I'm trying to find the volume of a wedge of an ellipsoid. The ellipsoid is the standard form: $$\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1 \leftarrow(Eqn 1)$$ The first bounding plane is the $xy$ plane, or in other words: $$z=0\leftarrow(Eqn 2)$$ The second bounding plane is: $$z=m*x+z_0 \leftarrow(Eqn 3)$$
I've looked around and found the volume of an ellipsoid sliced by a given plane. And I've found the volume of an ellipsoid sliced by two parallel planes, but I can find no examples of an ellipsoid sliced by two non-parallel planes. I have thought about this problem from many different angles, and feel that I keep getting wrapped around the axle on the various methods. I have investigated the following: Transforming into Spherical Coordinates - As suggested in this question: Volume of the smaller region of ellipsoid cut by plane, however, this results in some area below $xy$ plane.
Integrating the Ellipsoid Equation - However, I can't figure out what the limits of integration should be. I think that the integral should be a triple integral of the ellipsoid equation, except solved for z, $z=c*\sqrt(1-\frac{x^2}{a^2}-\frac{y^2}{b^2})$. So the integral should be $\int^E_F \int^C_D \int^A_B z dz dy dx$. The bounds $A$ and $B$ should be from 0 to the maximum location that the plane intersects the ellipsoid, i.e. solving for when Equation 1 and 2 intersect and y = 0. I get stuck at that point.
Any help thinking through this would be much appreciated.