Answer
We verify that ${\bf{F}} = \nabla f$.
$\mathop \smallint \limits_{\cal C} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = - 2$
Work Step by Step
1. We have ${\bf{F}}\left( {x,y} \right) = \left( {\cos y, - x\sin y} \right)$ and $f\left( {x,y} \right) = x\cos y$.
Evaluate
$\nabla f = \left( {\frac{{\partial f}}{{\partial x}},\frac{{\partial f}}{{\partial y}}} \right) = \left( {\cos y, - x\sin y} \right)$
Since ${\bf{F}}\left( {x,y} \right) = \left( {\cos y, - x\sin y} \right)$, so ${\bf{F}} = \nabla f$.
2. For the upper half of the unit circle centered at the origin, oriented counterclockwise, the path is from the point $P = \left( {1,0} \right)$ to the point $Q = \left( { - 1,0} \right)$.
Since ${\bf{F}} = \nabla f$, by definition ${\bf{F}}$ is conservative. By Theorem 1, the line integral of ${\bf{F}}$ over the given path is
$\mathop \smallint \limits_{\cal C} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = f\left( Q \right) - f\left( P \right) = f\left( { - 1,0} \right) - f\left( {1,0} \right)$
Since $f\left( {x,y} \right) = x\cos y$, so
$\mathop \smallint \limits_{\cal C} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = - 1 - 1 = - 2$
