By stoke's theorem

$\int\limits_{C} \bar{F} \cdot d\bar{r} = \int\int\limits_{S} \bar{N} \cdot (\bar{V} \times \bar{F})ds$

$\therefore$ in the xy plane r = xi + yj + 0k

$\therefore \partial r = \partial x i + \partial y j$

$\bar{F} \cdot d\bar{r} = xydx + xy^2 dy$ we get

$\bar{F} = xyi + xy^2 j + 0k$

$\therefore \bar{V} \times \bar{F} = \begin{vmatrix} i & j & k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial }{\partial z} \\ xy & xy^2 & 0 \end{vmatrix} \\ = i\bigg[\frac{}{\partial y}(0) - \frac{\partial}{\partial z}(xy^2)\bigg] - j\bigg[\frac{\partial}{\partial x}(0) - \frac{\partial}{\partial z}(xy)\bigg] + k\bigg[\frac{\partial}{\partial x}(xy^2) - \frac{\partial}{\partial y}(xy) \bigg] \\ = k[y^2 - x]$

Now $\bar{N} = k$ and ds = dxdy $\int\int\limits_{S}\bar{N} \cdot (\bar{V} \times \bar{F})ds = \int\int\limits_{S}(y^2 - x)dxdy$

Now for square what we will do is we just calculate the surface area in the first quadrant and multiply it by 4

So we will get for the whole square

Thus for the portion in first quadrant. Now this position is bounded by x = 0 y = 0 and the slant line equation can be found out by two point form

i.e. $\therefore \frac{x - x1}{y - y1} = \frac{x - x2}{y -y2}$ (1,0) (0,1) -> $(x_{1}, y_{1}), (x_{2}, y_{2})$

$\therefore \frac{x - 1}{y - 0} = \frac{1 - 0}{0 - 1} \\ \therefore \frac{x - 1}{y - 0} = \frac{1 - 0}{0 - 1} \\ \therefore x - 1 = -y \\ \therefore x + y = 1$

$\int\int\limits_{S} (y^2 - x)dxdy$

$\therefore$ Now the horizontal strip will slide from 0 to 1 in horizontal direction and this is the outer integration limits, whereas for inner integration limits upper limits is the slant line equation which can be written as y = 1 - x and lower limits is y = 0

$\therefore \int\int\limits_{S} (y^2 - x)dxdy = \int_{x = 0}^{1}\int\limits_{y = 0}^{y = 1 - x} (y^2 - x)dxdy \\ = \int\limits_{x = 0}^{1} \bigg[\frac{y^3}{3} - yx\bigg]dx \\ = \int\limits_{!}^{0}\bigg[\frac{(1 - x)^3}{3} - x(1 - x)\bigg] \\ = \int\limits_{0}^{1} \bigg[\frac{1 - 3x + 3x^2 - x^3}{3} + x^2 - x\bigg]dx \\ = \frac{1}{3} \int\limits_{0}^{1} (1 - 6x + 6x^2 - x^3)dx \\ = \frac{1}{3} \bigg[x - \frac{6x^2}{2} + \frac{6x^3}{3} - \frac{x^4}{4} \bigg]_{0}^{1} \\ = \frac{1}{3} \bigg[ 1 - 3 + 2 - \frac{1}{4}\bigg] = \frac{-1}{12}$

For complete square multiplying it by 4 we get

$\therefore \frac{-1}{12} \times 4 = \frac{-1}{3}$