Since there does not appear to be a complete derivation of the surface area of an ellipsoid readily available on the web, consider the manual evaluation in all its lovely detail. The concept of volume element indicates that surface area can be calculated much like volume, as an integral over the square root of the determinant of the metric, and that is how the evaluation will proceed.

A typical parametrization of an ellipsoid is

$\mathbf{r}=[asin\theta cos\phi ,bsin\theta sin\phi ,ccos\theta ]$

Forming the space-like interval on this surface, the corresponding metric tensor is

$\left[\begin{array}{cc}(a^2{cos}^2\phi +b^2{sin}^2\phi ){cos}^2\theta +c^2{sin}^2\theta & (b^2-a^2)sin\theta cos\theta sin\phi cos\phi \\ (b^2-a^2)sin\theta cos\theta sin\phi cos\phi & (a^2{sin}^2\phi +b^2{cos}^2\phi ){sin}^2\theta \end{array}\right]$

which looks rather complicated, but the determinant simplifies significantly:

$detg=a^2{b}^2{sin}^2\theta {cos}^2\theta +c^2(a^2{sin}^2\phi +b^2{cos}^2\phi ){sin}^4\theta$

Evaluating an integral of the square root of this expression can be done with expansions of binomials. The overall factor of a sine in the square root of this determinant makes the choice $x=cos\theta$ natural for reducing factors in the integrand, and the other angular variable can be left in place. With the abbreviations

$\beta =1-\frac{a^2}{b^2}\gamma =1-\frac{a^2}{c^2}$

and taking advantage of symmetries in the integrand, the surface area integral is

$S=\underset{0}{\overset{\pi }{\int }}d\theta \underset{0}{\overset{2\pi }{\int }}d\phi \sqrt{detg}=8bc\underset{0}{\overset{1}{\int }}dx\sqrt{1-\gamma x^2}\underset{0}{\overset{\pi /2}{\int }}d\phi \sqrt{1-\beta \frac{1-x^2}{1-\gamma x^2}{sin}^2\phi }$

Given the generalized binomial expansion

$\begin{array}{l}(1-x{)}^r=1-rx+\frac{r(r-1)}{2!}{x}^2-\frac{r(r-1)(r-2)}{3!}{x}^3+\cdots \\ \phantom{(1-x{)}^r}=\sum _{k=0}^{\infty }\frac{\Gamma (-r+k)}{\Gamma (-r)}\frac{x^k}{k!}\end{array}$

as well as a trigonometric integral, the integration over φ is

$\begin{array}{l}\underset{0}{\overset{\pi /2}{\int }}d\phi \sqrt{1-\beta \frac{1-x^2}{1-\gamma x^2}{sin}^2\phi }=\underset{0}{\overset{\pi /2}{\int }}d\phi \sum _{k=0}^{\infty }\frac{\Gamma (-\frac{1}{2}+k)}{\Gamma (-\frac{1}{2})}\frac{{\beta }^k}{k!}(\frac{1-x^2}{1-\gamma x^2}{)}^k{sin}^{2k}\phi \\ =\frac{\Gamma \left(\frac{1}{2}\right)}{2}\sum _{k=0}^{\infty }\frac{\Gamma (-\frac{1}{2}+k)}{\Gamma (-\frac{1}{2})}\frac{\Gamma (k+\frac{1}{2})}{\Gamma (k+1)}\frac{{\beta }^k}{k!}(\frac{1-x^2}{1-\gamma x^2}{)}^k\\ =\frac{\pi }{2}\sum _{k=0}^{\infty }\frac{\Gamma (\frac{1}{2}+k)}{\Gamma \left(\frac{1}{2}\right)}\frac{\Gamma (-\frac{1}{2}+k)}{\Gamma (-\frac{1}{2})}\frac{{\beta }^k}{(k!{)}^2}(\frac{1-x^2}{1-\gamma x^2}{)}^k\end{array}$

As an aside, this is the expansion of a complete elliptic integral of the second kind. With one more application of the binomial expansion, the surface area becomes

