Here, let us fix a Borel measurable function $f:\mathbb{R}^n\to\mathbb{C}$. Then we can evaluate its Lebesgue integration via spherical polar coordinates by the following way :
$\textbf{Notation & Setting :}$
$\bullet$ The spherical polar coordinates for $x\in\mathbb{R}^n\setminus\{0\}$ are $ r=|x|\in(0,\infty)$, $\hspace{.1cm}$ $x'=\frac{x}{|x|}\in\mathbb{S}^{n-1}$ ; where $\mathbb{S}^{n-1}$ is the unit sphere $\{x\in\mathbb{R}^n:|x|=1\}$.
(Remark : $\mathbb{S}^{n-1}$ is a hypersurface in $\mathbb{R}^{n}$.)
$\bullet$ The map $\Phi(x)=(r,x')$ is a homeomorphism from $\mathbb{R}^n\setminus\{0\}$ onto $(0,\infty)\times\mathbb{S}^{n-1}$.
$\bullet$ We denote by $m_{\ast}$ the Borel measure on $(0,\infty)\times\mathbb{S}^{n-1}$ induced by $\Phi$ from the Lebesgue measure $m$ on $\mathbb{R}^n$, that is, $m_{\ast}(E)=m(\Phi^{-1}(E))$.
$\bullet$ We also define the measure $\rho$ on $(0,\infty)$ by $d\rho=r^{n-1}dr$.
$\textbf{Theorem :}$ There exists a unique Borel measure $\sigma=\sigma_{n-1}$ on $\mathbb{S}^{n-1}$ such that $m_{\ast}=\rho\times\sigma$. If $f\geq0$ or $f\in L^1(\mathbb{R}^n,m)$, then $$ \int_{\mathbb{R}^n}f(x)dx=\int_0^{\infty}\int_{\mathbb{S}^{n-1}}f(rx')r^{n-1}d\sigma(x')dr. $$ (One can find all such details and the proof of the above theorem in Folland's book.)
$\textbf{My question :}$
Do we have the same kind of theorem for an arbitrary compact smooth hypersurface $S$ in $\mathbb{R}^n$ ? In other words, can we have the expression $$ \int_{\mathbb{R}^n}f(x)dx=\int_0^{\infty}\int_{x'\in S}f(rx')(\cdot)d\sigma(x')dr ? $$ ; where $r$ is an appropriately defined distance function, $\sigma$ is a surface measure on our hypersurface $S$ and $(\cdot)$ is/are some mathematical term(s) that justifies the above equality.
Any suggestion of a reference book is also welcomed.