Given an ellipsoid $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}} =1$ and a plane $z=k$ where $0<k<c$, how do I find the volume bounded by a 60° wedge of said ellipsoid and the plane? I'm not sure how to formulate the triple integral in spherical coordinates (or is it better to compute in cartesian?) and I'm planning to use Python to compute the volume.
Here's a sketch in the z-x and z-y axis: I would like to find the volume (yellow) bounded by the ellipsoid wedge (black) and the plane (dark red)
The volume integral can be expressed as,
$$ V= \int_{A(x,z)}2b\sqrt{1-\frac{x^2}{a^2}-\frac{z^2}{c^2}}dxdz $$
The $xz$-integral is over the shaded area of the graph on the right, for which $y=0$.
Rescale the variables as $u=\frac xa$ and $v=\frac zc$. The ellipsoid equation for $y=0$ becomes $u^2+v^2=1$, the wedge lines $x=\pm\frac1{\sqrt3}z$ becomes $u = \pm\frac a{\sqrt3c} v$ and the cut line $z=k$ becomes $v=\frac kc$. As a result, the volume integral becomes,
$$V = 2abc \int_{A(u,v)} \sqrt{1-u^2-v^2}dudv$$
Now, reexpress the $uv$-integral in its polar coordinates,
$$V= 2abc\int_{-\beta}^{\beta}\int_{\frac k{c\cos\theta}}^1 \sqrt{1-r^2} \>rdr\>d\theta= \frac43abc\int_0^{\beta} \left(1-\frac{k^2}{c^2}\sec^2\theta\right)^{\frac32}d\theta$$
where $ \beta= \tan^{-1}\frac a{\sqrt3c}$. In general, for arbitrary value of $\frac kc $, the last integral over $\theta$ would need to be performed numerically.