Consider $I=\int_C\ \bar F.\bar {dr}$ where $\bar F=x^2i-xyj\ and\ c \ is\ the \ square\ in\ plane\ with\ x=0, \ y=0, \ x=a\ and\ y=a.$

By Stoke's Theorem

$\int_C\ \bar F.\ \bar{dr}=\iint_S\ \bar N.\nabla \times F\ ds$

Hence $ds=dx\ dy\ and\ \bar N=k$

Consider $\nabla \times \bar F=\left[\begin{matrix} i&j&k \\ \dfrac{\partial }{\partial x} & \dfrac{\partial }{\partial y}& \dfrac{\partial }{\partial z}\\ x^2\ &\ -xy\ &\ 0 \end{matrix}\right]$=$=i(0-0)-j(0-0)+k(-y-0)=-ky$

$\bar N=k$

$\therefore \bar N.\nabla\times \bar F=-y $

We have

$\int_C\bar Fdr=\int_0^a\int_0^a-ydydx--\int_0^a\Bigg[\dfrac{y^2}{2}\Bigg]_0^adx=\dfrac{-a^2}{2}\int_0^adx=\dfrac{-a^2}{2}\big[x\big]_0^a=\dfrac{-a^3}{2}$

$\int_c\ \bar F\ dr=\dfrac{-a^3}{2}$