# Separating the volume integral into surface integral and a radial integral

Let $B(x,r)$ denote a ball in $\mathbb{R}^n$. I know that $$\int_{B(x,r)} {f(x)dx} = \int_0^r ds \int_{\partial B(x,s)} f(x) d\sigma$$ where $d\sigma$ denotes the surface measure on $\partial B(x,s)$.

My question is that can this equality be extended to other types of surfaces than a ball? More specifically, if $\Omega$ is bounded, then does the follow equality hold? $$\int_{B(x,r)} {f(x)dx} = \int_0^r ds \int_{\partial (s\Omega)} f(x) d\sigma$$ Where $s\Omega = \{sx:x\in \Omega\}$.

More generally, can the relation be somehow extended to surfaces that "continuously" transform? Let $\phi(s,x):[0,1]\times D \rightarrow \mathbb{R}^n$ (where $D\subset \mathbb{R}^{n-1}$ is compact) be continuous and 1-1 so that $x\mapsto \phi(s,x)$ is a different surface in $\mathbb{R}^{n-1}$ for every $s\in [0,1]$. Then would the following equality hold in some sense? $$\int_{\phi([0,1]\times D)} {f(x)dx} = \int_0^1 ds \int_{\phi(s,D)} f(x) d\sigma$$ Any references that may clear up my question would also be very helpful.