Proof of $\int_{B(x,R)}f(y)\,dy=\int_{\mathbb{S}^{n-1}}\int_0^Rf(x+r\sigma)r^{n-1}\,dr\,d\sigma$.

Question

I have sometimes applied the following result: $$\int_{B(x,R)}f(y)\,dy=\int_{\mathbb{S}^{n-1}}\int_0^Rf(x+r\sigma)r^{n-1}\,dr\,d\sigma,$$ where $\mathbb{S}^{n-1}=\{x\in\mathbb{R}^n:\,|x|=1\}$. Could you provide me a proof (or at least an idea) of this result?

Answer 1

Sketch. The integral on the left coincides with $\mathrm{B}(x, R) - \{0\}.$ Define the map $(0, R) \times \Bbb S^{n - 1} \to \mathrm{B}(x, R) - \{0\}$ by $(r, \sigma) = r \sigma.$ This is a bijection. Impose image sigma field and image measure of Lebesgue on the product space and use Lebesgue-Fubini theorem.