I am asked to compute the moment of inertia about the $z$-axis of a rugby ball in terms of its total mass.
A rugby ball surface is given by the ellipsoid:
$$\frac{x^2}{4} + \frac{y^2}{4} + \frac{z^2}{9} = 1$$
The momentum of inertia of the ball about the z-axis is defined as
$ I=\int_V \rho r^2_\perp\space dV = \rho \int_V (x^2+y^2) \space dV$
where $\rho$ is the constant density of the ball and $r_\perp$ is the orthogonal distance from a point to the $z$-axis, $\sqrt{x^2+y^2}$.
I am confused by a few things in this question. Is the moment of inertia the same as the momentum of inertia about a certain axis of a certain object? I also know that the moment of inertia is defined as $I = \int l^2 dM$, where $l$ is the distance of a mass element $dM$ from the axis, am I supposed to somehow relate this equation to the momentum of inertia equation stated above. Very lost here, all help is greatly appreciated!
The volume has a circular cross section perpendicular to the $z$ axis, so you can consider it to be the volume of revolution formed by rotating the ellipse $\frac{y^2}{4}+\frac{z^2}{9}=1$ about the $z$ axis.
Considering disc-like elements of thickness $\delta z$ and radius $y$, each element has volume $\pi y^2\delta z$, and hence moment of inertia $\frac 12\pi \rho y^4\delta z$. This is quoting the standard formula for the MI of a disc of radius $r$ and mass $m$ about an axis through the centre perpendicular to the plane of the disc, namely $\frac 12 mr^2$.
Therefore you need to evaluate the integral $$\frac 12\int_{-3}^{3}\pi\rho y^4dz,$$ where $\rho$ is the mass per unit volume, and $m=\rho V$.
There is a standard formula for the volume of an ellipsoid, or you can evaluate $$\int_{-3}^{3}\pi\rho y^2dz$$