Answer
$\mathop \smallint \limits_{{C_1}}^{} y{\rm{d}}x + {x^2}y{\rm{d}}y = \frac{1}{4}\left( {81 - 9\pi } \right)$
Work Step by Step
Write ${\bf{F}}\left( {x,y} \right) = \left( {y,{x^2}y} \right)$.
Since $d{\bf{r}} = {\bf{i}}dx + {\bf{j}}dy$, the integral $\mathop \smallint \limits_{{C_1}}^{} y{\rm{d}}x + {x^2}y{\rm{d}}y$ becomes
$\mathop \smallint \limits_{{C_1}}^{} y{\rm{d}}x + {x^2}y{\rm{d}}y = \mathop \smallint \limits_{{C_1}}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}}$
We can consider the curve ${C_1}$ consists of 3 segments: along the $x$-axis, along the part of the circle, and along the $y$-axis. Thus, the line integral along ${C_1}$ is the sum of the line integrals of these segments:
$\mathop \smallint \limits_{{C_1}}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_{x - axis}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} + \mathop \smallint \limits_{circle}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} + \mathop \smallint \limits_{y - axis}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}}$
1. along the $x$-axis
Here $y=0$ and $dy=0$, so
$\mathop \smallint \limits_{x - axis}^{} y{\rm{d}}x + {x^2}y{\rm{d}}y = 0$
2. along the circle of radius 3 for $0 \le t \le \frac{\pi }{2}$
The circle can be parametrized by ${\bf{r}}\left( t \right) = \left( {3\cos t,3\sin t} \right)$.
Evaluate:
${\bf{F}}\left( {{\bf{r}}\left( t \right)} \right) = \left( {3\sin t,27{{\cos }^2}t\sin t} \right)$
${\bf{r}}'\left( t \right) = \left( { - 3\sin t,3\cos t} \right)$
By Eq. (8) in Section 17.2:
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_a^b {\bf{F}}\left( {{\bf{r}}\left( t \right)} \right)\cdot{\bf{r}}'\left( t \right){\rm{d}}t$
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_0^{\pi /2} \left( {3\sin t,27{{\cos }^2}t\sin t} \right)\cdot\left( { - 3\sin t,3\cos t} \right){\rm{d}}t$
$ = \mathop \smallint \limits_0^{\pi /2} \left( { - 9{{\sin }^2}t + 81{{\cos }^3}t\sin t} \right){\rm{d}}t$
Since $\cos 2x = 1 - 2{\sin ^2}x$, so
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_0^{\pi /2} \left[ {\frac{9}{2}\left( {\cos 2t - 1} \right) + 81{{\cos }^3}t\sin t} \right]{\rm{d}}t$
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \frac{9}{2}\mathop \smallint \limits_0^{\pi /2} \left( {\cos 2t - 1} \right){\rm{d}}t + 81\mathop \smallint \limits_0^{\pi /2} {\cos ^3}t\sin t{\rm{d}}t$
Write $v = \cos t$. So, $dv = - \sin tdt$. The integral becomes
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \frac{9}{2}\mathop \smallint \limits_0^{\pi /2} \left( {\cos 2t - 1} \right){\rm{d}}t - 81\mathop \smallint \limits_1^0 {v^3}{\rm{d}}v$
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \frac{9}{2}\left( {\frac{1}{2}\sin 2t - t} \right)|_0^{\pi /2} - \frac{{81}}{4}\left( {{v^4}|_1^0} \right)$
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \frac{9}{2}\left( { - \frac{\pi }{2}} \right) + \frac{{81}}{4} = \frac{1}{4}\left( {81 - 9\pi } \right)$
3. along the $y$-axis
Since $x=0$ and $dx=0$, so
$\mathop \smallint \limits_{y - axis}^{} y{\rm{d}}x + {x^2}y{\rm{d}}y = 0$
Finally,
$\mathop \smallint \limits_{{C_1}}^{} y{\rm{d}}x + {x^2}y{\rm{d}}y = \mathop \smallint \limits_{{C_1}}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}}$
$\mathop \smallint \limits_{{C_1}}^{} y{\rm{d}}x + {x^2}y{\rm{d}}y = 0 + \frac{1}{4}\left( {81 - 9\pi } \right) + 0 = \frac{1}{4}\left( {81 - 9\pi } \right)$
So, $\mathop \smallint \limits_{{C_1}}^{} y{\rm{d}}x + {x^2}y{\rm{d}}y = \frac{1}{4}\left( {81 - 9\pi } \right)$.