$\begin{array}{l}S=4\pi bc\underset{0}{\overset{1}{\int }}dx\sqrt{1-\gamma x^2}\sum _{k=0}^{\infty }\frac{\Gamma (\frac{1}{2}+k)}{\Gamma \left(\frac{1}{2}\right)}\frac{\Gamma (-\frac{1}{2}+k)}{\Gamma (-\frac{1}{2})}\frac{{\beta }^k}{(k!{)}^2}(\frac{1-x^2}{1-\gamma x^2}{)}^k\\ \phantom{S}=4\pi bc\underset{0}{\overset{1}{\int }}dx\sum _{k,l=0}^{\infty }\frac{\Gamma (\frac{1}{2}+k)}{\Gamma \left(\frac{1}{2}\right)}\frac{\Gamma (-\frac{1}{2}+k)}{\Gamma (-\frac{1}{2})}\frac{\Gamma (k-\frac{1}{2}+l)}{\Gamma (k-\frac{1}{2})}\frac{{\beta }^k{\gamma }^l}{(k!{)}^{2}l!}{x}^{2l}(1-x^2{)}^k\\ S=4\pi bc\sum _{k,l=0}^{\infty }\frac{\Gamma (\frac{1}{2}+k)}{\Gamma \left(\frac{1}{2}\right)}\frac{\Gamma (k-\frac{1}{2}+l)}{\Gamma (-\frac{1}{2})}\frac{{\beta }^k{\gamma }^l}{(k!{)}^{2}l!}\underset{0}{\overset{1}{\int }}dx{x}^{2l}(1-x^2{)}^k\end{array}$

The last binomial has not been expanded because now one can take advantange of a binomial integral,

$\begin{array}{l}\underset{0}{\overset{1}{\int }}dx{x}^{2l}(1-x^2{)}^k=\frac{1}{2l+1}{}_{2}F_1(-k,l+\frac{1}{2};l+\frac{3}{2};1)\\ =\frac{1}{2l+1}\frac{\Gamma (l+\frac{3}{2})\Gamma (k+1)}{\Gamma (k+l+\frac{3}{2})\Gamma \left(1\right)}=\frac{\Gamma (l+\frac{1}{2})}{\Gamma \left(\frac{1}{2}\right)}\frac{\Gamma \left(\frac{3}{2}\right)}{\Gamma (k+l+\frac{3}{2})}k!\end{array}$

where the last line uses the Gauss summation formula. Putting this into the previous sum and reordering arguments slightly gives

$\begin{array}{l}S=4\pi bc\sum _{k,l=0}^{\infty }\frac{\Gamma (-\frac{1}{2}+k+l)}{\Gamma (-\frac{1}{2})}\frac{\Gamma (\frac{1}{2}+k)}{\Gamma \left(\frac{1}{2}\right)}\frac{\Gamma (\frac{1}{2}+l)}{\Gamma \left(\frac{1}{2}\right)}\frac{\Gamma \left(\frac{3}{2}\right)}{\Gamma (\frac{3}{2}+k+l)}\frac{{\beta }^k}{k!}\frac{{\gamma }^l}{l!}\\ \phantom{S}=4\pi bc{F}_1(-\frac{1}{2};\frac{1}{2},\frac{1}{2};\frac{3}{2};\beta ,\gamma )\\ S=4\pi bc{F}_1(-\frac{1}{2};\frac{1}{2},\frac{1}{2};\frac{3}{2};1-\frac{a^2}{b^2},1-\frac{a^2}{c^2})\end{array}$

which is a single Appell double hypergeometric function. This single Appell function form for the surface of an ellipsoid does not seem to be widely known. While a should be less than both b and c in accordance with the intermediate elliptic integral expansion, this form does not impose an ordering on b and c due to symmetry in the second and third parameters.

