About a differential mass element when $rho = rho(r)$

Question

I'm confused by the fact that, if
$$rho(r) = m(r)/V(r)$$
then
$$m(r) = rho(r) V(r)$$
so
$$frac{dm}{dr} = frac{drho}{dr} V(r) + frac{dV}{dr} rho$$
expanding
$$frac{dm}{dr} = frac{drho}{dr} V(r) + 4pi r^2 rho$$
finally
$$boxed{dm = V(r) drho + 4pi r^2 rho dr}$$
why is that, e.g., in the context of stars (and assuming a spherical geometry), we say that
$$boxed{frac{dm}{dr} = 4pi r^2 rho}$$
considering:

$$dm_{mathrm{box.}} = (dA dr rho) = (r^2 dOmega dr) rho$$
$$dm(r) = int_{Omega} dm_{mathrm{box.}} = 4pi r^2 dr rho$$

Question: is it because $rho equiv frac{dm}{dV} neq frac{m}{V}$ ?

J. Murray · Accepted Answer

Your starting assumption is wrong.  When you write $rho(r) = m(r)/V(r)$, what are the functions $m(r)$ and $V(r)$ supposed to mean?
The density $rho(mathbf r)$ is implicitly defined to be the quantity which you integrate over space to obtain the mass:
$$M = int rho(mathbf r) dV$$
The right way to think about it is to imagine a small volume $delta V$ centered at the point $mathbf r$.  If that small volume has mass $delta m$, then you can calculate the ratio $frac{delta m}{delta V}$, and this is what we call the density $rho(mathbf r)$.  In other words,
$$delta m = rho(mathbf r) delta V$$
In general, there is not a "mass function" which you differentiate with respect to volume (it's not clear what that even means) to obtain the density.  Instead, the density is the thing you integrate to get the mass.

About a differential mass element when $rho = rho(r)$

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