Answer
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \frac{{2\pi }}{{{R^2}}}$
Work Step by Step
We have the vector field ${\bf{F}}\left( {x,y} \right) = \left( {\frac{{ - y}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}},\frac{x}{{{{\left( {{x^2} + {y^2}} \right)}^2}}}} \right)$.
The part of circle of radius $R$ with center at the origin, can be parametrized by
${\bf{r}}\left( t \right) = \left( {R\cos t,R\sin t} \right)$, ${\ \ \ }$ for $0 \le t \le \pi $
So, we obtain
${\bf{F}}\left( {{\bf{r}}\left( t \right)} \right) = \left( {\frac{{ - \sin t}}{{{R^3}}},\frac{{\cos t}}{{{R^3}}}} \right)$
$d{\bf{r}} = {\bf{r}}'\left( t \right)dt$
$d{\bf{r}} = \left( { - R\sin t,R\cos t} \right)dt$
Evaluate the dot product ${\bf{F}}\left( {{\bf{r}}\left( t \right)} \right)\cdot{\bf{r}}'\left( t \right)dt$:
${\bf{F}}\left( {{\bf{r}}\left( t \right)} \right)\cdot{\bf{r}}'\left( t \right)dt = \left( {\frac{{ - \sin t}}{{{R^3}}},\frac{{\cos t}}{{{R^3}}}} \right)\cdot\left( { - R\sin t,R\cos t} \right)dt$
$ = \left( {\frac{{{{\sin }^2}t}}{{{R^2}}} + \frac{{{{\cos }^2}t}}{{{R^2}}}} \right)dt$
$ = \frac{1}{{{R^2}}}dt$
Evaluate $\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}}$ using Eq. (8):
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_a^b {\bf{F}}\left( {{\bf{r}}\left( t \right)} \right)\cdot{\bf{r}}'\left( t \right){\rm{d}}t$
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \frac{1}{{{R^2}}}\mathop \smallint \limits_0^{2\pi } {\rm{d}}t = \frac{{2\pi }}{{{R^2}}}$
