Calculate moment of inerta of cylinder along central diameter using inertia tensor 
how do I find the moment of inerta of cylinder along central diameter which is the one that is $(h^2 + 3r^2)M /12$ ??? 
I used triple integral and I get
 
Which I I'm not sure what I done wrong..Any help will be appreciated. Thanks

By definition of moment of inertia $I_{xx}$,
$$
I_{xx} = \int (y^2+z^2) dm = \int (r^2 \sin^2 \theta +z^2) \frac{M}{V} dV = \frac{M}{V}  \int_{-H/2}^{H/2} \int_0^{2\pi} \int_0^R (r^2 \sin^2 \theta +z^2) r dr d\theta dz
$$
what you doing wrong is that the limit of integration that you choose for $dz$ and the factor $\frac{M}{V}$. The integration on $z$ must run from $-H/2$ to $H/2$ by symmetry. And by substitute $V = \pi R^2 H$, you'll have the desired result. 
