Surface Area of a Hypercube I am interested in computing the surface area of an $n$-dimensional hypercube and am interested in a reference or an answer which defines the notion of surface area for higher dimensional polytopes as I am trying to compute the surface area of an infinite family of duoprisms and knowing the surface area of an $n$-dimensional hypercube would be very useful to my understanding.
That is, what is the surface area of an $n$-dimensional hypercube with side length $s$, and how can you think about surface area of higher dimensional polytopes in general?
EDIT: In regarding as to whether I am referring to "surface area" or "surface volume", I am interested in understanding any $k$-dimensional version of surface hyper(area/volume) for an $n$-dimensional polytope. I think it makes sense that the $n$-dimensional version of this quantity is the volume, the $(n-1)$-dimensional version would be surface "area", and there are $(n-2),...,1$-dimensional versions of this idea. Is there any way to understand this with differential forms perhaps?
 A: Expand the polynomial $(x+2)^n$.  For example, $(x+2)^3=1x^3+6x^2+12x+8$.  Read off the coefficients: A cube has 1 cube, 6 faces, 12 edges and 8 corners.  
A: In $n$ dimensions, one thinks of solid space as having $n$ dimensions, and surface area as having $n-1$ dimensions.  It corresponds to spaces defined by 0 and 1 equal-signs.  A surface has one equal sign, eg $x=0$ gives a point in 1D, a line in 2D, a 2d surface in 3D.  Area is then portion of this space.
The volume of a Sphere is given by $C_n = 2\pi r^2 C_{n-2}/n$, with $C_0 = 1, \; C_1 = 2$.  The value of $S_n = n C_n / r$.  
So, the volume of a sphere, relative to its radius, is  $C_2 = \pi r^2i$, $C_3 = 4\pi r^3/3$ $C_4 = \pi^2 r^4/2$, $C_5 = 8\pi^2 r^5 / 15$,  $C_6 = \pi^3 r^6 / 6$, and so forth.
For the cube, one might note that a cube has $2n$ faces, and thus its surface area is $2ne^{n-1}$.
Note that i have chosen to follow the terminology of the polygloss, where the measure by a specific dimension follows the sequence 1D = lineage or length,  2D = hedrage, 3D = chorage, 4D = terage, freeing 'surface' and 'volume' to stand for the covering and content of a solid figure.  
The volume of a polytope, in general, is the moment of surface.  For example, $V = \frac 1n \sum  r \cdot dS$, integrated over the full surface.  
A: A nice trick to go from $n$-dimensional volume to $(n-1)$-dimensional whatever is to differentiate the formula. Intuitively, if we were to subtract small cube from big cube, what is left is roughly surface-area of cube times the difference of side-lengths. The only thing to remember is that we need to shrink figures symmetrically, that is the formula should be dependent not on side-length, but half of side-length (like radius for balls instead of diameter).
For example, sphere volume is $\frac{4}{3}\pi r^3$, sphere surface area is $4\pi r^2$, cube volume is $(2r)^3 = 8r^3$, cube area is $6(2r)^2 = 24r^2$, total length of cube segments is $\frac{24(2r)=48r}{2} = 12(2r) = 24r$, because each edge is common to two faces. Analogously, each point is common to 3 sides, so there are $\frac{24}{3} = 8$ points.
This could be generalized to $n$-th dimension, I hope it helps ;-)
A: The answer is $$2 \frac{d}{dx}(V_{n}(L))$$
or, equivalently:
$$2nx^{n-1}$$
where $V_{n+1}(L)$ is the Volume of an n-dimensional sphere.
As to why, think about slightly increasing the size of a cube for example. How much new volume has been created? it's going to approach 3x$^2$ times the tiny change in x, because each face of the cube will be expanded, and all the other bits will converge to zero.
It's hard to explain, but here's a good video to watch to get a feel for what I'm talking about.
https://www.youtube.com/watch?v=S0_qX4VJhMQ
However, this only covers half of what the surface area of these shapes is, because it's only increasing with respect to half of the total sides. So, multiplying by two will give your final answer.
This similar tactic can be used to find the surface areas of other shapes too!
for instance, if you imagine slightly increasing the radius of a sphere, and make them both concentric, then subtract the overlapping area, what you're left with is a sort of shell, which as the slight nudge becomes smaller, becomes close to the true surface area, times the size of that small nudge.
