A thin disc of mass \(M\) and radius \(R\) has mass per unit area \(\sigma (r) = kr^2\) where \(r\) is the distance from its centre. Its moment of inertia about an axis going through its centre of mass and perpendicular to its plane is

\({I_{Disc}} = \int\limits_0^R {\left( {k{r^2}2\pi rdr} \right)} {r^2}\,\,\,\,\,\,\,\,\,Mass\,of\,disc\)

\({I_{Disc}} = 2\pi k\int\limits_0^R {{r^2}dr\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,M = } \int\limits_0^R {2\pi rdr\,k{r^2}} \)

\({I_{Disc}} = 2\pi k\left( {\frac{{{r^6}}}{6}} \right)_0^R\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,M = 2\pi k\int\limits_0^R {{r^3}dr} \)

\({I_{Disc}} = 2\pi k\frac{{{R^6}}}{6}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,M = 2\pi k\left. {\frac{{{r^4}}}{4}} \right|_0^R\)

\({I_{Disc}} = \frac{{\pi k{R^6}}}{3} = \left( {\frac{{\pi k{R^4}}}{2}} \right)\frac{{{R^2}2}}{3}\,M = 2\pi k\left. {\frac{{{r^4}}}{4}} \right|_0^R\)

\({I_{Disc}} = \frac{{M2{R^2}}}{3}\,\,\,\,\,;\,\,\,\,\,\,\,\,{I_{Disc}} = \frac{2}{3}M{R^2}\)