Surface Area of a Hypercube

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.

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.