homework and exercises - Why doesn't $vec{E} =frac{1}{4piepsilon_0} intfrac{rho hat{r};dxdydz}{r^2}$ blow up at $r=0$, when $rho$ is finite?

Electric field at $(x,y,z)$ produced by a continuous distribution of charges is given by:

$$\mathbf{E}(x,y,z) =\dfrac{1}{4\pi\epsilon_0} \int\dfrac{\rho(x',y',z') \mathbf{\hat{r}} \;\mathrm{d}x'\mathrm{d}y'\mathrm{d}z'}{r^2}.$$ Now, as Edward Purcell in his book writes :

This equation can be used to find the field at any point within the distribution. The integrand doesn't blow up at $r=0$ because the volume element in the numerator is in that limit proportional to $r^2 \mathrm{d} r$. That is to say, so long as $\rho$ remains finite everywhere, even in the interior or of a charge on the boundary of a charge distribution.

Why does the equation doesn't blow up at $r=0\;?$ Why does the volume become $r^2 \mathrm{d}r$ at limit?