$\displaystyle \bar{F}=x^2\bar{i}+xy\bar{j}+0\bar{k\ }and \\[2ex] \displaystyle d\bar{r}=dx\bar{i}+dy\bar{j}+dz\bar{k\ } \\[2ex] \displaystyle \therefore{}\bar{F}\bullet{}\ d\bar{r}=x^2dx+xydy \\[2ex] \displaystyle Part\ I:\ Consider, \\[2ex]$

$\displaystyle \int_c\bar{F}\bullet{}\ d\bar{r}=\int_0^2x^2dx+xydy\\[2ex] \displaystyle Equation:\ y=0\\[2ex]Along\ side\ OA\\[2ex] \therefore{}dy=0\\[2ex] \displaystyle \therefore{}\int_{OA}\bar{F}\bullet{}\ d\bar{r}=\int_0^2x^2dx+0\\[2ex] \displaystyle ={\left[\frac{x^3}{3}\right]}_0^2\\[2ex] \displaystyle =\frac{2^3}{3}-0\\[2ex] \displaystyle =\frac{8}{3}\\[3ex] $

$Along\ side\ AB\ \\[2ex] Equation:\ x=2\\[2ex] \displaystyle \therefore{}dx=0\\[2ex] \displaystyle \therefore{}\int_{AB}\bar{F}\bullet{}\ d\bar{r}=\int_0^30+2ydy{\\[2ex] \displaystyle =\left[\frac{{2y}^2}{2}\right]}_0^3=3^2-0=9\\[3ex]Along\ side\ BC\ \\[2ex] Equation:\ y=3\\[2ex] \therefore{}dy=0\\[2ex] \displaystyle \therefore{}\int_{BC}\bar{F}\bullet{}\ d\bar{r}=\int_2^0x^2dx+0{\\[2ex] \displaystyle =\left[\frac{x^3}{3}\right]}_2^0\\[2ex] \displaystyle =0-\frac{2^3}{3}\\[2ex] \displaystyle =\frac{-8}{3}$

$Along\ side\ COEquation:\\[2ex] x=0\\[2ex] \therefore{}dx=0\\[2ex] \displaystyle \therefore{}\int_{CO}\bar{F}\bullet{}\ \displaystyle d\bar{r}=\int_b^00+0=0\\[2ex] Hence\\[2ex] \therefore{}\int_c\bar{F}\bullet{}\ \displaystyle d\bar{r}=\int_{OA}\bar{F}\bullet{}\ d\bar{r}+\int_{AB}\bar{F}\bullet{}\ \displaystyle d\bar{r}+\int_{BC}\bar{F}\bullet{}\ d\bar{r}+\int_{CO}\bar{F}\bullet{}\ d\bar{r}\\[2ex] =\frac{8}{3}+9+\frac{-8}{3}+0\\[2ex] =9\ \ \ $

(\rightarrow{}\left(i\right)Part\ 2:\[3ex]Consider\[3ex] \displaystyle \iint_R\bar{N}\bullet{}\left(\nabla{}\times{}\bar{F}\right)ds\[2ex] \therefore{}\ \bar{F}=x^2\bar{i}+xy\bar{j}\[2ex] \therefore{}\nabla{}\times{}\bar{F}=\left[\begin{array}{ ccc} \bar{i} & \bar{j} & \bar{k} \ \frac{\partial{}}{\partial{}x} & \frac{\partial{}}{\partial{}y} & \frac{\partial{}}{\partial{}z} \ x^2 & xy & 0 \end{array}\right])

$\displaystyle =\bar{i}\left(0-0\right)-\bar{j}\left(0-0\right)+\bar{k}\left(y-0\right)\\[2ex] \displaystyle =y\bar{k}\\[2ex] In\ XY\ plane\ unit\ normal\ is,\\[2ex] \vec{N}=\vec{k} \ And \ ds=dxdy\\[2ex] \displaystyle \therefore{}\bar{N}\bullet{}\left(\nabla{}\times{}\bar{F}\right)=\vec{k}\bullet{}(y\vec{k})=y\\[2ex] \displaystyle \iint_R\bar{N}\bullet{}\left(\nabla{}\times{}\bar{F}\right)ds\\[2ex] \displaystyle =\iint_Rydxdy=\int_0^3\int_0^2ydxdy\\[2ex] \displaystyle =\int_0^3{\left[yx\right]}_0^2dy\\[2ex] \displaystyle =\int_0^3\left[2y-0\right]dy\\[2ex] ={\left[\frac{2y^2}{2}\right]}_0^3\\[2ex]=3^2-0\\[2ex] =9\ \ \ \ \rightarrow{}\left(ii\right)\\[2ex] $

$\therefore{}From\ \left(i\right)and\ \left(ii\right)\\[2ex] \displaystyle \int_c\bar{F}\bullet{}\ d\bar{r}=\iint_R\bar{N}\bullet{}\left(\nabla{}\times{}\bar{F}\right)ds\\[4ex] \therefore{}Stoke^{'}sTheorem\ is\ verified $