I am evaluating the integral over all space
$$\int \delta \left(r^2 - R^2\right) d \vec r$$
At first, I did this:
$$\int \delta \left(r^2 - R^2\right) d \vec r = 4 \pi \int_0^\infty \delta \left(r^2 - R^2\right) r^2 dr = 4 \pi R^2 $$
But then someone made me notice that we can use the property
$$\delta[f(r)] = \sum_i \frac{\delta (r-r_i)}{\mid f'(r_i) \mid} $$
where $r_i$ are the roots of $f$. The only root we have to consider, since $r\geq0$, is $+R$; I thus obtained
$$\delta \left(r^2 - R^2\right) = \frac{\delta(r-R)}{2R}$$
which holds the result
$$ \int \delta \left(r^2 - R^2\right) d \vec r = \frac{4 \pi}{2R} \int \delta(r-R) r^2 dr = 2 \pi R$$
Which result is the right one?
Update
I am starting to think that the first result is wrong because I am basically assuming that
$$\int_0^\infty \delta[f(x)] g(x) dx = g(x_i)$$
where $x_i$ is the only root of $f(x)$ contained in the interval $[0,\infty)$. But this is wrong because the normalization ($\mid f'(x_i) \mid^{-1}$) is missing...
Suppose you make the substitutions $u=r^2$, $U=R^2$. Then $du=2r\,dr$ and so $$4\pi \int_0^\infty \delta(r^2-R^2)\,r^2\,dr =4\pi \int_0^\infty \delta(u-U)\,\frac{\sqrt{u}}{2}\,du=2\pi \sqrt{U}=2\pi R$$ in accordance with the normalized formula. So a substitution which linearizes the argument of the delta reduces the integral to the standard case.