homework and exercises - Moment of inertia of a cylinder about its base

I've tried to find the moment of inertia of a cylinder rotating about an axis parallel to its base (i.e about the 'End diameter') as one can see here . But when I checked my results with different references ,I've found that it's incorrect!.I need a help to figure out where I did it wrong.

since $$I=\int\limits x^2.dm$$

$$dm = \rho.dv$$ where

$$dv= r.d\theta . dr .dh$$ &

$$\rho=\frac{M}{\pi R^2 H}$$

$M$:cylinder total mass. $H$:total height of cylinder. $R$:the radius of the cylinder.

The distance of each infinitesimal element from the axis of rotation would be :

$$\sqrt{r^2+h^2}$$

Therfore,

$$I= \frac{2M}{ R^2 H}\int\limits_{0}^{R} \int\limits_{0}^{H}\, r(r^2+h^2)dh\,dr$$

and this gives a result of : $$\frac{M~H^2}{3}+\frac{M~R^2}{2}$$

Which is obviously wrong since the second term should be multiplied by a factor of $1/2$.