# Moment of Inertia of half of a cylinder

Let's say I have a solid cylinder of uniform mass density, radius $R$ and height $h$. I know that the moment of inertia of this cylinder rotating about the axis parallel to the height and passing through the center of mass is $\frac{MR^2}{2}$. How would the Moment of Inertia (about the same axis) change if I were to cut this cylinder in half? (The cut goes along the length of the cylinder)

$$I=\rho\int_{z_1}^{z_2}\int_0^{2\pi}\int_0^R r^3drd\theta dz$$ With
$$\rho=\frac{M}{V}=\frac{M}{h\pi R^2}$$
Half of the cilinder means that $\theta$ goes from $0$ to $\pi$. Also you conserve the density.
• So it still remains $\frac{MR^2}{2}$. Just that this value is half because the mass is half. – nootnoot Oct 14 '16 at 2:39