln \left( {xy} \right)$, so
$\mathop \smallint \limits_C {\bf{F}}\cdot{\rm{d}}{\bf{r}} = 9\cdot\ln \left( {{{\rm{e}}^9}} \right) - 1\cdot\ln \left( {{{\rm{e}}^3}} \right) = 78$
(c) Since ${\bf{F}}$ is conservative, the line integral over any path from $P = \left( {\frac{1}{2},4,2} \right)$ to $Q = \left( {2,2,3} \right)$ has the value
$\mathop \smallint \limits_C {\bf{F}}\cdot{\rm{d}}{\bf{r}} = f\left( Q \right) - f\left( P \right) = f\left( {2,2,3} \right) - f\left( {\frac{1}{2},4,2} \right)$
Since $f\left( {x,y,z} \right) = z\ln \left( {xy} \right)$, so
$\mathop \smallint \limits_C {\bf{F}}\cdot{\rm{d}}{\bf{r}} = 3\cdot\ln \left( 4 \right) - 2\cdot\ln \left( 2 \right) = 6\ln 2 - 2\ln 2 = 4\ln 2$
(d) Because the definitions of ${\bf{F}}\left( {x,y,z} \right) = \left( {{x^{ - 1}}z,{y^{ - 1}}z,\ln \left( {xy} \right)} \right)$ and $f\left( {x,y,z} \right) = z\ln \left( {xy} \right)$ contain the logarithmic function $\ln \left( {xy} \right)$, they are not defined for $x \le 0$ and $y \le 0$. Therefore, it is necessary to specify that the path lies in the region where $x$ and $y$ are positive.
