Use equation [tex]I=\int r^2dm[/tex] to calculate the moment of inertia of a uniform, solid disk with mass M and radius R for an axis perpendicular to the plane of the disk and passing through its center.

We need to integrate over the volume of the disc, which is not smart enough, so we brake the disc into rings and integrate from the center r=0 to r=R, as shown in attachment

Since we know the moment of inertial of a ring to be

I=mr²

We cannot integrate the r² with respect to m, so we are going to write dm as a function of r

dm

So we assume that

The ratio of the area of the rings to the whole disc is equal to ratio of their masses.

dm/M=2πrdr/πR²

Therefore,

dm=2rM/R² dr

So we are going to replace dm in the integral

I= ∫r²dm

I= ∫r²(2rM/R²) dr. From 0 to R

Where 2, M and R are constant,

I= 2M/R² ∫r³dr. From r=0 to r=R

I= 2M/R² (r⁴/4) From r=0 to r=R

I=2M/R² (R⁴/4 - 0)

I=2M/R² × R⁴/4

I=MR²/2.

This is the moment of inertial of the disc and it doesn't depend on the thickness of the disc.