Answer
$0$
Work Step by Step
Stokes' Theorem states that $\iint_{S} curl F \cdot dS=\iint_{C} F \cdot dr$
where, $C$ corresponds to the boundary of the surface oriented counter-clockwise.
and $dS=\sqrt {1+(z_x)^2+(z_y)^2} dx dy$
We are given that the boundary of the surface and the bottom phase of the cube is same and all the points lie on the surface, thus, we have: $z=-1$
This means that $\iint_{S} curl F \cdot dS=\iint_{C} F \cdot dr=\iint_{H } F \cdot dS $ and $C$ is the common boundary , that is, the square in the plane is: $z=-1$
Now, we have: $F=xyz i+xy j+x^2 yz k \implies \ curl \ F=\iint_{H} y-xz dS$
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 $dS=\sqrt {1+0+0} \ dx \ dy= \ dx \ dy$
Thus, we have: $$\iint_{H } F \cdot dS=\iint_{S } F \cdot dS =\int_{-1}^1 \int_{-1}^1 [y+x] dx dy \\=\int_{-1}^{1}[yx +\dfrac{x^2}{2}]_{-1}^{1} dy \\=\int_{-1}^{1} 2y dy \\= 2(y^2/2)_{-1}^1 \\=1-1 \\=0$$
