Today, I decided I would try to generalize a formula to find area and volume of a hypercube. The formulae I came up with are as follows:
$n$ = number of dimensions
$L$ = side length
Surface area = $2nL^{n-1}$
Volume = $L^n$
So, naturally, I decided to make a table of surface area and volume of hypercubes, and using the formula, I found that the volume of a one-dimensional shape is two. This seems illogical, but then again, to me it's hard to conceptualize things in dimensions other than 2 or 3.
Can someone help me out here, and maybe update my formulae if needed?
Edit: I figured out the whole one-dimensional thing, but now I'm having trouble with the surface area in two dimensions. If $L = 2$, then shouldn't $A = 4$? I only get 8 using the equation.
Here's the table of values for $L = 1, 2$ and $n = 1, 2, 3$. $$\begin{array}{c|c|c|c} L & n & \mbox{Area} = 2nL^{n-1} & \mbox{Volume} = L^n & \\\hline 1 & 1 & 2 & 1 \\ 1 & 2 & 4 & 1 \\ 1 & 3 & 6 & 1 \\ 2 & 1 & 2 & 2 \\ 2 & 2 & 8 & 4 \\ 2 & 3 & 24 & 8 \\ 3 & 1 & 2 & 3 \\ 3 & 2 & 12 & 9 \\ 3 & 3 & 72 & 27 \\ \end{array} $$
So for example, with $L = n = 2$ (the case in your "Edit") you get a surface area of $8$, because the sum of the lengths of the $4$ sides of a square with side length $2$ is $8$.
Note that the surface areas when $n = 1$ need to be taken with a pinch of salt - we are pretending for convenience that the $0$-dimensional volume of a point is $1$.