Prove $\int_{\mathbb R^{d}}f(|y|)d\lambda^{d}(y)=C_{d}\int_{[0,\infty[}r^{d-1}f(r)d\lambda^{1}(r)$

Question

Let $f:[0,\infty[ \to \bar{\mathbb R}$ measurable, $d \in \mathbb N$ while $E_{d}:=\{x \in \mathbb R^d: |x| \leq 1\}$

Prove $\int_{\mathbb R^{d}}f(|y|)d\lambda^{d}(y)=C_{d}\int_{[0,\infty[}r^{d-1}f(r)d\lambda^{1}(r)$

and $C_{d}$ is constant with $C_{d}:=2\int_{E_{d-1}}\frac{1}{\sqrt{1-|x|^2}}d\lambda^{d-1}(x)$

Normally, I would write down a few of my ideas but I have no idea where to begin. I have only just acquainted myself with the $\lambda^{d}$ transformation formula, but do not see its applicability in this instance.

Any help is greatly appreciated

Answer 1

It is the polar coordinate change of the Lebesgue measure. To prove the validity of the formula, we can first test it for $f$ of the form $$ f(r) = 1_{(a,b)}(r). $$ Then by the volume formula of $d$-dimensional ball, we can check that it's true. Since it is true for any interval $(a,b)$, the result can be extended to any measurable set $E$ by approximation, and hence to any non-negative measurable $f$. This proves the formula for general integrable $f$.

Answer 2

Change to polar coördinates. If $\sigma$ is surface area measure on $S^{d - 1}$ and $s_{d-1}$ is the surface area of $S^{d-1}$, you have the following. $$\int_{\mathbb{R}^d}f(|x|)\,dx = \int_0^\infty \int_{S^{d-1}} f(r)\,d\sigma\, dr = s_{d-1} \int_0^\infty r^{d-1}f(r)\, dr$$