I'm looking for a part of the volume of an oblate ellipsoid ($a=b>c$) $$ \frac{x^2}{a^2} + \frac{y^2}{a^2} + \frac{z^2}{c^2} =1 $$
The part is formed by a cylinder $ x^2 + y^2 = r^2 $ and two couples of planes:
- couple 1: $ y = \frac{L}{2} $ and $ y = \frac{-L}{2} $
- couple 2: $ z = -c\sqrt{1-\frac{r^2}{a^2}} $ and $ z = -c\sqrt{1-\frac{r^2}{a^2}} + H $ with $H>0$ (attention: this is not symmetrical as in couple 1)
I'm only interested in the part of the volume in the positive $x$-axes (positive a) lying between the cylinder (core drilled out from the ellipsoid) and the rest of the ellipsoid. I already tried to calculate this volume by a triple integration $$\int_{-L/2}^{L/2}\int_{-c\sqrt{1-\frac{r^2}{a^2}}}^{-c\sqrt{1-\frac{r^2}{a^2}} + H}\int_{\sqrt{r^2-y^2}}^{a\sqrt{1-\frac{y^2}{a^2}-\frac{z^2}{c^2}}}dxdzdy$$
but the integral becomes to complicated in Cartesian coordinates. Is there another way (perhaps via spherical coordinates or by splitting up in different sections)?
You're integrating over a rectangle, so a coordinate change can't really help you here. You can split it into two parts, but that won't make it much easier.
If you make some simple linear substitutions, and ignore the easier $\sqrt{r^2 - y^2}$ term, you can reduce the integral to the volume bounded by an arbitrary rectangle and the unit sphere: $$\int \int \sqrt{1 - x^2 - y^2} dx dy.$$ This is an elementary function but it's still too long to be practical. Mathematica insists on expressing it with logs and complex numbers, but I managed to simplify it to this:
$$\frac{1}{6} \left(2 x y \sqrt{-x^2-y^2+1}-y \left(y^2-3\right) \tan ^{-1}\left(\frac{x}{\sqrt{-x^2-y^2+1}}\right)-x \left(x^2-3\right) \tan ^{-1}\left(\frac{y}{\sqrt{-x^2-y^2+1}}\right)+\tan ^{-1}\left(\frac{-x^2-y+1}{x \sqrt{-x^2-y^2+1}}\right)-\tan ^{-1}\left(\frac{-x^2+y+1}{x \sqrt{-x^2-y^2+1}}\right)\right)$$