Mass moment of inertia of a spherocylinder

[UPDATE 22/08/2014: Corrected a misprint in the formula for $I_x$. See comment below.]
Using the moments of inertia calculated in the previous post for the hemisphere (and taking for granted those of the cylinder), we can now determine those of a spherocylinder.

The height of the cylinder is $h$, while its radius (and that of the spherical caps) is $R$.

$I_z$ is easy to determine by summing the corresponding moments of the cylinder and of the two hemispheres: $I_z= \rho \, m_c \frac{R^2}{2} + 2 \rho \, m_h \frac{2}{5} R^2$. Developing $m_c$ and $m_h$ and introducing the aspect ratio $\gamma = 1+ \frac{h}{2R}$ yields:
$$I_z = \pi \rho R^5 \left [ (\gamma -1) + \frac{8}{15} \right ]$$

$I_x$ is the sum of the $x$ moment for the cylinder and of twice the moment of a hemisphere around an axis distant from its center of mass by $h/2 + z_{\text{CM}}$, calculated using the parallel axis theorem (see previous post).

$$I_x = \pi \rho R^5 \left \lbrace \frac{\gamma -1}{6} \left [ 3+4 (\gamma - 1)^2) \right ] + \frac{4}{3}  \left [  \frac{83}{320} +\left ( (\gamma - 1) +\frac{3}{8} \right )^2  \right ] \right \rbrace$$ $Oxyz$ is the principal reference frame on symmetry grounds.