Dimension of the boundary of a polytope for applying divergence theorem in $\Bbb{R}^n$ Let's use the unit hypercube in $\Bbb{R}^n$ as an example. The unit $n$-cube is formed by the intersection of $2n$ half-spaces, $n$ of them being defined by $x_i \ge 0$ and the other $n$ of them being defined by $x_i \le 1$. I want to use the divergence theorem to calculate the following integral:
$$ \int_V \vec{\nabla} \cdot \vec{F} \, \mathrm{d}V = \int_{\partial V} \vec{F} \cdot \hat{n} \, \mathrm{d}S $$
where $V$ is the 'volume' of the $n$-cube and $dS$ denotes an integral over the 'surface,' its boundary. My question is this:
What dimension is the boundary, $\partial V$?
Here's my attempt at answering it:
Since we are evaluating an $n$-polytope, every facet (of dimension $n-1$), is determined by the polytope's intersection with a supporting hyperplane. In our example, this region is where the weak inequality constraint imposed by one half-space now holds with equality. Put another way, the hyperplane is the $(n-1)$ dimensional set where $x_i = 1$ or $x_i = 0$. As a result, the boundary is given by the union of these facets, and so is $(n-1)$-dimensional.
The only reason I ask the question is my initial intuition was that the boundary was the collection of all $2$-faces of the polytope, because it is often said that 'the boundary of the boundary is empty'. However, the divergence operator only reduces the dimensionality by one, so it seems more reasonable to have an $(n-1)$-dimensional boundary. Does this then mean that the boundary has its own, non-empty, $(n-2)$-dimensional boundary?
Thoughts, comments, and intuition appreciated.
Substantial Edit:
To summarize answers, I am being told that repeated application of the divergence theorem will  lead to things canceling in such a way that the integral will be zero. I believe there may be an exception using strategically chosen degenerate vector fields, according to the below example in $\Bbb{R}^3$. Can you show me where this is wrong?
Consider attempting to integrate the function $f(x,y,z) = xyz$ over the domain of $[0,1]^3$. We can evaluate this integral directly as a check:
$$ \int_0^1 \int_0^1 \int_0^1 xyz \,\mathrm{d}x\mathrm{d}y\mathrm{d}z = \int_0^1 \int_0^1 \frac{1}{2}yz \,\mathrm{d}y\mathrm{d}z = \int_0^1 \frac{1}{4}z \,\mathrm{d}z = \frac{1}{8}.$$
Now for the counterexample. Let $\vec{F} = (\frac{1}{2}x^2 yz,0,0)$ and let $V = [0,1]^3$. Then,
$$\iiint_V f(x,y,z) \mathrm{d}V = \iiint_V \vec{\nabla} \cdot \vec{F} \mathrm{d}V = \iint_S \vec{F} \cdot \hat{n} \mathrm{d}S = \sum_{i=1}^6 \iint_{S_i} \vec{F} \cdot \hat{n}_i \mathrm{d}S $$
where $i$ denotes each face of the cube, $S_i$. The faces are described as follows: 
$$S_1: x = 0, S_2: y = 0, S_3: z = 0, S_4: x = 1, S_5: y = 1, S_6: z = 1. $$
The normal vectors are given by, e.g., $(-1,0,0)$ for $S_1$ and $(1,0,0)$ for $S_4$. Since $\vec{F}$ is zero outside the first element, all but $S_1$ and $S_4$ are zeroed out, so we obtain:
$$\sum_{i=1}^6 \iint_{S_i} \vec{F} \cdot \hat{n}_i \mathrm{d}S = - \iint_{S_1} \frac{1}x^2 y z  \mathrm{d}S + \iint_{S_4} \frac{1}x^2 y z  \mathrm{d}S. $$
Here is the trick: in order to apply the divergence theorem again, we cannot generate a new vector field by integrating w.r.t. $x$. However, we can do so with respect to $y$ and things will not cancel. Let $\hat{G} = (0, \frac{1}{4}x^2 y^2 z, 0)$. Then for each face, 
