Answer
$I = \mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = 2\pi $
We verify that $I$ changes sign when $C$ is oriented clock-wise.
Work Step by Step
We have the vector field ${\bf{F}}\left( {x,y} \right) = \left( {\frac{{ - y}}{{{x^2} + {y^2}}},\frac{x}{{{x^2} + {y^2}}}} \right)$.
The circle $C$ of radius $2$ centered at the origin can be parametrized by
${\bf{r}}\left( t \right) = \left( {2\cos t,2\sin t} \right)$, ${\ \ \ }$ for $0 \le t \le 2\pi $
So, we obtain
${\bf{F}}\left( {{\bf{r}}\left( t \right)} \right) = \left( {\frac{{ - \sin t}}{2},\frac{{\cos t}}{2}} \right)$
$d{\bf{r}} = {\bf{r}}'\left( t \right)dt$
$d{\bf{r}} = \left( { - 2\sin t,2\cos t} \right)dt$
Evaluate the dot product ${\bf{F}}\left( {{\bf{r}}\left( t \right)} \right)\cdot{\bf{r}}'\left( t \right)dt$:
${\bf{F}}\left( {{\bf{r}}\left( t \right)} \right)\cdot{\bf{r}}'\left( t \right)dt = \left( {\frac{{ - \sin t}}{2},\frac{{\cos t}}{2}} \right)\cdot\left( { - 2\sin t,2\cos t} \right)dt = dt$
Calculate $I = \mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}}$ using Eq. (8) when $C$ is oriented counterclockwise:
$I = \mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_a^b {\bf{F}}\left( {{\bf{r}}\left( t \right)} \right)\cdot{\bf{r}}'\left( t \right){\rm{d}}t$
$I = \mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_0^{2\pi } {\rm{d}}t = 2\pi $
If $C$ is oriented clock-wise than the integral above becomes
$I = \mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_{2\pi }^0 {\rm{d}}t = - 2\pi $
Thus, it is verified that $I$ changes sign when $C$ is oriented clock-wise.
