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I have a question about the process to find the mass of a sphere with a varying radial density in respect to the radius.

It's something really simple, but I would like someone to explain it me.

Say that the density varies with: $\rho(r)=a-br$

So when you go from $m=\rho V$

Why is it that it changes to $dm=\rho (r) dV$ and not $dm=Vd\rho $?

I thought the density varies only?

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An intuitive explanation: $$dm=\rho\,dV$$ says that a little bit of mass near a point is (approximately) equal to the density at that point times a little bit of volume. This is correct: it is pretty much just the definition of density.

On the other hand, $$dm=V\,d\rho$$ would be saying that a little bit of mass is the entire volume times a little bit of density. This would be wrong (1) because when the density is different in different places, it makes no sense to multiply the whole volume by some specific density; (2) because "a little bit of density" doesn't mean anything anyway.

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  • $\begingroup$ It is true that $dm=\rho dV$ is the more physically intuitive equation, but you actually can say $dm=Vd\rho$ if you interpret $V$ correctly, see my answer. :) $\endgroup$ Commented Dec 16, 2015 at 0:52
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The rate of change of the mass as the volume varies is given by the formula p(r)=a-br in other words dm/dV=p(r).So the density will vary with the radius that vary with volume.

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There actually is a sense in which you can say $dm=Vd\rho$ instead of $dm=\rho dV$.

In the usual interpretation, $dm=\rho dV$ means the infinitesimal bit of mass is equal to the density of an infinitesimal volume, times that extra little bit of volume. The infinitesimal volumes are added up to reach the total volume, and the infinitesimal masses are added up to reach the total mass. This is the more intuitive interpretation, since the (position-dependent) density of a continuous body is physically defined by $\rho=dm/dV$.

For the non-standard $dm=Vd\rho$, instead of slicing up the volume into infinitesimal amounts, you are doing this to the density. Each infinitesimal amount of mass is equal to an infinitesimal amount of density, times the total volume that carries that extra bit of density.

To make this more clear, consider that the whole volume of the body has at least a density of $0$. A smaller (or equal) volume has density at least $d\rho_1$ for some small amount of density. A smaller volume again has density at least $d\rho_1+d\rho_2$, and so on. If you break the maximum density into $n$ small amounts $\rho_{max}=\rho_n=\sum_{i=1}^n d\rho_i$ and define $\rho_k=\sum_{i=1}^k d\rho_i$, you should be able to convince yourself that you can add up the mass of the body by adding up small amounts of mass $dm_i=V(\rho_i)d\rho_i$ where $V(\rho_i)$ is the total volume with at least density $\rho_i$.

So in the continuum limit you can say $dm=V(\rho)d\rho$ and integrate over the density range to find the mass if $V(\rho)$ is the volume of the body that has at least density $\rho$.

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  • $\begingroup$ @A.S., I'm not sure that the interpretation of that relation ($\rho dV=d(\rho V)-Vd\rho$) has much physical relevance. The product $\rho V$ would be $\rho$ times the volume that has density at least $\rho$, and $dV$ would be the difference between $V(\rho)$ and $V(\rho+d\rho)$. The $\rho dV$ in this equation is not the same as the usual $\rho dV$ that the OP is asking about. $\endgroup$ Commented Dec 16, 2015 at 1:09
  • $\begingroup$ Which is basically integration by parts. $\endgroup$
    – A.S.
    Commented Dec 16, 2015 at 1:09
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    $\begingroup$ The only difference between $\rho dV$'s is the sign. $\rho dV=-\rho dV'=V'd\rho -d(\rho V')$ where $V$ is in definition of the OP and $V'$ in you definition. The last term integrates out to zero. Alternatively you can view it as interchange of order of integration. $\endgroup$
    – A.S.
    Commented Dec 16, 2015 at 1:16

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