Answer
$0$
Work Step by Step
Stokes' Theorem can be defined for this surface as $\iint_{S} curl F \cdot dS=\iint_{C} F \cdot dr$
The boundary of the surface and the bottom phase of the cube is the same and all the points lie on the surface; this means that $z=-1$
Thus, we can say that $\iint_{S} curl F \cdot dS=\iint_{C} F \cdot dr=\iint_{H } F \cdot dS $ and $C$ is the common boundary, which implies that the square is in the plane $z=-1$
Now, we have: $F=xyz i+xy j+x^2 yz k$
and $curl F=\iint_{H} y-xz dS$
Since all the points lie on the surface, this means that $z=-1$, so
$\iint_{H} y-xz dS=\iint_{H} y+x dS$
and $dS=\sqrt {1+(z_x)^2+(z_y)^2} dx dy=\sqrt {1+0+0} dx dy= dx dy$
Therefore, $\iint_{H } F \cdot dS=\iint_{S } F \cdot dS =\int_{-1}^1 \int_{-1}^1 [y+x] dx dy$
or, $=\int_{-1}^{1}[yx +\dfrac{x^2}{2}]_{-1}^{1} dy $
or, $=\int_{-1}^{1} 2y dy$
or, $=0$
