Proving that $\frac{\sigma_{n-1}}{\omega_n} = n$ in $\mathbb{R}^n$

Question

If $\sigma_{n-1}$ was the surface area of the unit sphere in $\mathbb{R}^n$ and $w_{n}$ was the area of the unit ball in $\mathbb{R}^n$, my lecture notes prove that $$\frac{\sigma_{n-1}}{\omega_n} = n$$ by the proof $$\omega_n = \int_{B_1(0)} dx$$$$=\int_0^1\sigma_{n-1} z^{n-1} dz$$$$=\sigma_{n-1}\int_0^1z^{n-1}dz$$$$=\frac{\sigma_{n-1}}{n}$$ but I don't understand how we got from the 1st line to the second line, namely that $$ \int_{B_1(0)} dx =\int_0^1\sigma_{n-1} z^{n-1} dz$$ could someone please explain this to me? Thanks!

Answer 1

It just has a polar transformation

Answer 2

As @robjohn said, fill the unit ball in $\mathbb R^n$ up with concentric shells of radius $r$, $0<r\le 1$. The "surface area" of the sphere of radius $r$ is $\text{area}(S_r)=r^{n-1}\omega_n$, and so $$\text{volume}(B_1(0)) = \int_0^1 \text{area}(S_r)\,dr = \int_0^1 r^{n-1}\omega_n\,dr\,.$$ (Note that $k$-dimensional volume scales by a factor of $r^k$ when you stretch the object by a factor of $r$.)