Answer
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \frac{9}{2}$
Work Step by Step
We have the vector field ${\bf{F}}\left( {x,y} \right) = \left( {4,y} \right)$.
The quarter circle ${x^2} + {y^2} = 1$ with $x \le 0$, $y \le 0$, can be parametrized by
${\bf{r}}\left( t \right) = \left( {\cos t,\sin t} \right)$, ${\ \ \ }$ for $\pi \le t \le \frac{{3\pi }}{2}$
So, we obtain
${\bf{F}}\left( {{\bf{r}}\left( t \right)} \right) = \left( {4,\sin t} \right)$
$d{\bf{r}} = {\bf{r}}'\left( t \right)dt$
$d{\bf{r}} = \left( { - \sin t,\cos t} \right)dt$
Evaluate the dot product ${\bf{F}}\left( {{\bf{r}}\left( t \right)} \right)\cdot{\bf{r}}'\left( t \right)dt$:
${\bf{F}}\left( {{\bf{r}}\left( t \right)} \right)\cdot{\bf{r}}'\left( t \right)dt = \left( {4,\sin t} \right)\cdot\left( { - \sin t,\cos t} \right)dt = \left( { - 4\sin t + \cos t\sin t} \right)dt$
Evaluate $\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}}$ using Eq. (8):
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_a^b {\bf{F}}\left( {{\bf{r}}\left( t \right)} \right)\cdot{\bf{r}}'\left( t \right){\rm{d}}t$
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_\pi ^{3\pi /2} \left( { - 4\sin t + \cos t\sin t} \right){\rm{d}}t$
$ = - 4\mathop \smallint \limits_\pi ^{3\pi /2} \sin t{\rm{d}}t + \mathop \smallint \limits_\pi ^{3\pi /2} \cos t\sin t{\rm{d}}t$
$ = 4\cos t|_\pi ^{3\pi /2} + \frac{1}{2}\mathop \smallint \limits_\pi ^{3\pi /2} \sin 2t{\rm{d}}t$
$ = 4 - \frac{1}{4}\left( {\cos 2t|_\pi ^{3\pi /2}} \right)$
$ = 4 - \frac{1}{4}\left( { - 1 - 1} \right) = \frac{9}{2}$
So, $\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \frac{9}{2}$.
