## Essential University Physics: Volume 1 (4th Edition) Clone

$\frac{1}{2}MR^2$
We know the following value for $dm$: $dm = \frac{2Mrdr}{R^2}$ Thus, we find: $I = \int_{0}^{R}r^2dm=I = \int_{0}^{R}r^2 \frac{2Mr}{R^2}dr = \frac{1}{2}MR^2$