How can I find the volume of this 4-dimensional ball? 
*

*Let $B$ be the $4$-dimensional ball
$$
\left\{(x, y, z, w): x^2 + y^2 + z^2 + w^2 \leq R^2 \right\}
$$
in $\mathbb{R}^4$. I want to make the change of variables
$$
x=r \cos \phi, \quad y=r \sin \phi, \quad z=\rho \cos \theta, \quad w=\rho \sin \theta
$$
to compute the volume of $B$, can you help? Thanks...
$\bf{Added:}$ The substitution above means that $x^2 + y^2 = r^2$, $z^2 + w^2 = \rho^2$, so $r^2 +\rho^2 = R^2$.  Therefore $r = R \cos u$, $\rho = R \sin u$, where $u \in [0, \pi/2]$, since $r$, $\rho\ge 0$. Also $\phi$, $\theta \in [0, 2 \pi]$.
 A: I've taught this: the main thing is to do induction by two dimensions, using polar coordinates. So: we know the volume of the $n$ ball of radius $R$ is $c_n R^n.$ Next, the unit ball in dimension $n+2.$
So, on the unit disc in dimension 2, at a point $(r, \theta)$ we integrate the $n$ volume of a ball with radius $\sqrt {1-r^2},$ which becomes $c_n (1-r^2)^{n/2}$
Our new number, called $c_{n+2},$ becomes
$$ c_{n+2} =  \int_0^{2 \pi}  \int_0^1 c_n (1-r^2)^{n/2}  r dr d\theta = 2 \pi c_n  \int_0^1  (1-r^2)^{n/2}  r dr$$
An antiderivative is
$$ \frac{-1}{n+2} \left( 1 - r^2 \right)^{\frac{n+2}{2}} \; , \; \; $$
and this is to be evaluated between $0$ and $1.$ This vanishes at endpoint $1,$ what remains is the negation of $\frac{-1}{n+2}.$ Put back the coefficients, we reach the simple
$$  c_{n+2} = \frac{2 \pi c_n}{n+2}  $$
For example, $c_1 = 2$ for the line segment of "radius" 1, or length 2. Then
$$ c_3 =  \frac{2 \pi 2}{3} = \frac{4 \pi }{3}, $$
and the volume of the ball with radius $R$ is $ \frac{4 \pi R^3}{3}$
Well, $c_2 = \pi$ for the area of the unit disc, $\pi R^2 \; . \;$ So
$$ c_4 =  \frac{2 \pi \pi}{4} = \frac{ \pi^2 }{2}, $$
and the 4-ball of radius $R$ has 4-volume $\frac{\pi^2 R^4}{2}$
