Consider a thin uniform disc of mass M and radius R.
I am interested in the calculation of moment of inertia about an axis passing through the center of mass, perpendicular to the disc, specifically using area elements in the integral that are thin rings of radius $r$ and width $dr$.
If you would like to see the exact calculation I am following, it is this one from an MIT OCW course, page 4. It is a physics course, but I am interested in the mathematical aspect of the integral.
We want to calculate
$$\int_{disc}{r_{dm}^2 dm}$$
where
$$dm=\frac{M}{\pi R^2}da_{ring}$$
My question regards $da_{ring}$
$$da_{ring}=\pi(r+dr)^2-\pi r^2 = 2\pi r dr+\pi(dr)^2$$
In the limit that $dr\to 0$ the term proportional to $(dr)^2$ can be ignored and the area is $da=2\pi r dr$
What is the technical reason why we can ignore the term proportional to $(dr)^2$?