I'm a non-mathematician and am trying to understand intuitively what happens to the volume and surface area of an n-ball as the size and dimensionalty increases. I've tried looking at similar topics in this forum but nothing answers my question (at least not in a way that I can understand).
What is puzzling me is that the volume of an $n$-ball peaks at $n = 5$ and then declines again (which seems odd in itself) but only for radius $= 1.$ For radius $2$ it's $n=24,$ for $3$ it's $n=56…$ the figure shows a graph of peak against radius for dimensions $1-350$ and it has a rather odd shape.
What I'm struggling with is why the radius should matter – it seems like the peak volume should be a property of the shape and shouldn't depend on whether the radius is in cm or m etc. Can anyone explain in non-mathematical terms? Many thanks!
Best Answer
My take on this is that there is no such thing as "peak volume" unless you pick the unit of measurement, so you can treat all sizes as dimensionless.
Let's look at the case where the radius is $1 cm$. The volumes, for dimension $n$, are:
$$\begin{array}{rrl}\text{dimension}&\text{size}&\text{unit}\\\hline 1&2.000&cm\\2&3.142&cm^2\\3&4.189&cm^3\\4&4.935&cm^4\\5&5.264&cm^5\\6&5.168&cm^6\\\text{etc.}\end{array}$$
(as per https://en.wikipedia.org/wiki/Volume_of_an_n-ball). Notice you can only compare the numbers in the second column if you can ignore the units. Otherwise, those are just values with different units and cannot be compared at all. (What is bigger: $2 cm$ or $3.142 cm^2$)?