I try to calculate the surface of an n-sphere by induction. For a sphere of radius $r$ in $n$ dimensions the surface should be: $$S_{n-1}(r)=\frac{2\pi^{n/2}}{\Gamma(n/2)}r^{n-1}$$
So I think I should be able to calculate it is by taking all possible choices for spheres with one dimension less: $$S_{n-1}(r)=\int_{-r}^r dx S_{n-2}(\sqrt{r^2-x^2})$$
Let's try: $$\int_{-r}^r dx \frac{2\pi^{n/2-1/2}}{\Gamma(n/2-1/2)}\sqrt{r^2-x^2}^{n-2}=\frac{2\pi^{n/2-1/2}}{\Gamma(n/2-1/2)}r^{n-1}\int_{-1}^1 dy \sqrt{1^2-y^2}^{n-2}$$
by change of variable $x=ry$,$dx=rdy$. So far looks good. Let's evaluate the integral using Beta function using another change $y=\sqrt{t}$ $dy=t^{-1/2}/2$: $$\int_{-1}^1 dy \sqrt{1^2-y^2}^{n-2}=\int_{0}^{1}dtt^{-1/2}\left(1-t\right)^{\left(n-2\right)/2}=\frac{\sqrt{\pi}\Gamma(n/2)}{\Gamma(n/2+1/2)}$$
so I don't get the correct result...
The problem with the above derivation is using the wrong measure on $x\in[-r,r]$. Using the correct measure: $$S_{n-1}(r)=\int_{-r}^r \frac{r dx}{\sqrt{r^2-x^2}} S_{n-2}(\sqrt{r^2-x^2})$$
This measure is derived using polar coordinates where $\theta\in[-\pi/2,\pi/2]$; in this representation $x=r\sin\theta$, $\sqrt{r^2-x^2}=r cos\theta$ and $dx=r \cos\theta d\theta$.
Now we have: $$\int_{-r}^r \frac{r dx}{\sqrt{r^2-x^2}} \frac{2\pi^{n/2-1/2}}{\Gamma(n/2-1/2)}\sqrt{r^2-x^2}^{n-2}=\frac{2\pi^{n/2-1/2}}{\Gamma(n/2-1/2)}r^{n-1}\int_{-1}^1 dy \sqrt{1^2-y^2}^{n-3}$$
by change of variable $x=ry$,$dx=rdy$. Let's evaluate the integral using Beta function using another change $y=\sqrt{t}$ $dy=t^{-1/2}/2$: $$\int_{-1}^1 dy \sqrt{1^2-y^2}^{n-3}=\int_{0}^{1}dtt^{-1/2}\left(1-t\right)^{\left(n-3\right)/2}=\frac{\sqrt{\pi}\Gamma(n/2-1/2)}{\Gamma(n/2)}$$
so that collecting all the terms we have the wanted result.