Why $\frac{m}{V}$ gives average density when density varies?

Question

This might be a very basic question, and for years I have gone by simply accepting it as a fact. That whenever the density is uniform the formula $\frac{m}{V}$ gives that uniform value of density and when the density varies the same formula gives the average value of density. However, I don't understand why the latter is true. Why $\frac{m}{V}$ gives the average value of density when it is not uniform?

I have worked something out but I'm not really sure if that's the right way to go about it.

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My Work: Let there be a quantity of matter of mass m kg and volume V$mm^3$. The density is non-uniform that is every unit volume (say 1$mm^3 $) doesn't have the same mass. Let $\rho_1 , \rho_2, \rho_3...$ be the masses of unit volumes taken in the quantity of matter. We can write,

$$\rho_1 kg +\rho_2 kg+\rho_3 kg+.........+\rho_n kg=m kg \quad \quad (1)$$ and $$n \text{ mm}^3= Vmm^3 \quad \quad (2)$$

The average density will be, $$\rho_{avg} \frac{kg}{mm^3}=\frac{\rho_1 + \rho_2 +\rho_3 +.....+ \rho_n}{n} \frac{kg}{mm^3}$$ Dividing (1) and (2),

$$\frac{\rho_1 kg +\rho_2 kg+\rho_3 kg+.........+\rho_n kg}{n \text{ mm}^3}=\frac{m kg}{Vmm^3}$$

$$\rho_{avg} \frac{kg}{mm^3}= \frac{m}{V} \frac{kg}{mm^3}$$

$$\rho_{avg} = \frac{m}{V}$$

Answer 1

Think the similarities between density and the velocity. Both of them can be expressed as a quantity over another quantity.

When you are travelling from city A to B with a car your velocity is very seldom constant. However for many practical uses and purposes the average velocity is a meaningful measure. The average velocity, will not be the most commonly encounter velocity during your travel, however it will give in one measure salient information about the trip.

Of course you could calcualate the instantaneous velocity (distance travelled over a very small time duration divided by the small time duration $\left.\frac{\Delta L}{\Delta t}\right|_{\Delta_t \rightarrow 0}$) and that will give you the information about the velocity at a very specific point in your trip. (however this is not very useful for the entire trip).

Hopefully, it is clear that the average density in composite materials is not equal to the density of any of the phases (at least in the majority of the cases), however the average density is a useful metric that can be used when performing calculations (in a similar manner to the velocity above).

Answer 2

Your solution approach was a good start, you just need to take it all the way to infinity. In what follows $\boldsymbol{x}$ is an arbitrary point in the body under consideration.

First, unpack what you mean when you say "average density". It's the average of $\rho(\boldsymbol{x})$, the pointwise density.

Then, the average of a scalar field is given by $$ \rho_{\text{avg}}=\frac{\int_\Omega \rho(\boldsymbol{x})\, dV}{\int_\Omega dV} $$

And working out each term you have $$ \begin{aligned} \int_\Omega \rho(\boldsymbol{x})\, dV = m \\ \int_\Omega dV = V \end{aligned} $$

And you get your expression $$ \rho_{\text{avg}}=\frac{m}{V} $$