Answer
$\mathop \smallint \limits_{\cal C} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = 0$
Work Step by Step
We have $f\left( {x,y,z} \right) = xy\sin \left( {yz} \right)$. Since ${\bf{F}} = \nabla f$, by definition ${\bf{F}}$ is conservative.
Thus, by Theorem 1, the line integral over any path from $\left( {0,0,0} \right)$ to $\left( {1,1,\pi } \right)$ has the value
$\mathop \smallint \limits_{\cal C} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = f\left( {1,1,\pi } \right) - f\left( {0,0,0} \right) = 0 - 0 = 0$
