"Polar" coordinates for balls given by any norm

Question

So, for the usual euclidean norm we have $$ \int_{B_{R}(0)} f(x) dx = \int_{0}^{R}\int_{\mathbb{S}^{n-1}} f(r\omega)r^{n-1}d\omega dr, $$ where the first integral is the surface integral on the sphere, so I was wondering if maybe there was a more general formula, say, given a norm $\lVert \cdot \rVert$ in $\mathbb{R}^n$, you have the ball given by this norm $$ B_{R}^{\lVert \cdot \rVert}(0) = \left\{x \in \mathbb{R}^n : \lVert x \rVert \leq R\right\} $$ and the unit sphere $$ \mathbb{S}^{n-1}_{\lVert \cdot \rVert} = \left\{x \in \mathbb{R}^n : \lVert x \rVert =1\right\}. $$ Could it be that we have $$ \int_{B_{R}^{\lVert \cdot \rVert}(0)} f(x) dx = \int_{0}^{R}\int_{\mathbb{S}^{n-1}_{\lVert \cdot \rVert}} f(r\omega)g(r)d\omega dr, $$ for some $g$ that depends on the norm? If so, how would the $g$ would look like?

Answer 1

If function $f$ is L1 integrable and norm is Lipchitz we can try to use coarea formula with $u(x) = \|x\|$. $$\int_{\Omega} f(x)|\nabla u(x)| d x=\int_{\mathbb{R}}\left(\int_{u^{-1}(t)} f(x) d H_{n-1}(x)\right) d t$$ where $H_{n-1}$ is Hausdorff measure with dimension $n-1$.