Answer
$\mathop \smallint \limits_C^{} y{\rm{d}}x - x{\rm{d}}y = - \frac{8}{3}$
Work Step by Step
Write the vector field ${\bf{F}}\left( {x,y} \right) = \left( {y, - x} \right)$ and the path ${\bf{r}} = \left( {x,y} \right)$, such that
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_C^{} y{\rm{d}}x - x{\rm{d}}y$
The parabola can be parametrized by ${\bf{r}}\left( t \right) = \left( {t,{t^2}} \right)$ for $0 \le t \le 2$. So,
${\bf{F}}\left( {{\bf{r}}\left( t \right)} \right) = \left( {{t^2}, - t} \right)$
$d{\bf{r}} = {\bf{r}}'\left( t \right)dt$
$d{\bf{r}} = \left( {1,2t} \right)dt$
Using Eq. (8), the vector line integral becomes
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_C^{} y{\rm{d}}x - x{\rm{d}}y = \mathop \smallint \limits_0^2 {\bf{F}}\left( {{\bf{r}}\left( t \right)} \right)\cdot{\bf{r}}'\left( t \right){\rm{d}}t$
$\mathop \smallint \limits_C^{} y{\rm{d}}x - x{\rm{d}}y = \mathop \smallint \limits_0^2 \left( {{t^2}, - t} \right)\cdot\left( {1,2t} \right){\rm{d}}t$
$\mathop \smallint \limits_C^{} y{\rm{d}}x - x{\rm{d}}y = - \mathop \smallint \limits_0^2 {t^2}{\rm{d}}t = - \frac{1}{3}{t^3}|_0^2 = - \frac{8}{3}$
So, $\mathop \smallint \limits_C^{} y{\rm{d}}x - x{\rm{d}}y = - \frac{8}{3}$.
