Why is volume of a high-D ball concentrated near its surface? I came across the following sentence while reading a book on applied math:

Volume of a high dimensional unit ball is concentrated near its
  surface and is also concentrated at its equator.

This is from book's introduction, and I believe the sentence will be explained at some later point in the book. However, it is hard to me to accept it on intuition level. Can some of you explain this sentence to me, in a sort of laymen-style, if possible?
 A: The book probably means that the volume of the unit ball in $\mathbb{R}^n$ has an unexpected behaviour as a function of $n$: it is given by
$$ V_n = \frac{\pi^{n/2}}{\Gamma\left(1+\frac{n}{2}\right)} $$
and the surface area is given by
$$ A_n = n V_n = \frac{n\pi^{n/2}}{\Gamma\left(1+\frac{n}{2}\right)}. $$
In particular, $\frac{V_n}{V_{n-1}}$ (that is a measure of "concentration around the equator") behaves like $\sqrt{\frac{\pi}{n}}$ and $\frac{V_n}{A_n}$ (that is a measure of "concentration around the boundary") is exactly $\frac{1}{n}$. Additionally, by the central limit theorem the behaviour of
$$ A(\tau)=\mu\{(x_1,\ldots,x_n)\in\mathbb{R}^n: x_1+\ldots+x_n = \tau, x_1^2+\ldots+x_n^2\leq 1\} $$
is approximately gaussian and more and more concentrated around $\tau=0$ as $n\to +\infty$.
A: It's true of a low-dimensional ball too! In a unit disk, consider all points within 0.1 of the boundary; compare the area of that to all points within 0.1 of the center point. It's much larger. 
For a sphere, the effect is even greater. The power of $r^n$ in the formula for volume in each dimension amplifies this effect as the dimension grows. 
A: The volume of an n dimensional sphere is proportional to $r^n$. For example the area of a circle (2-sphere) is $\pi r^2$ and the volume of a 3-sphere is $\frac43 \pi r^3$.
This means that the volume grows with a high power of radius, and there is more of that volume near the boundary than near the centre. For a circle, $\frac34$ of the area is closer to the circumference than the centre; for a sphere this becomes $\frac78$ and this increases closer to 1 as the number of dimensions increases. 
With enough dimensions, the outer 10% (by radius), or any small proportion you like, will have more volume than the remainder. Go high enough and the outer 1% of the n-sphere will be responsible for over 99% of the volume.
A: The volume of an $n$-dim ball with radius $r$ is of the form $A_n r^n=V_{n,r}$ (see the expression by @Jack).
Let's fix the radius of the ball to unity, and plot what fraction of volume (of the ball with radius 1) is within a smaller (concentric) ball with radius $r$ ($0\leq r \leq 1$).
This fraction is $\frac{V_{n,r}}{V_{n,1}} = \frac{A_n r^n}{A_n 1^n} = r^n$. For different $n$ the plots looks like the following ($y$-axis is $\frac{V_{n,r}}{V_{n,1}}$).

The observation is that for large $n$ (see bottom-most red plot) the fraction is almost zero until $r$ gets closer to 1. This means that the most of the volume of the ball is within a thin shell closer to its surface.
