Originally Posted by

**hollywood** It's hard to describe without drawing pictures, but here goes:

Let $\displaystyle h$ be the height of the cone and $\displaystyle r$ the radius of the base. Position the cone so that the apex is at the center of the coordinate system and the center of the base is at $\displaystyle \phi=0$.

As you said, it is clear that $\displaystyle \theta$ goes from $\displaystyle 0$ to $\displaystyle 2\pi$.

The variable $\displaystyle \phi$ goes from the central axis $\displaystyle \phi=0$ to the lateral surface of the cone $\displaystyle \phi=\phi_\text{max}$. Setting up a triangle gives $\displaystyle \tan(\phi_\text{max})=\frac{r}{h}$, so the limits are $\displaystyle 0$ and $\displaystyle \tan^{-1}\frac{r}{h}$.

The variable $\displaystyle \rho$ goes from the apex $\displaystyle \rho=0$ to $\displaystyle \rho=\rho_\text{max}$ at the base. Setting up a different triangle gives $\displaystyle \cos{\phi}=\frac{h}{\rho_\text{max}}$. So the limits are $\displaystyle 0$ and $\displaystyle \frac{h}{\cos{\phi}}$.

To get the moment of inertia, we integrate the square of the distance from the axis to the volume element. Yet another triangle gives this distance as $\displaystyle \rho\sin{\phi}$.

So the integral is $\displaystyle \int_0^{2\pi} \int_0^{\tan^{-1}\frac{r}{h}} \int_0^{\frac{h}{\cos{\phi}}} (\rho\sin{\phi})^2 \rho^2\sin{\phi}\,d\rho\,d\phi\,d\theta$.

Integrating this gives the correct value $\displaystyle \frac{1}{10}\pi r^4h$.

Your integral is correct. In your analysis, R should be the perpendicular distance from the mass element to the axis of rotation (instead of from the axis of rotation to the slant height).

- Hollywood