Answer
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \frac{1}{3}$
Work Step by Step
We have ${\bf{F}}\left( {x,y} \right) = \left( {2xy,{x^2} + {y^2}} \right)$.
The path of the unit circle in the first quadrant oriented counterclockwise can be parametrized by
${\bf{r}}\left( t \right) = \left( {\cos t,\sin t} \right)$ for $0 \le t \le \frac{\pi }{2}$
Evaluate:
${\bf{F}}\left( {{\bf{r}}\left( t \right)} \right) = \left( {2\cos t\sin t,{{\cos }^2}t + {{\sin }^2}t} \right) = \left( {2\cos t\sin t,1} \right)$
${\bf{r}}'\left( t \right) = \left( { - \sin t,\cos t} \right)$
By Eq. (8) in Section 17.2:
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_a^b {\bf{F}}\left( {{\bf{r}}\left( t \right)} \right)\cdot{\bf{r}}'\left( t \right){\rm{d}}t$
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_0^{\pi /2} \left( {2\cos t\sin t,1} \right)\cdot\left( { - \sin t,\cos t} \right){\rm{d}}t$
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_0^{\pi /2} \left( { - 2\cos t{{\sin }^2}t + \cos t} \right){\rm{d}}t$
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = - 2\mathop \smallint \limits_0^{\pi /2} \cos t{\sin ^2}t{\rm{d}}t + \mathop \smallint \limits_0^{\pi /2} \cos t{\rm{d}}t$
Write $v = \sin t$. So, $dv = \cos tdt$. The integral becomes
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = - 2\mathop \smallint \limits_0^1 {v^2}{\rm{d}}v + \mathop \smallint \limits_0^{\pi /2} \cos t{\rm{d}}t$
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = - \frac{2}{3}\left( {{v^3}|_0^1} \right) + \left( {\sin t} \right)|_0^{\pi /2} = - \frac{2}{3} + 1 = \frac{1}{3}$
So, $\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \frac{1}{3}$.