A cylinder of length 2a and radius ‘a’ has the x-axis as its axis. Its two ends (plane surfaces) are at x=a and x=3a respectively. Point charges +q and -q are located at x=2a and x=0 respectively on the axis of cylinder. The electric flux through the curved surface of cylinder is (nearly)
 

${\varphi }_{curved}={\varphi }_{curved,q}+{\varphi }_{curved,-q}$

${\varphi }_{curved,q}$= electric flux leaving the cylinder (due to $q$ ) minus 

                     electric flux leaving the flat surfaces (due to $q$ ).

$\Rightarrow {\varphi }_{curved,q}=\frac{q}{{\epsilon }_0}-{\varphi }_{flat,q}$

$=\frac{q}{{\epsilon }_0}-2\times \left[\frac{q}{{\epsilon }_0}\frac{2\pi \left(1-\cos 45°\right)}{4\pi }\right]=\frac{q}{{\epsilon }_0}-\frac{q}{{\epsilon }_0}\left(1-\cos 45°\right)$

$\Rightarrow {\varphi }_{curved,q}=\frac{q}{\sqrt{2}{\epsilon }_0}$     (flux leaving the curved surface)

${\varphi }_{curved,-q}$= electric flux leaving the left flat surface (due to $-q$ ) minus 
                        electric flux entering the right flat surface (due to $-q$ ).

$\Rightarrow {\varphi }_{curved,-q}=\left[\frac{q}{{\epsilon }_0}\frac{2\pi }{4\pi }\left(1-\cos 45°\right)\right]-\left[\frac{q}{{\epsilon }_0}\frac{2\pi }{4\pi }\left(1-\cos \left\{{\tan }^{-1}\left(\frac{1}{3}\right)\right\}\right)\right]$

$\Rightarrow {\varphi }_{curved,-q}=\frac{q}{2{\epsilon }_0}\left[\left(1-\frac{1}{\sqrt{2}}\right)-1+\frac{3}{\sqrt{10}}\right]$

$\Rightarrow {\varphi }_{curved,-q}=\frac{q}{2{\epsilon }_0}\left[\frac{3}{\sqrt{10}}-\frac{1}{\sqrt{2}}\right]$

$(Notice that this flux is entering the curved surface)$

$\Rightarrow {\varphi }_{curved}=\frac{q}{\sqrt{2}{\epsilon }_0}-\left[\frac{q}{2{\epsilon }_0}\left(\frac{3}{\sqrt{10}}-\frac{1}{\sqrt{2}}\right)\right]$

$\Rightarrow {\varphi }_{curved}=\frac{q}{{\epsilon }_0}\left[\frac{1}{\sqrt{2}}-\frac{3}{2\sqrt{10}}+\frac{1}{2\sqrt{2}}\right]$

$\Rightarrow {\varphi }_{curved}=\frac{q}{{\epsilon }_0}\left[\frac{3}{2\sqrt{2}}-\frac{3}{2\sqrt{10}}\right]$

$\Rightarrow {\varphi }_{curved}=\frac{q}{{\epsilon }_0}\left[1.06-0.47\right]=0.59\frac{q}{{\epsilon }_0}\approx 0.6\frac{q}{{\epsilon }_0}$