Calculate the flux of $G(x,y,z)=x(1-z)\hat{k}$ through the cube with the vertices $(0, 0, 0), (0, 0, 1), (0, 1, 0), (0, 1, 1), (1, 0, 0), (1, 0, 1), (1, 1, 0), (1, 1, 1)$ with an outward pointing normal vector.
I don't really have an idea what to do, so far I just calculated the normal vector, $\hat{N}=\frac{\nabla G}{|\nabla G|}$=$\frac{(0,0,-x)}{x}=(0,0,-1)$.
So $\int\int_s F \bullet dS$=$\int\int_D(F_1 \frac{\partial(y,z)}{\partial(u,v)}+F_2 \frac{\partial(x,z)}{\partial(u,v)}+F_3 \frac{\partial(y,x)}{\partial(u,v)}) du dv$
Break the integral up in to six separate integrals over the 6 faces of the cube, where the faces are indicated for convenience by $x=0$ etc. on the integral signs.
$$\int \int_{x=0} \mathbf{G}.\mathbf{(-i)}dS + \int \int_{x=1} \mathbf{G}.\mathbf{i}\,dS + \int \int_{y=0} \mathbf{G}.\mathbf{(-j)}dS +\int \int_{y=1} \mathbf{G}.\mathbf{j}\,dS $$ $$+\int \int_{z=0} \mathbf{G}.\mathbf{(-k)}dS + \int \int_{z=1} \mathbf{G}.\mathbf{k}\,dS$$
The dot products pick out the components of $\mathbf{G}$ and it is only the $z$ component which is non-zero so the first 4 integrals are zero. Of the remaining 2 integrals, $$\int \int_{z=1} \mathbf{G}.\mathbf{k}\,dS$$ is zero because $z=1$ on that face. That leaves one double integral to perform on the $z=0$ face where $\mathbf{G}=x\mathbf{k}$ and $dS=dx\,dy$
$$\int \int_{z=0}(x\mathbf{k}).(-\mathbf{k})\,dS = \int_{x=0}^{x=1} \int_{y=0}^{y=1} (-x) \,dx\,dy$$
Also note that the Divergence Theorem could be considered as an alternative approach to solving this.