Simple Way to Calculate the Volume of solid enclosed by Quadric (E.g. Ellipsoid) Surface I am trying to apply ellipsoid specific fitting to a set of points with measured coordinates with respect to a reference point. Following that, I need to estimate the volume of the fitted ellipsoid.
How should I proceed with the volume calculation after I have acquired a set of coefficients according to the general from of a quadric surface ?
$$A{x^2} + B{y^2} + C{z^2} + Dxy + Exz + Fyz + Gx + Hy + Iz + J = 0$$ (https://tutorial.math.lamar.edu/classes/calciii/quadricsurfaces.aspx)
Any advice on fitting is also appreciated !
Best regards,
 A: If, as suggested in the comments, you follow the method used in https://math.stackexchange.com/a/2805519/1257 you end up with an axis-aligned ellipsoid of the form
$$
A{x^2} + B{y^2} + C{z^2} +  Gx + Hy + Iz + J = 0.
$$
Completing the square for each of $x,y,z$ this can be rewritten as
$$
A\left(x+\frac{G}{2A}\right)^2 -  \frac{G^2}{4A} + 
B\left(y+\frac{H}{2B}\right)^2 -  \frac{H^2}{4B} + 
C\left(z+\frac{I}{2C}\right)^2 -  \frac{I^2}{4C} + 
J  = 0,
$$
or, letting $J'=J -  \dfrac{G^2}{4A}  
 -  \dfrac{H^2}{4B} 
 -  \dfrac{I^2}{4C},$
$$
A\left(x+\frac{G}{2A}\right)^2 + 
B\left(y+\frac{H}{2B}\right)^2 + 
C\left(z+\frac{I}{2C}\right)^2 + 
J'  = 0.
$$
This is the same shape and size as the origin-centered ellipsoid
$$
Ax^2+By^2+Cz^2+J'=0,
$$
from which the the axis lengths $a,b,c$ can be read off (e.g. $a=2\sqrt{-J'/A}$).
The volume of the ellipsoid, as mentioned in the comments, is then $V=4/3\cdot \pi abc.$
As for ellipsoid fitting, this is a big topic that you can search for online and perhaps pose as a separate specific question either on this site or Stackoverflow.com.
