Moment of inertia tensor of hollow cone

[dpqb29]

I am having trouble right now with the same problem (finding Ixx and Iyy).
\begin{equation}
I_{yy} = \int(x^2 + z^2)dm
\end{equation}
where
\begin{equation}
dm = \frac{2M}{R^2 + H^2} q dq
\end{equation}
and q is my generalized coordinate that is measured from the origin down the length of the cone. I am able to integrate z^2 since it can simply be related to q by
\begin{equation}
z = \frac{Hq}{\sqrt{R^2 + H^2}} ,
\end{equation}
but I am unable to simply relate x to q. I know that
\begin{eqnarray}
\rho^2 = x^2 + y^2\\
\rho = \frac{Rq}{\sqrt{R^2 + H^2}}
\end{eqnarray}
by the way.

 

[member 553083]

How do I integrate x^2 with respect to q?You can use the Pythagorean theorem to find the relation between x and q as follows:\begin{equation} x^2 = \rho^2 - y^2 = \frac{R^2q^2}{R^2 + H^2} - \frac{H^2q^2}{R^2 + H^2} = \frac{R^2H^2}{(R^2 + H^2)^2}q^2 \end{equation}Therefore, you can integrate $x^2$ with respect to q using the following equation:\begin{equation} I_{yy} = \int \left(\frac{R^2H^2}{(R^2 + H^2)^2}q^2 + \frac{H^2q^2}{R^2 + H^2}\right) \frac{2M}{R^2 + H^2} q dq\end{equation}