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 $1/2$.
 A: 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.
