# The intuition behind volume and surface area of an $n$-ball

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!

• The volume of an $n$-ball of radius $r$ is $r^n$ times the volume of the unit $n$-ball. Jan 2, 2020 at 21:06
• That downturn looks suspicious to me. The approximation $\ln \Gamma(n)\sim n\ln n$ implies the measure-maximising $n$ should be approximately exponential in $n$. For large $n$, your code is probably buggy.
– J.G.
Jan 2, 2020 at 23:11
• You're right, thank you. The numbers got too big and I didn't think to check. So it's just exponential to infinity I assume - that makes a lot more sense. Jan 3, 2020 at 12:56

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 $$3.142 cm^2$$)?