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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.

zanestoy
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  • Have you tried a change of coordinates that turns the ellipsoid into a sphere? That might make working out the bounds easier. – amd Oct 16 '18 at 19:19
  • Yes, that was one of the first things I tried, the link I posted above mentioned it. However, I could not figure out what the bounds would be for the inclined plane. – zanestoy Oct 16 '18 at 19:27
  • There are basically two types of regions to consider: ones for which $z$ is given by the unit sphere and the other where it’s given by the transformed plane. This plane intersects the sphere in a circle, which will be an ellipse when projected onto the $x$-$y$ plane. The unit circle, this ellipse and the line of intersection of the plane with the $x$-$y$ plane give you the boundaries of these regions. – amd Oct 16 '18 at 19:43
  • Thanks @amd that has given me a good direction to go. I pulled out my old Calculus textbook last night, and I think that I understand how to apply it now. I will try to get it all pulled together today and will post my results. – zanestoy Oct 17 '18 at 14:53

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