[10000truths]

> Just like almost all of the volume of a reasonably round convey body is near its surface.

I’d say that’s pretty intuitive for anyone who can see a pattern in surface area to volume ratios.

1D ball: 2 / (2 * r) = 1/r

2D ball: (2 * pi * r) / (pi * r^2) = 2/r

3D ball: (4 * pi * r^2) / (4/3 * pi * r^3) = 3/r

nD ball: ... = n/r

[atq2119]

Most people don't find arguing from formulas intuitive unless the formulas themselves are intuitive. If you truly believe they are, I'd be curious to know why.

[krastanov]

There is an intuitive version of this. Volume in n dimensions is C*r^n (C is some constant) and surface is the first derivative, leading to a ratio of n/r (the C constant cancels out). Hmm... Maybe not that intuitive

[kazinator]

But the former formula already tells you that most of the volume is near the high range of r: that which was to be shown.

The surface area to volume concept adds nothing.

Because the volume of a sphere is proportional to r cubed, you know there is much more volume between r in [0.9, 1.0] than in the same sized interval of r [0.0, 0.1].

You can find the break-even point almost in your head. At what r value is half the volume of a R = 1.0 sphere below that value? Why, that's just the cube root of 1/2 ~= 0.794. So almost half the volume is within 20% of the radius from the surface.

That's far from the claim that almost all the volume is near the surface: half is not almost all, and 20% isn't all that near. However, you can see how it gets nearer and nearer for higher dimensions.

For a ten dimensional sphere, the tenth root of 1/2 is ~ 0.933. So over half the volume of a ten dimensional sphere is within the 7% depth.

[10000truths]

The surface area to volume ratio is just a limit of the shell volume to total volume ratio as the shell thickness goes to zero. So both should asymptotically scale with higher dimensions in the same way.