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.

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),

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

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

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

  • $\begingroup$ I think this question might be better addressed in the Mathematics SE - I think this is a question of why is an "average" defined as the sum of a set of values divided by the number of elements in the set. $\endgroup$
    – J. Ari
    Commented Aug 30, 2022 at 14:33
  • $\begingroup$ @J.Ari I would like to point out, I don't want to know why average is defined as sum of the values divided by the number of values. I'm okay with accepting it, what I want to know is why the average density is m/V and not say M/2V or 6M/V or M/3V? why m/V itself? That's what I have tried proving as well, in the question. I'm not sure however that it is the right way of doing it, it does lead to the same result though. $\endgroup$ Commented Aug 30, 2022 at 14:49
  • $\begingroup$ @J.Ari I think I get you now. May be answering why averages are defined that way in maths will help to know why this average density is defined as m/V. I must ask a similar question in math stack exchange. $\endgroup$ Commented Aug 30, 2022 at 18:04

2 Answers 2


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).

  • $\begingroup$ Hi NMech. I do understand the importance of average quantities and the similarities between density and velocity. I was specifically interested in knowing why when any quantity say density varies, $\frac{m}{V}$ will give me average density. Or in the case of velocity why $\frac{\Delta L}{\Delta t}$ give me the average velocity of time $\Delta t$, when velocity varies. How did we know that the average of all those instantaneous velocities would be equal to $\frac{\Delta L}{\Delta t}$? $\endgroup$ Commented Aug 30, 2022 at 12:03
  • $\begingroup$ NMech, I've made some edits. $\endgroup$ Commented Aug 30, 2022 at 13:24

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

  • $\begingroup$ Thanks, given that we know calculus this is a good approach to finding the average density. However, I think calculus came after average values of quantities were defined. One approach that I found by asking a question on math SE was that we first define what average density is. We define it as that value of density that if were to remain uniform would've given me the same total mass. In this way $\rho_{avg} V=m$ and we will get our result. $\endgroup$ Commented Aug 31, 2022 at 8:00

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