Moment of inertia

Here is the solution finally.

Viewed from a point of view parallel to the axis, we can consider the disc as an ellipse of semi-axes $R$ and $R/2$ , with equation $4x^2+y^2=R^2$ . Hence, considering the axis as the $z$ axis, a point of mass $\mathrm{d}m$ of the "ellipse" on this plane is at distance $r=\sqrt{x^2+y^2}$ from the axis. This way, its moment of inertia can be calculated as

$I=\int \int \mathrm{d}m (x^2+y^2).$

In this plane, being the disc of homogeneous mass, we can write the proportion

$\frac{\mathrm{d}m}{\mathrm{d}x\mathrm{d}y}=\frac{M}{\pi R^2/2},$

(the area of an ellipse with semi/axes $a$ and $b$ is $\pi ab$ ). So we can end up calculating the moment of inertia as

$I=\frac{2M}{\pi R^2}\int_{-R}^{R} \int_{-\sqrt{\frac{R^2-y^2}{4}}}^{\sqrt{\frac{R^2-y^2}{4}}}(x^2+y^2)\mathrm{d}x\mathrm{d}y=\frac{5}{16}MR^2.$

The limits of the integral are given by the equation of the ellipse.

5/16(MR^2)