How to derive moment of inertia for a thin disc?

Question

Wikipedia's article on moment of inertia provides examples of how to derive the moments of inertia for different bodies:

$$I_P=\iiint_Q p(x,y,z)||\mathbf{r}||^2 dV$$

It gives an example of how to derive the moment of inertia for a thin rod:

$$I_{C,rod}=\iiint_Q p(x,y,z)||\mathbf{r}||^2dV=\int_{-\frac{l}{2}}^{\frac{l}{2}} px^2s dx=\left.px\frac{x^3}{3}\right\rvert_{-\frac{l}{2}}^\frac{l}{2}=\frac{ps}{3}\left(\frac{l^3}{8}+\frac{l^3}{8}\right)=\frac{ml^2}{12}$$

That was pretty straightfoward, but I don't understand why you need a double integral for the example that follows, the derivation for a thin disk:

$$I_{C,disk}=\iiint_Q p(x,y,z)||\mathbf{r}||^2dV=\int_0^{2\pi}\int_0^Rpr^2sr\, dr\, d\theta=2{\pi}ps\frac{R^4}{4}=\frac{1}{2}mR^2$$

I know it is because $V({r,\theta})$ is a function of two variables, but I still don't understand where those limits come from. Could someone please explain?

Answer 1

When you calculate the moment of inertia for a thin rod, you reduce the problem to one dimension because you're only considering it's length.

A disk is a 2D object, hence you need those two spatial coordinates to characterize it, the radial and angular components. Their origin becomes straighforward once you learn about the "polar coordinate system", which I encourage you to investigate. As for the limits of integration, think of a straight line of length R with an edge at the origin. Then rotate that line 2π radians around the origin, a full 360° turn that is. This will draw the disk/circle you are trying to parameterize.