This is a bit physics-y, but I think the question is more of a mathematical bent so I thought it'd be more appropriate to ask it here. To find the moment of inertia of a uniform disk with radius $a$, one notes that $I$ is defined globally as:

$$I = \sum_i^n mr_i^2$$

Where $r$ is the distance of each particle from the axis of rotation, which is perpendicular to and at the center of the disk. In the case that the axis of rotation goes through the center, the moment of inertia can be expressed, in a more handy way, as:

$$\int\int \ r^2 \ dA$$

The area element will be $r \ dr \ d\theta$ since integration of a shape this form will fill the area up.

The limits should be pretty straightforward, and the order doesn't matter, since we're essentially integrating a circle.

$$\int_{\theta = 0}^{2\pi} \int_{r=0}^a \rho \ r^3 \ dr \ d\theta$$

I'm not really too sure on why, but the areal density must be added into the integral. As for intuitively, I'm not exactly sure, but it's a constant.

The integral will end up being:

$$ I = \rho \ (\pi /2) \ a^4$$

Now, the main question I have is, what exactly are we doing when integrating here? When the integrand is $1$, it's the area of a circle, but when there is an integrand, the entire approach all the way up to integration is the exact same, so how different is it really then? I've heard that when there is an integrand, you're finding the volume under the surface, so we're finding the volume under the surface of $r^2$? In polar $r=\sqrt{\theta}$ has a spiral shape that expands out less for greater $\theta$.

My main confusions are as follows:

  • Why including an areal density $\rho$ is necessary.

  • What is meant by double integrating a function $\ne 1$ (so that we're not just finding area)

  • Why this, all put together, shows the moment of inertia of the surface.


I feel I've gained, perhaps a better guess at what double integrating a function implies. The area element is like a selection marker for a function, for lack of a better word, and then integrating finds the volume of the function between that selection and the surface. Is that somewhat correct?

  • $\begingroup$ You didn't specify that the axis of rotation is perpendicular to the plane of the disc, and isn't contained in the plane or at any arbitrary angle. Of course I can easily deduce from the equations you wrote, but it's always better to define the problem precisely. $\endgroup$ – CiaPan Nov 20 '17 at 14:26
  • $\begingroup$ Oh okay, I'll edit it to provide this. $\endgroup$ – sangstar Nov 20 '17 at 14:38
  • $\begingroup$ I personally opine that the grammar of this could use some proofreading, particularly near the beginning. Other than that, wonderful question $\ddot\smile$ $\endgroup$ – gen-z ready to perish Nov 21 '17 at 5:43

As an answer to your second question, double integrating a function (not equal to one) is the continuum analogue of the double sum

$ \sum_{i=1}^n\sum_{j=1}^n f(i,j) .$

In fact, the integral is defined as the limit of sums of this kind, so you are just adding up the total amount of stuff (in this case, mass) in the region you are integrating over.

Were you looking for anything more than this? Reading the last part of your question, it seems like you have more or less arrived at this answer yourself too.


I’m actually a bit confused by the integral you provided. If the discrete case is $I=\sum_i r_i^2m_i$ (which is definitely true), then the analogue for a body of continuous mass should be $$I=\int r^2 \, dm$$ (where I’ve left the limits implied). What you’ve cited makes no mention of mass.

So why is it necessary to include an areal mass density $\rho$? Well, that’s how you introduce the mass for consideration!

The areal mass density $\rho$ is defined as the amount of mass per little piece of area, $\rho=dm/dA$. (There are a couple of ways to make this derivative definition work in practice, but that’s tangential.) This term may be constant throughout the entire mass, or it may vary. If it does, we need to integrate to get an exact, un-averaged value.

To do this, you take $dm=\rho\,dA$, giving you $$I=\int r^2\rho\,dA$$

Now you can introduce your conversion to polar form via the area element substitution, giving you $$I = \iint r^3 \rho \,dr\,d\theta$$

(Sorry if my conversion from a single to a double integral was a bit shady.)

What does it mean to double integrate a function $\neq1$? If I understand you correctly, you’re asking what double integration is. Well, the answer is that we’re considering a piece of mass that’s teeny-tiny in two ways (teeny-tiny in the radial dimension and teeny-tiny in the $\theta$ dimension). Therefore, to arrive a real, not-teeny-tiny answer, we have to add up the values in both directions. If we didn’t integrate, what we would get might be a mere average.

By the way, you can tell each of our steps is accurate, because we have $\dim I = \mathsf{L}^2 \, \mathsf{M}^1$ throughout, which is appropriate for rotational moment of inertia.

Why does this give you the actual moment of inertia? Because we started from an accurate definition that has been empirically proven and manipulated it using valid calculus.


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