Find the moment of inertia of a circular disk or solid cylinder of radius `R` about the axis through the centre and perpendicular to the flat surface.

Explanation

The figure shows that the appropriate mass element is a circular ring of radius `r` and width `dr`.
Its area is `dA=2pi rdr` and its mass is `dm=sigmadA`.

where `sigma=M/A` is the real mass density. The moment of inertia of this element is
`dI=dm r^(2)=2piar^(3)dr`
For the whole body.
`I=2pisigma int_(0)^Rr^(3) dr=1/2 pisigmaR^(4)`
The mass of whole disck or cylinder is
`M=sigmaA=sigmaspiR^(2)`, and so `I=1/2MR^(2)`