For a symmetric ellipsoid with $a=b$ the surface area reduces to a Gauss hypergeometric function that can be related to circular functions,

$\begin{array}{l}{S}_{a=b}=4\pi ac{}_{2}F_1(-\frac{1}{2},\frac{1}{2};\frac{3}{2};\gamma )\\ \phantom{S_{a=b}}=4\pi ac[\frac{1}{2}{}_{2}F_1(\frac{1}{2},\frac{1}{2};\frac{3}{2};\gamma )+\frac{1}{2}{}_{2}F_1(-\frac{1}{2},\frac{1}{2};\frac{1}{2};\gamma )]\\ S_{a=b}=2\pi ac[\frac{{sin}^{-1}\sqrt{\gamma }}{\sqrt{\gamma }}+\sqrt{1-\gamma }]\end{array}$

where the second step applies a contiguous relation and the third identifies two special cases. The two options for parameters are

$\begin{array}{l}{S}_{\mathrm{prolate}}=2\pi a^2+\frac{2\pi ac}{\sqrt{1-\frac{a^2}{c^2}}}{sin}^{-1}\sqrt{1-\frac{a^2}{c^2}},a<c\\ S_{\mathrm{oblate}}=2\pi a^2+\frac{2\pi ac}{\sqrt{\frac{a^2}{c^2}-1}}{sinh}^{-1}\sqrt{\frac{a^2}{c^2}-1},a>c\end{array}$

using the fact that the inverse sine or inverse hyperbolic sine of an imaginary argument is the other function multiplied by the imaginary unit. For $c\to a$ both expressions simplify to

$\begin{array}{l}{S}_{\mathrm{sphere}}=4\pi a^2\end{array}$

since both the inverse sine and and inverse hyperbolic sine have a leading linear term in their expansions.

To relate the single Appell function form to Legendre elliptic integrals, recall the expansions near the beginning of this presentation:

$\begin{array}{l}F\left(\phi \right|m)=sin\phi F_1(\frac{1}{2};\frac{1}{2},\frac{1}{2};\frac{3}{2};{sin}^2\phi ,m{sin}^2\phi )\\ E\left(\phi \right|m)=sin\phi F_1(\frac{1}{2};\frac{1}{2},-\frac{1}{2};\frac{3}{2};{sin}^2\phi ,m{sin}^2\phi )\end{array}$

The result for the surface area can be expressed as a linear superposition of the Appell functions appearing here, along with a third function to be determined:

$F_1(-\frac{1}{2};\frac{1}{2},\frac{1}{2};\frac{3}{2};\beta ,\gamma )=A{F}_1(\frac{1}{2};\frac{1}{2},\frac{1}{2};\frac{3}{2};\beta ,\gamma )+B{F}_1(\frac{1}{2};-\frac{1}{2},\frac{1}{2};\frac{3}{2};\beta ,\gamma )+f(\beta ,\gamma )$

Since the only one of these three Appell functions not symmetric in the second and third parameters is the third, its minus sign will be associated with the argument β for final consistency. The simplest way to establish the linear superposition appears to via an integral representation of the Appell function:

$\begin{array}{l}{F}_1(a;b_1,b_2;c;z_1,z_2)\\ =\frac{\Gamma \left(c\right)}{\Gamma \left(a\right)\Gamma (c-a)}\underset{0}{\overset{1}{\int }}dx{x}^{a-1}(1-x{)}^{c-a-1}(1-z_{1}x{)}^{-b_1}(1-z_{2}x{)}^{-b_2}\end{array}$

While this representation does not converge at the lower limit for the cases of parameters considered here, one can assume the infinities cancel from both sides of the equation and proceed anyway. With identifications

$\Gamma \left(\frac{3}{2}\right)=\frac{1}{2}\Gamma \left(\frac{1}{2}\right)=-\frac{1}{4}\Gamma (-\frac{1}{2})$

the linear superposition takes the form

$\begin{array}{l}-\frac{1}{4}\underset{0}{\overset{1}{\int }}dx\frac{(1-x)}{x^{3/2}\sqrt{1-\beta x}\sqrt{1-\gamma x}}\\ =\frac{A}{2}\underset{0}{\overset{1}{\int }}dx\frac{1}{\sqrt{x}\sqrt{1-\beta x}\sqrt{1-\gamma x}}+\frac{B}{2}\underset{0}{\overset{1}{\int }}dx\frac{\sqrt{1-\beta x}}{\sqrt{x}\sqrt{1-\gamma x}}+f(\beta ,\gamma )\end{array}$

