Why do volume and surface area of unit ball in $\mathbb{R}^d$ behave the way they do for $d \uparrow$?

Question

Is there any demonstrative / intuitive explanation for the behavior of the surface area and the volume of the unit ball as the dimension increases?

I sort of get that it tends to zero, because all the coordinates become smaller and smaller (although I'm not quite satisfied with this "explanation"). But why the maximum? And why is the maximum not at the same dimension for the two quantities?

There is probably some buzzword I should google, but I can't figure it out.

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edit: I just saw the related question Volumes of n-balls: what is so special about n=5? which is still "unanswered" and does not cover the topic of surface area.

Answer 1

First, why there is a maximum: I think the easy way to get the meaning of it is if you kook at the recursion relationship between $V_n$ and $V_{n-2}$. $$V_n(R)=\frac{2\pi R^2}{n}V_{n-2}(R)$$ Ignoring $R$ since it's $1$ in your problem. you see that the volume increases or decreases by a factor of $2\pi/n$. You can see that if $n\ge7$ that fraction is less than 1, so the volume decreases. You therefore get $V_1<V_3<V_5$ and then $V_5>V_7>V_9>...$ And similarly for even dimensions. You would need to go into details of the gamma function to show the relationship between odd and even dimensions. It's just a little more calculation involved.

The second question is why they have maxima at different dimensionality. If you remember, the volume of the $n$-dimensional sphere becomes surface area of the $n+1$ dimensional sphere. If you look at this link, you get $$A_{n+1}(R)=2\pi RV_n(R)$$ Therefore if you have a maxima at $5$ in volume, you will get a maxima at $6$ in surface area.

Answer 2

What is "volume" in $n$ dimensions? It is a certain quantity obtained by calculating a certain measure defined on $\mathbb{R}^{n}$. We know exactly which measure it is - it is the product measure which gives the box $\prod_{i=1}^{n}[a_i,b_i]$ the measure $\prod_{i=1}^{n}(b_i-a_i)$. The fact that we use this measure and not any other measure (there are lots of possibilities) as the "natural" measure on $\mathbb{R}^n$ is probably because

A. If is very easy to calculate this measure of boxes.

B. Many important and interesting subsets of $\mathbb{R}^n$ can be approximated very well by boxes.

We call this measure of sets the volume of sets, although a more proper name should be the content of sets, because volume, really, is a three-dimensional notion. Once we agree on this measure, we find that when we come to measure the volume of the unit-ball, i.e., the set $\{x\in\mathbb{R}^n:\sum_ix_i^2\leq 1\}$ it turns out that the largest copy of the unit-box (i.e. the set $\{x\in\mathbb{R}^n:\max_i|x_i|\leq 1\}$) that fits into the unit-ball has to be dilated by a factor of $n^{1/2}$ (i.e., the unit cube has to be multiplied by $n^{-1/2}$ in order to fit into the unit ball). Now this does not prove anything yet, but it clearly gives some recognition of the fact that the unit-ball in $\mathbb{R}^n$ does not occupy much volume, and that it is much "harder" to be inside a $100$-dimensional ball than inside a $3$-dimensional ball because there are $97$ more inequalities that have to be satisfied. Of course, all this is just wave-handing heuristics, but when you perform the calculation you find a number that goes to zero approximately like $n^{-1/2}$, validating some of the feeling of small volume we got by comparing to the cube - the basic building block of the underlying measure.

As for the question why is there a maximum - there is really a trivial answer: the volume starts somewhere, and then it decreases to zero, so it must have a maximum.

Regarding the third question - why does the surface area attain a maximum in a different dimension than the volume, I don't see any "naturally" good answer. Just as well, I could ask "why should it?"

Answer 3

All the "strangeness" in the exposed results is just due to the fact that the volume of a $n-$Ball of radius $R$ should "normally" be compared to the volume of a $n-$Cube of side $2R$.

Then $$ \upsilon _{\,n} = {{V_{\,n} (R)} \over {\left( {2R} \right)^{\,n} }} = {{\pi ^{\,n/2} } \over {2^{\,n} \Gamma \left( {1 + n/2} \right)}}\, = {{\Gamma \left( {1/2} \right)^{\,n} } \over {2^{\,n} \Gamma \left( {1 + n/2} \right)}} $$

and this is a monotonic decreasing function for $n \in \mathbb N$,
which tastes "natural", because at increasing $n$, the volume of the $n$-ball will be less than that of the cylinder based on the $(n-1)-$ball, which in turn is always less than that of the unitary cube.