$$ \iint_{S_i} \vec{\nabla}\cdot\vec{G} \mathrm{d}S = \oint_{C} \vec{G} \cdot \hat{n}_C \mathrm{d}\vec{r} = \sum_{j = 1}^4 \int_{e_j} \vec{G} \cdot \hat{n}_j \mathrm{d}\vec{r}. $$
The edges, $e_i$, are now given by $e_1: y = 0, e_2: z = 0, e_3: y = 1, e_4: z = 1$, and the normal vectors are, e.g., $\hat{n}_1 = (0,-1,0), \hat{n}_3 = (0,1,0)$. Once again, since $\vec{G}$ is zero outside the second element, we obtain:
$$\sum_{j = 1}^4 \int_{e_j} \vec{G} \cdot \hat{n}_j \mathrm{d}\vec{r} = - \int_{e_1} \frac{1}{4}x^2 y^2 z \mathrm{d}\vec{r} + \int_{e_3} \frac{1}{4}x^2 y^2 z \mathrm{d}\vec{r} = - \int_{e_1} \frac{1}{4}x^2 y^2 z \mathrm{d}z + \int_{e_3} \frac{1}{4}x^2 y^2 z \mathrm{d}z. $$
because the path of integration for these edges is with respect to $z$ only.
To summarize, we now have:
$$\iiint_V f(x,y,z) \mathrm{d}V = -\left( \left( - \int_{e_1} \frac{1}{4}x^2 y^2 z \mathrm{d}z \right)_{y=0} + \left(\int_{e_3} \frac{1}{4}x^2 y^2 z \mathrm{d}z \right)_{y=1} \right)_{x=0} + \left( \left( - \int_{e_1} \frac{1}{4}x^2 y^2 z \mathrm{d}z \right)_{y=0} + \left(\int_{e_3} \frac{1}{4}x^2 y^2 z \mathrm{d}z \right)_{y=1} \right)_{x=1}
$$
where each edge then goes from $z=0$ to $z=1$ using the same logical process as before. Everything goes to zero except the last term, so:
$$\iiint_V f(x,y,z) \mathrm{d}V = \int_0^1 \frac{1}{4}(1)^2 (1)^2 z \mathrm{d}z = \frac{1}{8}.
$$
Sorry this is so long, but I feel it does represent a legitimate counterexample. Any ideas? Basically, I am not arguing that this works in the general case, just that there may exist specific polytopes and functions for which it works. 
 A: The boundary of your $n$-dimensional cube is indeed $(n-1)$-dimensional.  As you said, in a comment, each of the $2n$ facets of your cube has a nonempty, $(n-2)$-dimensional boundary; that consists of $2(n-1)$ cubes (all of dimension $n-2$).  But each of these $(n-2)$-dimensional cubes $X$ is part of the boundary of two facets of your original cube, which fit together so that $X$ is not on the boundary of the boundary of the original cube.
If this is hard to imagine in $n$ dimensions, look at the case $n=2$.  You have a square $S$, whose boundary $B$ consists of $4$ edges. Each of those edges has two endpoints, at two of the corners of $S$. But each corner is an endpoint of two edges, and does not contribute to the boundary of $B$.  Indeed, $B$ is one single, continuous, closed curve, and it has no boundary.  
You might similarly consider the next case, $n=3$, where you can still see what's going on (as opposed to higher, unvisualizable dimensions).  
Also, note that the principle "boundary of boundary is $0$" refers to the chain complex notion of boundary, not the topological notion (closure minus interior) which depends on the ambient space.
A: Yes, the boundary of an $n$-dimensional object has dimension $n-1$. In your example, it is the union of the $(n-1)$-dimensional faces indeed.
But sometimes an object has no boundary at all (a circle, the surface of a sphere, the surface of a torus...). In particular, "the boundary of the boundary" is always empty indeed (including the boundary of your polytope, it has no boundary).
A: Perhaps you should check "integration over chains" (eg Spivak´s book). Indeed, the faces of a cube are lower dimensional cubes, which have boundaries, but their orientation is defined in such a way that their boundaries cancel out. 
