On solving IE Irodov 3.21,

A ball of radius $R$ is uniformly charged with the volume density $\rho$. Find the flux of the electric field strength vector across the ball's section formed by the plane located at a distance $r_{0} < R$ from the center of the ball.

From calculus obtained result is $$\boxed{\phi_1=\dfrac{\pi\rho r_o(R^2-r_o^2)}{3\epsilon_o}}$$

But if consider that flat surface as a flat surface of a cylinder then radius and height of such a cylinder will be $\sqrt{R^2-r_o^2}$ and $2r_o$ respectively, then from Gaus' Law flux through **whole of cylinder** is given by $$\boxed{\phi_2=\dfrac{\pi\rho (2r_o)(R^2-r_o^2)}{\epsilon_o}}$$

This is very similar to flux $\phi_1$ obtained for flat surface. More precisely $$\phi_1=\dfrac{\phi_2}{6}$$

So can we prove that flux through flat part of the cylinder is one-third of the total, and that through curved part is two-thirds of the total, in general?