Two-dimension recursion formula for computing volumes The two-dimension recursion formula for computing volumes of balls says:

A proof of the recursion formula relating the volume of the $n$-ball and an $(n -2)$-ball can be given using the proportionality formula above and integration in cylindrical coordinates. Fix a plane through the center of the ball. Let $r$ denote the distance between a point in the plane and the center of the sphere, and let $\theta$ denote the azimuth. Intersecting the $n$-ball with the $(n − 2)$-dimensional plane defined by fixing a radius and an azimuth gives an $(n − 2)$-ball of radius ...

I don't quite understand this. Would someone give an explanation for the case $n=3$, so that $n-2=1$? 
So when it says "Fix a plane through the center of the ball" that means a line through the center of the ball. 
"Let $r$ denote the distance between a point in the plane and the center of the sphere." But there are so many points on this "plane" (line). Which one should I take?
What would be all the "planes" taken in this case $n=3$?
 A: This might have taken longer than you would have liked...but I'll answer since I was recently wondering this same thing. 
In the $n=3$ case, you have a $3$-ball of radius R centered around the origin in $\mathbb R^3$. The proof says to fix a plane through the center of the ball. I will choose to use the $(x,y)$-plane.
The intersection of this plane with the $3$-ball creates a disk of radius $R$ centered around the origin on the $(x,y)$-plane. Let any point on this disk be defined by the coordinate $(r,\theta)$ where $0\leq r \leq R$ and $0 \leq \theta \leq 2\pi$.
(Note: $r$ represents the distance from the origin on the $(x,y)$-plane and $\theta$ represents the counterclockwise angle with respect to the $x$-axis...the "azimuth" they talk about).
For some arbitrary $(r,\theta)$ on this disk, you have a point on the surface of the $3$-ball directly above/below it. Take either point on the surface of the $3$-ball, and draw a little segment from the origin to this point, you have a segment of length $R$.
Now you have a right triangle with hypotenuse $R$, and with "adjacent" side $r$. The length of the third "opposite" side is $\sqrt{R^2-r^2}$. The main idea here is that this third "segment" can be thought of as the radius of a ball, a $1$-ball to be precise (this is the $n-2$ ball radius they're talking about).
So we discovered that an $n=3$ ball is actually composed of a bunch of $n=1$ balls, which is where the integration comes in. This same idea is generalized for higher dimensions on the Wikipedia explanation.
