Moment of inertia of cylinder How to calculate moment of inertia of a cylinder about an axis passing through its geometrical center and inclined at an angle (\theta) with vertical.
 A: From the point of view of its moments of inertia a body is completely described by its inertia tensor, whose components in a given reference frame are defined by
$$I_{ij} = \int dm \left( \delta_{ij} r^2 - x_i x_j \right)$$
where $x_i$ are the coordinates, $x_1=x$, $x_2=y$, $x_3=z$.
As the tensor is symmetric, there are six independent components, which are integrals of quadratic combinations of coordinates ($r^2=x_1^2+x_2^2+x_3^2$).
If you rotate the reference frame, you have three new coordinates which are linear functions of the old ones. It follows that the components of the inertia tensor in the new reference frame can be written as linear functions of the old ones (see the example below).
Choose your reference frame with the origin in the center of the cylinder, the z axis along the cylinder's one and the x and y axes along diameters. It is easy to see that the components with $i\neq j$ are zero owing to the symmetry of the cylinder. For example
$$I_{xy} = - \int dm x y$$
does not change if you reflect the cylinder in the $yz$ plane, by changing $x\rightarrow -x$. This means that $I_{xy}=-I_{xy}$ and so $I_{xy}=0$. The only nonzero components are $I_{zz}$ and $I_{xx}=I_{yy}$ (once again the equality is a consequence of the symmetry).
Now $$I_{zz}=\int dm \left( x^2+y^2 \right)$$is the moment of inertia of the cylinder about its axis and similarly $I_{xx}=I_{yy}$ is the moment of inertia of the cylinder about its diameter.
If we rotate the reference frame by an angle $\theta$ around its x axis we get the new coordinates
$$x^\prime = x$$
$$y^\prime = y \cos \theta - z \sin \theta$$
$$z^\prime = z \cos \theta + y \sin \theta$$
and the moment of inertia we want to evaluate is
$$I_{z^\prime z^\prime} = \int dm \left[ (x^\prime)^2+(y^\prime)^2 \right]$$
By substituting the unprimed coordinates you get
$$I_{z^\prime z^\prime} = \int dm \left[ (y \cos \theta-z\sin \theta)^2+x^2 \right]$$
and expanding and rearranging the terms
$$I_{z^\prime z^\prime} =  \int dm  \left[\left(x^2+y^2\right) \cos^2 \theta +
\left(x^2+z^2\right) \sin^2 \theta -2 y z \cos \theta \sin \theta  \right]
$$
which is
$$I_{z^\prime z^\prime} =  \cos^2 I_{zz}+\sin^2 \theta I_{yy} -2\cos \theta \sin \theta I_{yz}$$
but $I_{yz}=0$ and if you know the moment of inertia of the cylinder about its axis and its diameter you are done.
