Answer
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = 0$
Work Step by Step
We have the vector field ${\bf{F}}\left( {x,y} \right) = \left( {{x^2},xy} \right)$.
The part of circle ${x^2} + {y^2} = 9$ with $x \le 0$, $y \ge 0$, can be parametrized by
${\bf{r}}\left( t \right) = \left( {3\cos t,3\sin t} \right)$, ${\ \ \ }$ for $\frac{\pi }{2} \le t \le \pi $
So, we obtain
${\bf{F}}\left( {{\bf{r}}\left( t \right)} \right) = \left( {9{{\cos }^2}t,9\cos t\sin t} \right)$
$d{\bf{r}} = {\bf{r}}'\left( t \right)dt$
$d{\bf{r}} = \left( { - 3\sin t,3\cos t} \right)dt$
Evaluate the dot product ${\bf{F}}\left( {{\bf{r}}\left( t \right)} \right)\cdot{\bf{r}}'\left( t \right)dt$:
${\bf{F}}\left( {{\bf{r}}\left( t \right)} \right)\cdot{\bf{r}}'\left( t \right)dt = \left( {9{{\cos }^2}t,9\cos t\sin t} \right)\cdot\left( { - 3\sin t,3\cos t} \right)dt$
$ = \left( { - 27{{\cos }^2}t\sin t + 27{{\cos }^2}t\sin t} \right)dt$
$ = 0$
Evaluate $\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}}$ using Eq. (8):
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_a^b {\bf{F}}\left( {{\bf{r}}\left( t \right)} \right)\cdot{\bf{r}}'\left( t \right){\rm{d}}t$
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = 0$
