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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.

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closed as off-topic by Raskolnikov, Shailesh, Namaste, Saad, JMP Feb 16 '18 at 3:43

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    $\begingroup$ What values did you use for $L$ and $n$ to get $L^{n} = 2$? $\endgroup$ – Rob Arthan Feb 15 '18 at 21:00
  • $\begingroup$ L = 2, and n = 1. $\endgroup$ – Andrew Gordnier Feb 15 '18 at 21:04
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    $\begingroup$ Well the "1-dimensional volume" (i.e., length) of a line of length 2 is 2. What's the problem with that? $\endgroup$ – Rob Arthan Feb 15 '18 at 21:07
  • $\begingroup$ @RobArthan, nevermind. I figured it out, I was just over-thinking it. $\endgroup$ – Andrew Gordnier Feb 15 '18 at 21:10
  • $\begingroup$ In $n$-dimension, the 'surface area' you referred to is the $n-1$-dimension measure of the boundary of the hypercube. When $n = 2$, what you called 'surface area' is the perimeter of the solid square of side $2$ and has value $8$. $\endgroup$ – achille hui Feb 15 '18 at 21:36
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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 $1$.

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  • $\begingroup$ Thank you. If you didn't assume the the volume of a point is one, then the area would be zero for a one-dimensional shape, correct? $\endgroup$ – Andrew Gordnier Feb 15 '18 at 22:20
  • $\begingroup$ Taking the $0$-dimensional volume of a point as $1$ gives us the nice property that the area in the case $L = 1$ is the number of facets (i.e., $(n-1)$-cubes that make up the surface of the $n$-cube for all $n \ge 1$.We could adopt a different convention, but this one is convenient. $\endgroup$ – Rob Arthan Feb 15 '18 at 22:46

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