Volume of $n$-ball in terms of $n-2$ ball Let $V_n(R)$ be the volume of the ball with radius $R$ in $\mathbb{R}^n$. This page says 

$$V_n(R)=\int_0^{2\pi}\int_0^RV_{n-2}(\sqrt{R^2-r^2})r\,dr\,d\theta$$

I don't really understand the explanation given in there. Could someone explain it in the case $n=3$ (so $n-2=1$) how this integral is derived?
 A: Let's do the $n=2$ case first.  What is a zero-dimensional sphere, you ask?  Well, it's nothing but a point, and $0$-dimensional volume is, I claim, just cardinality.  So $V_0 (R) = \lvert\{\ast \}\rvert= 1$.
Now, how do we calculate $V_2 (R)$, the $2$-volume (i.e. area) of a $2$-ball (i.e. disc)?  Well, we want to add up the volume of all those teensy $0$-dimensional balls which make up the disc, parametrized by the radius and the angle.  First, imagine cutting the disc into very skinny rings.  We will look at one ring with inner radius $r$ and outer radius $r+dr$.  Next, we cut this ring into extremely thin sections.  We will focus on one such section, that sweeps out a very small angle, $d\theta$.
Call the area of this little piece $dA$.  Then the volume we want is exactly the sum of all those little pieces, which is exactly $\int dA$.  There are very general principles that tell us how to compute $dA$, but here is how to work it out from scratch.
Our little chunk is very very close to being a rectangle when $d\theta$ is very small.  It's so close, that we might as well pretend it is a rectangle (if this makes you uncomfortable, then try working out the error term and seeing how quickly it goes to zero, or else look at the rigorous proof of the multivariate change-of-variables formula).
And what are the side lengths of this rectangle?  The radial one is obvious: it's just $dr$.  The other one is a little more tricky.  Whatever it is, if we add it up all around the circle, we should get the circumference.  Since $\int_0^{2\pi} r d\theta = 2\pi r$, the length we want is $r d\theta$.
So we conclude that $dA = r \,dr\, d\theta$.  Since we're interested in adding up these areas over all $r$ and $theta$ with $0\leq r \leq R$ and $0\leq \theta \leq 2\pi$, we have $V_2 (R) = \int dA = \int r\,dr\,d\theta = \int_0^{2\pi} \int_0^R r\,dr\,d\theta$.
If we want, we can actually compute this out as a sanity check, and get $2\pi \left.\frac{1}{2} r^2 \right|_0^R = \pi R^2$.
Now that we've done $n=2$, we're more than ready for $n=3$.  Imagine that you're looking down on a sphere of radius $R$, so that it looks like a disc.  Now we want to chop up the disc exactly as before, except now, instead of tiny near-rectangular bits, these are very thin bars that go all the way through the sphere (picture an apple slicer).  Their volume is just the area $dA$ that we had before, times... hmm...
We need to break out the Pythagorean theorem to find the length of our rectangular bar.  Say that the length is $2s$.  Then, if you think for a minute, you should find that $s^2 + r^2 = R^2$, because the top of our bar lies on the sphere, which has radius $R$, and the line from the origin is the hypoteneuse of a right triangle that goes radially outwards (from our top-down perspective) a distance $r$, then up (towards us) a distance that we defined to be $s$.  So the length we want is $2s$—which just so happens to be $V_1 (s)$, the "volume" of the $1$-sphere of radius $s$.
So we have $V_3 (R) = \int 2s \cdot r\,dr\,d\theta = \int V_1 (s) \cdot r\,dr\,d\theta$ $= \int V_1 (\sqrt{R^2 - r^2})r\,dr\,d\theta$ $= \int_0^{2\pi} \int_0^R V_1 (\sqrt{R^2 - r^2}) r\,dr\,d\theta$.
One again, we can do a sanity check, and actually compute the thing to get $\frac{4}{3} \pi R^3$.
Now the general case should be clear.  We will dividing up the $n$-ball into pieces, making rectangular cuts.  The rectangles will have side lengths $dr$ and $rd\theta$, and they will cut out a ball of dimension $n-2$ and radius $s$, with $s^2 + r^2 = R^2$.
A: $$
V_3(R)=\int_{x^2+y^2+z^2\le R^2}dxdydz=\int_{x^2+y^2\le R^2}\Bigl(\int_{z^2\le R^2-(x^2+y^2)}dz\Bigl)dxdy.
$$
Now change to polar coordinates: $x=r\cos\theta$, $y=r\sin\theta$, to get
$$
V_3(R)=\int_0^{2\pi}\int_0^R\Bigl(\int_{z^2\le R^2-)r^2}dz\Bigl)r\,drd\theta=\int_0^{2\pi}\int_0^RV_1\bigl(\sqrt{R^2-r^2}\,\bigr)\,r\,drd\theta.
$$
