Surface Area of an Ellipsoid $AX^2 + BY^2+ CZ^2 + DXY + EXZ + FYZ - R_0^2= 0$

Question

I am attempting to determine the surface area of an ellipsoid described by the equation:

$$AX^2 + BY^2 + CZ^2 + DXY + EXZ + FYZ - R_{0}^2 = 0 \; .$$

When it comes to an ellipse aligned with its axes at the origin of a Cartesian system, as given by $\frac{x^2}{a^2} + \frac{y^2}{b^2}=1$, the process is straightforward, and we swiftly find that $A = \pi ab$, where $a$ and $b$ represent the semi-major and semi-minor axes respectively.

However, for an ellipse not meeting these conditions, like the example mentioned here, the task becomes more challenging but remains achievable through algebraic manipulation.

I attempted to apply a similar approach to the equation $AX^2 + BY^2 + CZ^2 + DXY + EXZ + FYZ - R_{0}^2= 0$ representing an ellipsoid, but I have not been able to solve it. Can someone provide assistance with this question, please?

Edit:

I do not have any other information such as eccentricity, length of the axes, etc... I only have the equation above.

Answer 1

• First find $\lambda_1, \lambda_2, \lambda_3$, the eigenvalues of

$$ \begin{pmatrix} A & \frac{D}{2} & \frac{E}{2} \\ \frac{D}{2} & B & \frac{F}{2} \\ \frac{E}{2} & \frac{F}{2} & C \end{pmatrix}$$

• For triaxial ellipsoid,

$$0<\lambda_1<\lambda_2<\lambda_3$$

• Semi principal axes are

\begin{align} a &= \frac{R_0}{\sqrt{\lambda_1}} \\ b &= \frac{R_0}{\sqrt{\lambda_2}} \\ c &= \frac{R_0}{\sqrt{\lambda_3}} \\ \end{align}

• Surface area $S$ for $a>b>c>0$

\begin{align} S &= 2\pi \left[ c^{2}+\frac{bc^{2}}{\sqrt{a^{2}-c^{2}}}\, F\left( \cos^{-1} \frac{c}{a}, k \right)+b\sqrt{a^{2}-c^{2}}\, E\left( \cos^{-1} \frac{c}{a}, k \right) \right] \\ k &=\frac{a}{b}\sqrt{\frac{b^{2}-c^{2}}{a^{2}-c^{2}}} \end{align}

• See also the journal article by Leo R. M. Maas here.

• See also my older post for spheroids here.

Answer 2

There is a paper written in 1979 that discusses methods for approximating the surface area of a scalene (triaxial) ellipsoid.