# How to derive moment of inertia for a thin disc?

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?

• Do you understand polar coordinates? Oct 12, 2023 at 19:25
• @Ghoster Yes, I do Oct 12, 2023 at 19:27
• What range of polar coordinates describes a thin disk of radius $R$, when the origin is at the center? Oct 12, 2023 at 19:29
• @Ghoster Would it be 0 to $2\pi$? Oct 12, 2023 at 19:35
• There are two polar coordinates. That’s the range for $\theta$. It’s not the range for $r$. Oct 12, 2023 at 20:07