## University Physics with Modern Physics (14th Edition)

$I = \frac{1}{2}MR^2$
Equation (9.20) states: $I = \int~r^2~dm$ Let $\rho$ be the area density of the disk. $I = \int_{0}^{R}~r^2~\rho~(2\pi~r~dr)$ $I = \rho\int_{0}^{R}2\pi~r^3~dr$ $I = \rho~(\frac{\pi~r^4}{2})\vert_{0}^{R}$ $I = \rho~(\frac{\pi~R^4}{2})$ $I = \frac{1}{2}~\rho~(\pi~R^2)~R^2$ $I = \frac{1}{2}~\rho~(Area)~R^2$ $I = \frac{1}{2}MR^2$