Moving integrals to the left-hand side of the equation, the combined integrand is

$-\frac{1}{2\sqrt{x}\sqrt{1-\beta x}\sqrt{1-\gamma x}}[\frac{1-x}{2x}+A+B(1-\beta x)]$

Since derivatives of linear square roots have one less power algebraically, a combination to consider is

$\frac{d}{dx}\frac{\sqrt{1-\beta x}\sqrt{1-\gamma x}}{\sqrt{x}}=-\frac{1}{2\sqrt{x}\sqrt{1-\beta x}\sqrt{1-\gamma x}}(\frac{1}{x}-\beta \gamma x)$

For the term inverse in x here to match that in the previous bracket, an overall factor of one half is needed and the other constants follow immediately:

$\frac{1}{2}(\frac{1}{x}-\beta \gamma x)=\frac{1-x}{2x}+A+B(1-\beta x)\to A=\frac{1-\gamma }{2},B=\frac{\gamma }{2}$

The remaining function, ignoring the singular lower limit, is now simply

$f(\beta ,\gamma )=\frac{1}{2}\underset{0}{\overset{1}{\int }}dx\frac{d}{dx}\frac{\sqrt{1-\beta x}\sqrt{1-\gamma x}}{\sqrt{x}}=\frac{1}{2}\sqrt{1-\beta }\sqrt{1-\gamma }$

and the linear relation among the three Appell functions is

$\begin{array}{l}{F}_1(-\frac{1}{2};\frac{1}{2},\frac{1}{2};\frac{3}{2};\beta ,\gamma )=\frac{1-\gamma }{2}{F}_1(\frac{1}{2};\frac{1}{2},\frac{1}{2};\frac{3}{2};\beta ,\gamma )+\frac{\gamma }{2}{F}_1(\frac{1}{2};-\frac{1}{2},\frac{1}{2};\frac{3}{2};\beta ,\gamma )\\ +\frac{1}{2}\sqrt{1-\beta }\sqrt{1-\gamma }\end{array}$

With the identifications

$\beta =m{sin}^2\phi \gamma ={sin}^2\phi 1-\gamma ={cos}^2\phi$

the surface area of the three-dimensional ellipsoid can be written

$S=2\pi a^2+\frac{2\pi bc}{sin\phi }\left[{cos}^2\phi F\right(\phi \left|m\right)+{sin}^2\phi E\left(\phi \right|m\left)\right]$

which is the expected result with a and c interchanged. This form reduces to the result for the prolate ellipsoid for $a=b$ , which for the traditional Legendre form means $m=0$ .

The single Appell function result above is patently simpler than the traditional form.

Using a relation of the Carlson completely symmetric integral of the second kind to Legendre elliptic integrals

$\begin{array}{l}{R}_G({cos}^2\phi ,1-m{sin}^2\phi ,1)\\ =\frac{{cos}^2\phi }{2sin\phi }F\left(\phi \right|m)+\frac{sin\phi }{2}E\left(\phi \right|m)+\frac{cos\phi }{2}\sqrt{1-m{sin}^2\phi }\end{array}$

which for the given parameter choices becomes

$R_G(1-\gamma ,1-\beta ,1)=\frac{S-2\pi a^2}{4\pi bc}+\frac{1}{2}\sqrt{1-\gamma }\sqrt{1-\beta }=\frac{S}{4\pi bc}$

the surface area of the ellipsoid can be written

$S=4\pi bc{R}_G(\frac{a^2}{b^2},\frac{a^2}{c^2},1)$

using the symmetry of the arguments of the Carlson elliptic integral. This expression has the same overall factors as the single Appell function form, so that this particular Carlson function is identical to the single Appell function.

Uploaded 2020.06.21 — Updated 2020.07.13 analyticphysics.com