Clarification on higher dimensional unit sphere/unit cube properties I've been learning probability lately and I have come across this neat result (in Rick Durrett's book):

Most of the volume of the unit cube in $\mathbb{R}^n$ comes from the set $A_{n,\epsilon} := \{x \in \mathbb{R}^n \, : \,(1-\epsilon)\sqrt{\frac{n}{3}} < |x| < (1+\epsilon)\sqrt{\frac{n}{3}} \},$ which is almost the sphere of radius $\frac{n}{3}.$

At the same time, we know the two basic facts that the volume of the unit ball in $\mathbb{R}^n$ goes to zero as $n$ grows and the volume of the unit cube remains the same, but most of the volume get concentrated in the corners of the cube.  
I realize that these two points are pretty different ideas altogether, but it seems strange to me that most of the volume in a high dimensional cube can be contained within a sphere and yet also be concentrated in the corners.  Could someone clarify what is really going on? Or perhaps my understanding is off.. I am not very experienced in geometry at all.  Thanks!
 A: I didn't know this result (+1), but here's how I'd interpret it geometrically:
For $0 < r < \sqrt{n}$, let $B(r)$ be the intersection of the closed ball of radius $r$ centered at the origin with the unit cube $[0, 1]^{n}$, and (for fixed positive $\epsilon \ll 1$) let $S(r)$ denote the shell $B(r + \epsilon) \setminus B(r - \epsilon)$.
Let $c_{n}$ denote the $(n - 1)$-dimensional volume of the unit sphere. When $r \leq 1 - \epsilon$, $S(r)$ is a thin spherical shell of radius $r$, whose volume is $2\epsilon c_{n} r^{n-1} + O(\epsilon^{2})$. When $r \approx \sqrt{n}$, $S(r)$ is a thin shell near the far corner $(1, 1, \dots, 1)$, whose volume is $O(\epsilon^{n})$. The volume $V(r)$ of $S(r)$ is (clearly?) a continuous function of $r$, and vanishes at the "endpoints", so it has a maximum; this happens to occur at $r = \sqrt{\frac{n}{3}}$.
More qualitatively, increasing $r$ beyond $1$ (where the sphere starts to stick out of the cube) causes the volume $V(r)$ to change for two reasons: increasing because the sphere's total volume is larger, and decreasing because more of the sphere sticks out of the cube.
(If instead you intersect the $n$-ball with the cube $[-1, 1]^{n}$, everything gets multiplied by $2^{n}$, but the ratios of volumes don't change: Both the ball and the cube comprise $2^{n}$ congruent orthants.)
