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?

$$
V_3(R)=\int_{x^2+y^2+z^2\le R^2}dxdydz=\int_{x^2+y^2\le R^2}\Bigl(\int_{z^2\le R^2-(x^2+y^2)}dz\Bigl)dxdy.
$$
Now change to polar coordinates: $x=r\cos\theta$, $y=r\sin\theta$, to get
$$
V_3(R)=\int_0^{2\pi}\int_0^R\Bigl(\int_{z^2\le R^2-)r^2}dz\Bigl)r\,drd\theta=\int_0^{2\pi}\int_0^RV_1\bigl(\sqrt{R^2-r^2}\,\bigr)\,r\,drd\theta.
$$
