Answer
$\mathop \smallint \limits_C^{} \left( {x - y} \right){\rm{d}}s = - 2$
Work Step by Step
We have $f\left( {x,y} \right) = x - y$.
The unit semicircle ${x^2} + {y^2} = 1$, $y \ge 0$ can be parametrized by
${\bf{r}}\left( t \right) = \left( {\cos t,\sin t} \right)$ for $0 \le t \le \pi $
So, we obtain
$f\left( {{\bf{r}}\left( t \right)} \right) = \cos t - \sin t$
$ds = ||{\bf{r}}'\left( t \right)||dt = \sqrt {\left( { - \sin t,\cos t} \right)\cdot\left( { - \sin t,\cos t} \right)} dt = dt$
By Eq. (4) in Section 17.2:
$\mathop \smallint \limits_C^{} f\left( {x,y} \right){\rm{d}}s = \mathop \smallint \limits_a^b f\left( {{\bf{r}}\left( t \right)} \right)||{\bf{r}}'\left( t \right)||{\rm{d}}t$
$\mathop \smallint \limits_C^{} \left( {x - y} \right){\rm{d}}s = \mathop \smallint \limits_0^\pi \left( {\cos t - \sin t} \right){\rm{d}}t$
$\mathop \smallint \limits_C^{} \left( {x - y} \right){\rm{d}}s = \left( {\sin t + \cos t} \right)|_0^\pi = - 2$
So, $\mathop \smallint \limits_C^{} \left( {x - y} \right){\rm{d}}s = - 2$.
