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We all know from school that density is defined as mass over volume $$\rho = \frac{m}{V}$$

I'm wondering what the mathematically correct definition of density is. I'm considering two options.

OPTION 1. Is density defined by the means of multiple integral? $$m = \iiint \rho \, \mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z$$ This way density is $\rho = \cfrac{\partial^3 m}{\partial x \partial y \partial z}$.

OPTION 2. Is density defined by the means of volume integral? $$m = \int \rho \, \mathrm{d}V$$ This way density is $\rho = \cfrac{dm}{dV}$.

If Option 2 is correct, then what is volume integral? In my engineering calculus course, I learnt about multiple and curvilinear integrals. Curvilinear integrals include line integrals (of two types) and surface integrals (of two types). I assume that volume integral is a curvilinear integral, but I'm not sure.

If volume integral is a curviulinear integral, then what is the way of calculating it in Decart (Cartesian) coordinates? I mean that there are formulas to calculate line integrals in Cartesian coordinates (one formula for each type of the line integral) and there are formulas to calculate surface integrals in Cartesian coordinates (again, each type of the surface integral has a formula to calculate it in Cartesian coordinates); is there an analogous formula for the volume integral?

If both options 1 and 2 are not correct, then what is the mathematically rigorous way to define density?

P.S. I did read this post. But I do still have my question unanswered. Is density a mass derivative over volume or a third mass derivative over three Cartesian coordinates? In other words, I'm trying to figure out is there such a thing as a coordinate along a volume (I know about coordinate along a line - hence, line integral; I know about coordinate along a surface - hence, surface integral; but is there volume integral in the same sense ...)

ANSWERS

  1. I found an answer to a half of my question here. I.e., the answer to the post I've cited basically validates that there's such a thing as volume integral and we can map it to Cartesian coordinates using Jacobian (the same way as we do for surface integrals).

  2. There's a very interesting answer by @CyclotomicField. See two CyclotomicField's comments bellow.

  3. The answer by @Othing calrifies it. Basically, the definition of density is $\rho = m / V$. Then I need to pick a MEASURE I want. The notion of measure is explained in the wikipedia article cited by @CyclotomicField.

  4. The comment by herb steinberg is very interesting as well.

  5. @Peek-a-boo comment is very good! Definitely, check the link provided in the comment.

Thank you very much for all your inputs!

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    $\begingroup$ Mathematically it depends on how you define volume which could be any measure. You just have to pick one. $\endgroup$
    – CyclotomicField
    Commented Mar 28, 2021 at 2:33
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    $\begingroup$ For physics usage $dV=dxdydz$ so it doesn't make much difference. $\endgroup$
    – herb steinberg
    Commented Mar 28, 2021 at 2:54
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    $\begingroup$ You have the definition of density as mass per unit volume. A measure is a means by which you define volume. en.wikipedia.org/wiki/Measure_(mathematics) $\endgroup$
    – CyclotomicField
    Commented Mar 28, 2021 at 3:13
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    $\begingroup$ @IvanNepomnyashchikhwait. My answer doesn't say option one is a definition of density, the definition is $\rho=m/V$. An object has a given mass and volume. You may use that (and an appropriate measure, as CyclotomicField said) to write that mass as an integral over a certain volume, and in that case you arrive at option 2 (of which 1 is a particular case). That integral must agree with the definition, not the other way around. If, say, your volume changes with time, than you must change the formula. If the integral is not defined for any reason, than the definition of density still holds. $\endgroup$
    – Othin
    Commented Mar 28, 2021 at 3:58
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    $\begingroup$ @IvanNepomnyashchikh don't worry about measures yet, think about the usual euclidean space where dV=dxdydz. You could take, say, small (but finite) cube about this point and measure its volume. You could also measure the mass inside that cube, so you may divide these numbers. If, in the limit where that volume goes to zero, your calculation of the density gives a well defined number, then you may say that "the density at $(x,y,z)$ equals that number". + $\endgroup$
    – Othin
    Commented Mar 28, 2021 at 21:06

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The definition of (mass) density is precisely the mass per unit volume. The mathematical formulation of this is the very first formula you wrote: $\rho=m/V$. If you take the mass of any object and divide by its volume, you end up with the density. The formula $$ m=\int_{V}\rho(\mathbf{r}) d^3x $$ is actually used to measure $\rho$, not to define it, so this is actually more of a physical discussion. This last formula implies that the density is a function of your coordinates, and its value at a point $\mathbf{r}$ is given by $\rho=dm/dV$, where $dV$ is the volume element at $\mathbf{r}$. In an experiment you usually can't measure $\rho$ directly. But you can measure mass and volume pretty accurately. So it is possible to measure the mass of a substance/object over a region with a very small (and known) volume. In the limit of an infinitesimal element of volume you would find the function $\rho(x,y,z)$. You'll never reach that limit of course, but you can make progressively better measurements.

Edit: To be clear the integral formula above is just a consequence of the definition of density, and it holds in an euclidean space. I wrote it that way because it is used for measurements, which are usually made in that space. If you change your volume form, the formula changes and there might be extreme cases where you can't define such a formula.

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  • $\begingroup$ @IvanNepomnyashchikh Sure $\endgroup$
    – Othin
    Commented Mar 28, 2021 at 3:07
  • $\begingroup$ Thank you for your response @Othin. I have several follow up questions if you allow. 1) If $m$ changes with $V$ not constantly then we can't use operation of division. We need to use differentiation instead. In this sense, can you clarify your statement that mathematical formulation of density is $m/V$? 2) What do you mean saying that $dV$ is the volume element at point $r$? Point doesn't have dimensions. Therefore, point $r$ should exist at a vertex of $dV$ or anywhere within $dV$. So, where is this point located in $dV$? 3) $m$ as a function of $x, y, z$ implies that $m$ is a field, correct? $\endgroup$
    – Ivan Nepomnyashchikh
    Commented Mar 28, 2021 at 3:12
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    $\begingroup$ @IvanNepomnyashchikh the mass is a property an object has. In the sense that most physicists use it, mass is just a number you attribute to a physical object. It does does not change within a volume. You can always divide them to find its density. Now the thing is, it's very useful to talk about the density of a substance/material or similar, because I don't want to measure mass and volume of every single object. So we may try to apply this definition to an object made of that specific material and which has a small volume. That's where the integral formula comes from. $\endgroup$
    – Othin
    Commented Mar 28, 2021 at 3:21
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    $\begingroup$ 2)For the reasons mentioned above, you want to calculate the density at a point P in your object. As each point of your object you should have a volume element. You may usually just consider the volume as the volume enclosed by your object's surface, and $dV$ will be the volume of the space you are be working on. This doesn't imply a point has any dimension, the volume element is a measure that can be rigorously defined in any orientable manifold. But really, just think of it as $dV=dxdydz$ for now. The differentials at a point are well defined, so $dV$ should be as well. $\endgroup$
    – Othin
    Commented Mar 28, 2021 at 3:36
  • $\begingroup$ 3)@IvanNepomnyashchikh do you mean $\rho$? $m$ does not depend on the coordinates in the definition I gave. In physics a field usually has a value at each point of space and at each instant of time. I suppose $\rho$ (the function inside the integral) could be thought of as a field if you define it to be zero at every point outside of $V$. $\endgroup$
    – Othin
    Commented Mar 28, 2021 at 3:43

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