Integration of radial functions?

Spherical coordinates in $\mathbb{R}^n$ is treated fully in Karl Stromberg's book "Introduction to Classical Real Analysis" p. 369-370, see here. Note that on p. 370, it shows the Jacobian of the coordinate transformation as $$r^{n-1} \prod_{j=1}^{n-2} (\sin\theta_j)^{n-j-1}$$ So if your function is radial, then \begin{align*}\int_{\mathbb{R}^n}f(|x|)\mathrm{d}x &=\int_{-\pi}^\pi\int_0^\pi\int_0^\pi\cdots\int_0^\infty f(r)r^{n-1}\prod_{j=1}^{n-2} (\sin\theta_j)^{n-j-1}\,dr \, d\theta_1\cdots d\theta_{n-2}\,d\theta_{n-1}\\&= \omega_{n-1}\int_0^\infty f(r) r^{n-1} \mathrm{d}r,\end{align*} where $$\omega_{n-1}=\int_{-\pi}^\pi\int_0^\pi\int_0^\pi\cdots\int_0^\pi \prod_{j=1}^{n-2} (\sin\theta_j)^{n-j-1}\,d\theta_1\cdots d\theta_{n-2}\,d\theta_{n-1}$$

Since $\mathbb{R}^n=\{0\}\cup (0,\infty)\times S_r^{n-1}$, its volume element $dV$ can be written as a the product of $dr$ times the volume element of the $n-1$ sphere of radius $r$. By dimensional grounds, the latter is equal to $r^{n-1}$ times the volume element of the unit $(n-1)$-sphere. For example, if $n=2$ one has $dV=r dr\wedge d\theta$, note that the integral of $d\theta$ over $S^1$ is $2\pi$, i.e. the "volume" of $S^1$; similarly for $n=2$, $dV=r^2 \sin\theta\, dr\wedge d\theta$ and $\int _{S^2} \sin\theta d\theta\wedge d\phi=4\pi$. Since your function is radial, you can factorize the integral into a radial and an angular part. The integral of the angular part, which you denote with $\omega_{n-1}$, is therefore the volume of the unit $(n-1)$-sphere. A nice way of calculating $\omega_{n-1}$ is presented here http://scipp.ucsc.edu/~haber/ph116A/volume_11.pdf

There are a couple good answers already, however I thought I would add a more elementary approach, avoiding the (direct) use of measures or spherical coordinates. The price we pay here is that we must assume $f$ to be continuous, as opposed to merely integrable. (We can get around this, since one can approximate integrable functions with continuous functions in the $L^1$-norm.)

Claim: If $f$ is a continuous real valued function on $\mathbb{R}$, then $$\int_{\mathbb{R}^n}f(|x|)\,dx=\omega_{n-1}\int_0^{\infty}f(r)r^{n-1}\,dr.$$

Proof: Fix $R>0$, set $B_R:=B(0,R)=\{|x|<R\}$. For each fixed integer $N$, one can partition $B_R$ into $N$ sub-annuli $B^i_R\subset B_R$, $1\le i\le N$, where each $B^i_R=\{r_{i-1}\le|x|<r_i\}$, and $r_0=0<r_1<\cdots<r_{N-1}<r_N=R$. We will make the partition size equal, and denote it by $\delta_N:=R/N$. Note that for each $i\in\{1,\dots,N\}$, the mean value theorem furnishes a number $s_i\in(r_{i-1},r_i)$ such that $$r_{i}^n-r_{i-1}^n=ns_i^{n-1}(r_{i}-r_{i-1}).$$ With this in hand (and keeping in mind we are working with the integral of a radial function), we define a new function $f_N:B_R\to\mathbb{R}$ by setting $$f_N(x)=f(s_i),$$ whenever $x\in B^i_R$. We wish to show that $f_N(\cdot)\to f(|\cdot|)$ uniformly on $B_R$ as $N\to\infty$. Fix $\epsilon>0$ and choose $N$ so large that $\delta_N<\delta$, where $\delta>0$ is such that $|f(r)-f(s)|<\epsilon$ whenever $r,s\in B_R$ with $|r-s|<\delta$ (this $\delta$ exists by the uniform continuity of $f$ on $B_R$.) Note then that for any $x\in B_R$, then $x\in B^i_R$ for some $i$ and we have $$|f_N(x)-f(|x|)|=|f(s_i)-f(|x|)|<\epsilon.$$ Therefore $f_N(\cdot)\to f(|\cdot|)$ uniformly on $B_R$ as $N\to\infty$ and hence we have $$\int_{B_R}f_N(x)\,dx\to\int_{B_R}f(|x|)\,dx$$ as $N\to\infty.$ Note that the volume of $B^i_R$ is equal to $\omega_{n-1}(r^n_i-r^n_{i-1})=\omega_{n-1}s_i^{n-1}(r_i-r_{i-1})=\omega_{n-1}s_i^{n-1}\delta_N$. Now, using this and the definition of Riemann integration, we can compute $$\int_{B_R}f_N(x)\,dx=\sum_{i=1}^Nf(s_i)\omega_{n-1}s^{n-1}_i\delta_N\to\omega_{n-1}\int_0^Rf(r)r^{n-1}\,dr,$$ as $N\to\infty$. Hence we have $$\int_{B_R}f(|x|)\,dx=\omega_{n-1}\int_0^Rf(r)r^{n-1}\,dr.$$ Letting $R\to\infty$ gives the desired result.