Let A=(1,0); B=(0,1); C=(-1,0) and D=(0,-1)

All points of square ABCD do not have z co-ordinate.

So square ABCD lies in XY plane.

Using two point form, equation of line AB is 

$\dfrac {y-y_1}{y_1 - y_2} = \dfrac {x-x_1}{x_1 - x_2}$

$\therefore \dfrac {y-0}{0-1}= \dfrac {x-1}{1-0}$

$\therefore y= -x+1$

Similarly, equation of line BC is y=x+1; line CD is y=-x-1; line DA is y=x-1;

Let $\overrightarrow {F} \cdot d\overrightarrow {r} = xy dx + x^2 ydy $

But, $d\overrightarrow {r}=dx\overrightarrow {i} + dy \overrightarrow {j} + dz\overrightarrow {k}$

$\therefore \overline{F} = (xy)\overrightarrow {i}+ x^2 y\overrightarrow {j}+0\overrightarrow{k}$

$\nabla \times \overline F = \begin{vmatrix} \overrightarrow {i} & \overrightarrow {j} & \overrightarrow {k} \\ \partial / \partial x & \partial / \partial y & \partial / \partial z \\ xy & xy^2 & 0 \end{vmatrix}$

$=\overrightarrow {i}[0-0] - \overrightarrow {j}[0-0]+ \overrightarrow {k} (y^2 - x)$

$=\overrightarrow {k} (y^2 -x) \ \to \ (2)$

For XY-plane, unit normal $= \overrightarrow{k}$ and ds=dx dy →(3)

∴ From (2) and (3)

$\overrightarrow {N}\cdot (\nabla \times \overrightarrow {F}) = \overrightarrow {k} \cdot \overrightarrow {k} (y^2-x)$

$= (y^2 -x) \to (4)$

By Stoke's Theorem,

$\displaystyle \int_c \overrightarrow {F} \cdot d\overrightarrow{r} = \iint_R \overrightarrow {N}\cdot (\nabla \times \overrightarrow {F})ds$

$\therefore \displaystyle \int_c (xy\ dx + x^2ydy) = \iint_R (y^2 - x)ds$

$\displaystyle = \int^0_{-1} \int^{x+1}_{-x-1} (y^2 - x)dydx + \int^1_0 \int^{-x+1}_{x-1} (y-2 - x)dy \ dx \ (from \ 4)$

$\displaystyle = \int^0_{-1} \left [ \dfrac {y^3} {3} - xy \right ] ^{x+1}_{-(x+1)} dx + \int^{1}_0 \left [ \dfrac {y^3}{3} - xy \right ]^{-(x-1)}_{x-1}dx $

$\displaystyle = \int^0_{-1} \left \{ \left [ \dfrac {1}{3}(+1)^3 - x(x+1)\right ] - \left [ \dfrac {-1}{3} (x+1)^3 + x(x+1) \right ] \right \}dx + \int^1_0 \left \{ \left [ \dfrac {-1}{3}(x-1)^3 + x(x-1) \right ] - \left [ \dfrac {1}{3}(x-1)^3 -x (x-1) \right ] \right \}dx$

$=\displaystyle \int^0_{-1} \left \{ \dfrac {1}{3}(x+1)^3 -x^2 -x +\dfrac {1}{3} (x+1)^3 - x^2 - x \right \}dx + \int^1_0 \left \{ \dfrac {-1}{3} (x-1)^3 + x^2 - x - \dfrac {1}{3} (x-1)^3 + x^2 -x \right \}dx$

$=\displaystyle 2 \int^0_{-1} \left \{ \dfrac {1}{3}(x+1)^3 - x^2 - x \right \}dx + 2\int^1_{0} \left \{ \dfrac {-1}{3} (x-1)^{3}+x^2 - x \right \}dx$

$=\displaystyle 2 \left [ \dfrac {1}{3} \times \dfrac {(x+1)^4}{4} - \dfrac {x^3}{3}- \dfrac {x^2}{2} \right ]^0_{-1} + 2 \left [ \dfrac {-1}{3} \times \dfrac {(x-1)^4} {4}+ \dfrac {x^3}{3}- \dfrac{x^2}{2} \right ]^1_0 $

$= 2 \left [ \left (\dfrac {1}{12}- 0-0 \right )- \left ( 0+ \dfrac {1}{3} - \dfrac {1}{2} \right ) \right ]+2 \left [ \left (0+ \dfrac {1}{3}- \dfrac {1}{2} \right )- \left ( \dfrac {-1}{12}\times (-1)^4+0-0 \right ) \right ]$

$=2 \left [ \dfrac {1}{12} - \dfrac {1}{3}+ \dfrac {1}{2} \right ]+ 2\left [ \dfrac {1}{3} - \dfrac {1}{2}+ \dfrac{1}{12} \right ]$

$= \dfrac {1}{6}+ \dfrac {1}{6}$

$= \dfrac {1}{3}$