Integral of a radial symmetric function.

I've found an article that essentially states that an integral of the form

$$I[f] = \int_{\mathbb{R}^N} f(x_1,\ldots,x_N) dx_1 \ldots dx_N = \int_{\mathbb{R}^N} f(r) dx_1 \ldots dx_N$$

where $$r = \sqrt{x_1^2 + \ldots x_N^2}$$, can always be split as product of two integrals, where the first integral is the surface of the $$n$$ dimensional sphere, say $$A_n$$ and the second integral is the radial integral

$$R[f] = \int_0^1 f(r) r^{N-1}dr$$

The article I've found only states the result, but it doesn't actually prove it, but assuming this results holds it proves that

$$A_N = \frac{2 \pi^{N/2}}{\Gamma\left( \frac{N}{2}\right)}$$

Is there a way to prove the statement which doesn't rely on generalized polar coordinates, which are a bit hard to remember and manipulate, if there's no other way is there an easy way to derive to generalized polar coordinates and again derive the result?

I would assume the proof is a calculus fact.

Since $$f$$ is radially symmetric, $$f(\vec{x})d^N\vec{x}=r^{N-1}f(r)drd\Omega$$, with $$d\Omega$$ an $$(N-1)$$-dimensional infinitesimal element over the angle coordinate(s). We don't need to work out how $$d\Omega$$ looks at all, so don't get out your $$\sin\theta$$s here. The claimed factorisation is immediate.
If you prefer a different proof that an $$(N-1)$$-sphere of radius $$R$$ has measure $$\frac{2\pi^{N/2}}{\Gamma(N/2)}R^{N-1}$$, let's prove instead that an $$N$$-ball of radius $$R$$ has volume $$\int_0^R\frac{2\pi^{N/2}}{\Gamma(N/2)}r^{N-1}dr=\frac{\pi^{N/2}}{\Gamma(N/2+1)}R^N$$. (To clarify, an $$n$$-ball is $$n$$-dimensional, but an $$n$$-sphere is the $$n$$-dimensional surface of an $$(n+1)$$-ball.) Equivalently, the unit $$N$$-ball has measure $$V_N:=\frac{\pi^{N/2}}{\Gamma(N/2+1)}$$. We'll proceed by induction.
The result is correct for $$N=0$$; a "$$0$$-sphere" is the single point in $$0$$-dimensional space, and the definition of measure is counting that point. (If that argument seems to strange, take $$N=1$$ as your base step, for which measure is a line segment's width, or failing that $$N=2$$, for which you just want a circle's area.) By slicing an $$N$$-ball into hypercylinders of infinitesimal thickness with $$(N-1)$$-ball cross sections of radius $$\sqrt{1-x^2}$$,$$V_N=\int_{-1}^1V_{N-1}(1-x^2)^{(N-1)/2}dx=2V_{N-1}\int_0^1(1-x^2)^{(N-1)/2}dx\\=V_{N-1}\int_0^1y^{-1/2}(1-y)^{(N-1)/2}dx=V_{N-1}\operatorname{B}\left(\frac12,\,\frac{N+1}{2}\right).$$(The intuition of this is easiest with slicing a sphere into cylinders, since you can visualise that.) So the inductive step is$$V_{N-1}=\frac{\pi^{(N-1)/2}}{\Gamma((N+1)/2)}\implies V_N=\frac{\pi^{(N-1)/2}}{\Gamma((N+1)/2)}\frac{\sqrt{\pi}\Gamma((N+1)/2)}{\Gamma(N/2)+1},$$which is what we needed.
• @user8469759 Well, that's just the polar form of $d^N\vec{x}$. It needs to be proportional to $d(r^N)$, because an $N$-ball of radius $R$ wouldn't otherwise have measure $\propto R^N$. – J.G. Oct 19 '19 at 18:24
• @user8469759 Yes, like when you write $f(x)dx=f(x)g^\prime(u)du$ if $x=g(u)$. – J.G. Oct 19 '19 at 18:30