About a Divergence Theorem proof I'm currently working through a proof of the divergence theorem for simple volumes. I've found some steps of the proof not formal enough, and I have not seen anything online that was of help. Here is a sketch of the proof (I'll get into details regarding the steps that aren't clear later):
Let $W$ be a simple volume, i.e. that can be described as:
$$W := \{(x,y,z) \in \mathbb{R}^3 : \ (x,y) \in D, \alpha(x,y) \leq z \leq \beta(x,y) \}$$ for a region $D \subseteq \mathbb{R}^2$ such that $\partial D$ is $C^1$  and $\alpha, \beta$ differentiable functions; and the same way for the other two axis: two coordinates belonging to a certain well behaved domain and the other one bounded by two differentiable functions. Let $F:\mathbb{R}^3 \to \mathbb{R}^3$ be a $C^1$ vector field. Then, if $\partial W^+$ is the boundary of $W$ with the outward orientation, we want to prove that
$$
\iint_{\partial W^+}F\cdot \textbf{dS} = \iiint_W\nabla F\cdot dV
$$ 
Now if $F \equiv(P,Q,R) $, it is enough to prove that 
$$
\iint_{\partial W^+}(0,0,R)\cdot \textbf{dS} = \iiint_W\frac{\partial R}{\partial z}\cdot dV
$$
$$
\iint_{\partial W^+}(0,Q,0)\cdot \textbf{dS} = \iiint_W\frac{\partial Q}{\partial y}\cdot dV
$$
$$
\iint_{\partial W^+}(P,0,0)\cdot \textbf{dS} = \iiint_W\frac{\partial P}{\partial x}\cdot dV
$$
All three proofs are simliar, so I'll just go with the first one since the notation is nicer to describe: at this point I'll ease on the formalities since I'm not too sure the proof is formal enough anyways. The idea is to separate $\partial W^+$ into three parts, the graphs of $\alpha$ and $\beta$ and the "vertical wall" that connects them. 
Now, here is the first step I don't get: every proof I've seen online says the vertical wall integral cancels out, handwaving that the normal vector points orthogonally to the z axis. In my notes I have something like trying to get a parametrization for $\partial D$ and with that a parametrization for the vertical wall, but it's not very clear to me what that parametrization is. It would be very helpful if you could shed some light on this part of the proof. 
The final problem I have is when summing the other two integrals. I get every step except the last one (I mean, it sure sounds true but not formal enough to me at least). Taking the trivial parametrizations for the graphs  of $\alpha$ and $\beta$ and summing the integrals, we get
$$
\iint_{D}R(x,y,\beta(x,y)) - R(x,y,\alpha(x,y)) \cdot dxdy = \\
= \iint_{D} \int_{\alpha(x,y)}^{\beta(x,y)}\frac{\partial R}{\partial z} \cdot dz \ dxdy = \iiint_W \frac{\partial R}{\partial z} dxdydz.
$$
I see why intuitively the last step must be true, but I'm not fully convinced. If $D$ were a type I / II / III region, I can see why this would work, but here it is not necessarily the case.
 A: First of all, most proofs for this theorem in introductory courses assume a symmetrical region, this means it can be described as either a type I, II or III, so the proof makes heavy use of this. Now following your line of thought: 
First let us calculate $$\iiint_W\frac{\partial R}{\partial z}\cdot dV$$
Since the region $W$ can be described as $W = \{(x,y,z) \in \mathbb{R}^3 : \ (x,y) \in D, \alpha(x,y) \leq z \leq \beta(x,y) \}$ the integral can be calculated as: 
$$\iint_D \int_{\alpha(x, y)}^{\beta(x, y)}\frac{\partial R}{\partial z}dzdxdy=\iint_D[R(x,y, \beta(x,y))-R(x,y, \alpha(x,y))]dxdy$$ You can think of the  LHS as just a triple iterated integral with a known order of integration and the equality holds using the Fundamental Theorem of Calculus.
Now let us go through the other part of the proof. For this part, since $W$ is symmetrical let us split the surface $\partial W$ into a convenient way. Let us define:
$S_1=\{(x,y,z)\in \mathbb R^3 :(x, y)\in D, z=\beta(x,y)\}$
$S_2=\{(x,y,z)\in \mathbb R^3 :(x, y)\in D, z=\alpha(x,y)\}$
$S_3=\{(x,y,z)\in \mathbb R^3 :(x, y)\in \partial D, \alpha(x,y)\le z\le\beta(x,y)\}$
Again, this can be done because $\partial W$ is a symmetrical closed surface. Using this we can calculate $\iint_{\partial W^+}(0,0,R)\cdot dS$ as:
$$\iint_{\partial W^+}(0,0,R)\cdot dS=\iint_{S_1^+}(0,0,R)\cdot dS+\iint_{S_2^+}(0,0,R)\cdot dS+\iint_{S_3^+}(0,0,R)\cdot dS$$ let us see what happens to $S_3$:
For $\partial D$, and using the fact that it is $C¹$,we can parametrize it as: $$\sigma : [a,b]\subset \mathbb R \rightarrow \partial D\subset \mathbb R^2 :\sigma(u)=(x(u), y(u))$$ It is then clear that the surface $S_3$ can be parametrized as (again, using the symmetry argument): $$\psi : [a,b]\times[c,d]\subset \mathbb R^2 \rightarrow S_3 : \psi(u,v)=(x(u),y(u),v)$$ now it is easy to see that taking the normal for the surface $S_3$ gives you: $\psi_u \times \psi_v=(y'(u),-x'(u),0)$ and from here is is clear that: $$\iint_{S_3^+}(0,0,R)\cdot dS=0$$ which is what you called the "vertical wall" component. For the remaining bits of the proof you just use the trivial parametrizations for $S_1$ and $S_2$ to show that: $$\iint_{\partial W^+}(0,0,R)\cdot dS=\iint_{S_1^+}(0,0,R)\cdot dS+\iint_{S_2^+}(0,0,R)\cdot dS=\iint_D [R(x,y, \beta(x,y))-R(x,y, \alpha(x,y))]dxdy$$ 
Bear in mind that the proof makes extensive use of the symmetry of region $W$, without this argument we couldn't have even begin the proof.
Hope this helps! E.J.J G