Reversing the view, the ratio you have considered is that of the ball vs. the cube with one vertex at the origin:
in $3$D this corresponds to the unit cube covering one octant of the ball.
So you are multiplying the ratio $\upsilon(n)$ above (decreasing) by the number of octants, which is $2^n$ (increasing).
The maximum is just the result of the combination of the two opposite rates.

Answer 4

Letting $V_n(r)$ be the volume of the sphere of radius $r$ in $\mathbb{R}^n$, then we get the recursion $$ V_n(r)=\int_{-r}^rV_{n-1}\!\left(\sqrt{r^2-t^2}\right)\mathrm{d}t\tag1 $$ Homogeneity tells us that $$ V_n(r)=\Omega_nr^n\tag2 $$ So let us try to show $(2)$ inductively using $(1)$. Suppose that $(2)$ is true for $n-1$, then $$ \begin{align} V_n(r) &=\int_{-r}^rV_{n-1}\!\left(\sqrt{r^2-t^2}\right)\mathrm{d}t\\ &=\int_{-r}^r\Omega_{n-1}\left(r^2-t^2\right)^{\frac{n-1}2}\mathrm{d}t\\ &=r^n\int_{-1}^1\Omega_{n-1}\left(1-t^2\right)^{\frac{n-1}2}\mathrm{d}t\\[9pt] &=\Omega_nr^n\tag3 \end{align} $$ where $$ \begin{align} \Omega_n &=\Omega_{n-1}\int_{-1}^1\left(1-t^2\right)^{\frac{n-1}2}\mathrm{d}t\\ &=\Omega_{n-1}\int_0^1(1-t)^{\frac{n-1}2}t^{-\frac12}\mathrm{d}t\\ &=\Omega_{n-1}\frac{\Gamma\!\left(\frac{n+1}2\right)\Gamma\!\left(\frac12\right)}{\Gamma\!\left(\frac{n+2}2\right)}\tag4\\ \end{align} $$ Since $V_1(r)=2r$, this completes the induction.

$\Gamma\!\left(\frac12\right)=\sqrt\pi$ and because Gamma is log-convex we have $$ \Gamma\!\left(\frac{n+1}2\right)\le\Gamma\left(\frac{n}2\right)^{\frac12}\Gamma\left(\frac{n+2}2\right)^{\frac12}\tag5 $$ and $$ \Gamma\!\left(\frac{n+2}2\right)\le\Gamma\left(\frac{n+1}2\right)^{\frac12}\Gamma\left(\frac{n+3}2\right)^{\frac12}\tag6 $$ Inequalities $(5)$ and $(6)$, along with $\Gamma(n+1)=n\Gamma(n)$, imply that $$ \sqrt{\frac2{n+1}}\le\frac{\Gamma\!\left(\frac{n+1}2\right)}{\Gamma\!\left(\frac{n+2}2\right)}\le\sqrt{\frac2n}\tag7 $$ Thus, $(4)$ and $(7)$ show that $$ \sqrt{\frac{2\pi}{n+1}}\le\frac{\Omega_n}{\Omega_{n-1}}\le\sqrt{\frac{2\pi}n}\tag8 $$ From $(8)$, we know that if $n\ge6$, then $\Omega_{n+1}\lt\Omega_n$ and if $n\le5$, $\Omega_{n-1}\lt\Omega_n$. However, this doesn't tell us which is greater, $\Omega_5$ or $\Omega_6$. To determine this, we can apply $(4)$ twice to get $$ \Omega_n=\frac{2\pi}n\Omega_{n-2}\tag9 $$ We know that $\Omega_1=2$ and $\Omega_2=\pi$. Equation $(9)$ yields that $$ \begin{array}{c|c|c} n&\Omega_n&\approx\Omega_n\\\hline 1&2&2.00000000000000\\ 2&\pi&3.14159265358979\\ 3&\frac{4\pi}3&4.18879020478639\\ 4&\frac{\pi^2}2&4.93480220054468\\ 5&\frac{8\pi^2}{15}&5.26378901391432\\ 6&\frac{\pi^3}6&5.16771278004997 \end{array}\tag{10} $$ Thus, $\Omega_5$ is the greatest.

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Since $$ \begin{align} S_{n-1}(r) &=\frac{\mathrm{d}}{\mathrm{d}r}V_n(r)\\ &=n\Omega_nr^{n-1}\\[5pt] &=2\pi\Omega_{n-2}r^{n-1}\\[6pt] &=\omega_{n-1}r^{n-1} \end{align} $$ we see that $\omega_6$ is greatest, but that is the (six-dimensional) surface area of the unit sphere in $\mathbb{R}^7$. So, we need to be clear about which dimension we are talking about. The dimension of the surface, or the dimension of the embedding space. In the chart in the question, it seems that the dimension is the dimension of the surface, not the dimension of the embedding space.