I want to point out I worked with polar coordinates:

$\iint_{D} f(x, y)dxdy = \iint_{D'} f(rcos\theta, rsin\theta)rdrd\theta$

$dxdy=dA=r drd\theta$

Let disk rotate about x axis

Being moment of inertia $I = mr^2$, $a$ the radius of the disk and $\gamma=\frac{m}{\pi a^2}$ then we know:

$$I=\int_{0}^{2\pi}\int_{0}^{a} \gamma r^3sin^2\theta drd\theta$$

Actually, this integral gives the correct answer: $\frac{a^2m}{4}$

However, obtaining the desired result does not mean I am right (even the great Euler experienced that).

That is why I want to discuss the method I used. I took as an infinitesimal length $dr$ and as an infinitesimal width $rd\theta $.

Therefore, $dA=r drd\theta$

The distance from the x axis to the infinitesimal 2D fragment is $ d=rsin\theta $

Then moment of inertia of the infinitesimal fragment: $ I=dm(rsin\theta )^2 $