Integral involving Dirac delta: two different results?

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…

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