Moment of inertia of a cylinder in unusual position

[24forChromium]

See image. It would be better if you can show me the calculation process.

EDIT: the axis of rotation goes right through the centre of the cylinder.

NEVER MIND! GOT IT!

 

[tommyxu3]

My opinion: for a thin disc whose radius is $r$ and mass is $m,$ with the axis perpendicular to the disc, the moment of inertia is $I_0=\frac{1}{2}mr^2,$ which may be easy to find. Then with the perpendicular axis theorem, if the axis is parallel to the disc and passes the center, the moment of inertia should be $I_i=\frac{1}{2}I_0=\frac{1}{4}mr^2.$
With this result and the parallel axis theorem, we may be able to solve the problem. For the disc away from the center $x$(based on your diagram), the disc's moment of inertia is $\frac{1}{4}mr^2+mx^2$(of course $m$ should be adjusted. Then, just integrate them we may get the answer.