I think I understand what you want. Let us call $z$ the axis along the cylinder and $x$, $y$ the other two directions. If the center of the rod is at point C and the end at A you want $I_{xx}^A = \int \rho ({y^2+z^2}) {\rm d} V$
With $m=\int \rho \,{\rm d}V =\rho \pi R^2 H $ and ${\rm d}V = r\,{\rm d}\theta{\rm d}r{\rm d}z$ and $(x,y,z)=(r \cos\theta, r \sin\theta,z)$
$$ \begin{align}
I_{xx}^A &= \frac{m}{\pi R^2 H} \int_0^H \int_0^R \int_{-\pi}^{\pi}\, ({r^2\sin^2\theta+z^2})\,r\,{\rm d}\theta\,{\rm d}r\,{\rm d}z \\
& = \frac{m}{\pi R^2 H} \int_0^H \int_0^R \,\pi r (2 z^2+r^2)\,{\rm d}r\,{\rm d}z \\
& = \frac{m}{ R^2 H} \int_0^H \, \frac{ R^2 (4 z^2+R^2)}{4}\,{\rm d}z \\
& = \frac{m}{ R^2 H} \left( \frac{ H^3 R^2}{3} + \frac{ H R^4}{4} \right) \\
& = m \frac{H^2}{3} + m \frac{R^2}{4}
\end{align}$$
which matches the hyperphysics answer.
NOTE: Any separation along the $x$ axis does not play into the integral because this is axis of rotation and it just represents a parallel displacement. BTW the parallel axis theorem is a misnomer as it should be the perpendicular axis theorem.
The $dm$ you have calculated is incorrect. The radius will vary. Which you have assumed constant. So ,
(https://i.stack.imgur.com/f4VjF.png)
[r1=x is the distance of each element from axis]
$$dm=\rho 2\pi x dx l$$.
$$\rho=\frac{M}{\pi R^2l}$$
$$dI=(dm) x^2$$
So,
$$I=\int_0^R \frac{2M}{R^2}x^3$$
$$I=\frac{MR^2}{2}$$
Best Answer
If you do the calculations, you get that the Moment of inertia of a cilinder it's
$$ I=\rho\int_{z_1}^{z_2}\int_0^{2\pi}\int_0^R r^3drd\theta dz $$ With
$$ \rho=\frac{M}{V}=\frac{M}{h\pi R^2} $$
Half of the cilinder means that $\theta$ goes from $0$ to $\pi$. Also you conserve the density.
Since there is no angular dependence, it's just the half of the Moment of inertia of the initial cilinder.