I have this question which states
Let S be the surface of the box bounded by the six planes $$x=0,x=2,y=0,y=4,z=0,z=1$$ $\vec{n}$ is the outward normal and $$\vec{F}(x,y,z) = (y^2-\sin(yz))\vec{i} + (y-z) \vec{j} + (x^2y\sin(x) + x^2 + y^2)\vec{k}$$ Pick $S_{1}$ to be one of the six faces of S. Parametrise $S_{1}$ and express $$\int\int_{S_{1}} \vec{F} \cdot \vec{n}dS$$ as a double integral.
Evaluate $$\int\int_{S}\vec{F} \cdot \vec{n}dS$$ where S and $\vec{F}$ are defined as above
So my initial thoughts let us take the top of the box and just parametrise is with (0,0,k) being the normal vector [for $S_{1}$] and then once we express it as a double integral we move to the second part where my question lies.
If we do the same process of using the normal which points outwards and do them in pairs we have a "pair" which adds to $0$ i.e.
$$\int\int_{S_{1}} \vec{F} \cdot \vec{n}dS + \int\int_{S_{1}}\vec{F} \cdot \vec{n}dS$$
and if we repeat this process for each face with each pair we obtain
$$\int\int_{S}\vec{F} \cdot \vec{n}dS = 0$$
I was hoping someone could either correct me or advise me how to continue with this problem.
Thank you