Derivation of Surface Area of an Ellipsoid

The Surface Area Of An Ellipsoid

A. Dieckmann, Universität Bonn, July 2003

This short note shows a way to the formula for the surface area of an ellipsoid. The result is given for example in the formula gallery of the Mathematica Book (Ref. 1), but with a typo. There was no instruction accessible to me, how it is obtained, and I had to try for myself. It proved harder than I first thought, so I think it is worthwhile to put the derivation on the internet.
To make a later integration valid, we have to name the semi-axes of the ellipsoid c > a > b > 0, so that b < c and $RowBox[{0, <, (k = c^( 2)/(c^2 - b^2) * (a^2 - b^2)/a^2), , <, 1.}]$ The axes may be renamed at the end.

Now calculate the area numerically – for reference:

$NIntegrate[Sin[θ] (a^2 b^2 Cos[θ]^2 + b^2c^2 Sin[θ]^2 Cos[ϕ]^2 + a^2 c^2 Sin[θ]^2Sin[ϕ]^2)^(1/2), {θ, 0, π}, {ϕ, 0, 2π}]$

Symmetry allows the reduction of the integration domain:

$NIntegrate[Sin[θ] (a^2 b^2 Cos[θ]^2 + b^2c^2 Sin[θ]^2 Cos[ϕ]^2 + a^2 c^2 S ... in[ϕ]^2)^(1/2), {θ, 0, π/2}, {ϕ, 0, π/2}, WorkingPrecision20] * 8$

Integrate symbolically with respect to θ:

$a=. ; b=. ; c=. ; Integrate[Sin[θ] (a^2 b^2 Cos[θ]^2 + b^2c^2 Sin[θ]^2 Cos[ϕ]^2 + a^2 c^2 Sin[θ]^2Sin[ϕ]^2)^(1/2), {θ, 0, π/2}] * 8//FullSimplify$

8 If[Im[1/(-1 + c^2 (Cos[ϕ]^2/a^2 + Sin[ϕ]^2/b^2))] ≠0 || 1/(1 + c^2 (-Cos[ ... + b^2 c^2 Cos[ϕ]^2 Sin[θ]^2 + a^2 c^2 Sin[θ]^2 Sin[ϕ]^2)^(1/2) θ]

After numerical integration with respect to φ the area comes out correctly:

$RowBox[{RowBox[{a, =, 1.2}], ;, RowBox[{b, =, 1.1}], ;, RowBox[{c, =, 4.7}], ;}] 2π a b + ... c^2 Sin[ϕ]^2))/(c^2 (b^2 Cos[ϕ]^2 + a^2 Sin[ϕ]^2))^(1/2)], {ϕ, 0, π/2}]$

Express ArcSinh by ArcTanh:

$2π a b + 4NIntegrate[(c^2 (b^2 Cos[ϕ]^2 + a^2 Sin[ϕ]^2))/(-b^2 c^2 Cos[ϕ]^ ... 2))^(1/2) ArcTanh[(1 - c^2 (Cos[ϕ]^2/a^2 + Sin[ϕ]^2/b^2))^(1/2)], {ϕ, 0, π/2}]$

And the integral can be reduced to a simpler form by substituting

${{ϕ -ArcCos[-(a (c^2 - b^2 x)^(1/2))/((a^2 - b^2)^(1/2) c)]}, {ϕArcC ... ))/((a^2 - b^2)^(1/2) c)]}, {ϕArcCos[(a (c^2 - b^2 x)^(1/2))/((a^2 - b^2)^(1/2) c)]}}$

The Jacobian dφ/dx is:

(a b^2)/(2 ((a - b) (a + b))^(1/2) c (b^2 (c^2 - a^2 x))/((-a^2 + b^2) c^2)^(1/2) (c^2 - b^2 x)^(1/2))

Check the new, 'simple' form :

$RowBox[{RowBox[{a, =, 1.2}], ;, RowBox[{b, =, 1.1}], ;, RowBox[{c, =, 4.7}], ;}] 2π a b - ... /2) (-c^2 + b^2 x)^(1/2)), {x, c^2/a^2, c^2/b^2}, MinRecursion10, MaxRecursion33]$

Then integrate by parts with

(2 (-(b^2 (-1 + x))/(b^2 - c^2))^(1/2) (-c^2/a^2 + x) (-c^2/b^2 + x)/(c^2/a^2 - c^2/b^2)^(1/2) ... ((1 - x)^(1/2) (-c^2/a^2 + x)/(-c^2/a^2 + c^2/b^2)^(1/2) (c^2 - a^2 x)^(1/2) (-c^2 + b^2 x)^(1/2))

uv[x_] := (-2ArcTanh[(1 - x)^(1/2)])/(a (b^2 - c^2)^(1/2)) ((1 - c^2/b^2) EllipticE[ArcSin[(-c ... EllipticF[ArcSin[(-c^2/b^2 + x)/(c^2/a^2 - c^2/b^2)^(1/2)], (c^2 (b^2 - a^2))/(a^2 (b^2 - c^2))])

$RowBox[{RowBox[{a, =, 1.2}], ;, RowBox[{b, =, 1.1}], ;, RowBox[{c, =, 4.7}], ;, uv[c^2/b^2] - uv[c^2/a^2]}]$

uv(upper limit) - uv(lower limit):

(2ArcTanh[(1 - c^2/a^2)^(1/2)])/(a (b^2 - c^2)^(1/2)) ((1 - c^2/b^2) EllipticE[(c^2 (b^2 - a^2))/(a^2 (b^2 - c^2))] - EllipticK[(c^2 (b^2 - a^2))/(a^2 (b^2 - c^2))])

$2π a b - 4a ArcTanh[(1 - c^2/a^2)^(1/2)] ((b^2 - c^2)^(1/2) EllipticE[(c^2 (b^2 - a^2))/( ... /b^2 + x)/(c^2/a^2 - c^2/b^2)^(1/2)], (c^2 (b^2 - a^2))/(a^2 (b^2 - c^2))], {x, c^2/a^2, c^2/b^2}]$

There are now two integrals left to attack, but luckily, both have already been done – about 130 years ago. The first is:

$NIntegrate[1/(x (1 - x)^(1/2)) EllipticE[ArcSin[(-c^2/b^2 + x)/(c^2/a^2 - c^2/b^2)^(1/2)], (c^2 (b^2 - a^2))/(a^2 (b^2 - c^2))], {x, c^2/a^2, c^2/b^2}]$

Solve[ArcSin[(-c^2/b^2 + x)/(c^2/a^2 - c^2/b^2)^(1/2)] z, x]

${{xc^2 (1/b^2 + Sin[z]^2/a^2 - Sin[z]^2/b^2)}}$

$a = 12/10 ; b = 11/10 ; c = 47/10 ; (2 (a^2 - b^2) b )/(a^2 (b^2 - c^2)^(1/2)) NIntegrate[(Ell ... b^2/c^2) Sin[z]^2) (1 - (c^2 (b^2 - a^2))/(a^2 (b^2 - c^2)) Sin[z]^2)^(1/2)) , {z, 0, π/2}]$

If the conditions cited at the beginning
• 0 < ( t = ArcCos[b/c] ) < π/2, or equivalently b < c, and
• 0 < k < 1
are all fulfilled, then this integral has a solution, found with Gradshteyn Ryzhik (6.123) (Ref.2) to be :

((b - a) b π)/(a (b^2 - c^2)^(1/2)) + 2ArcTanh[(1 - c^2/a^2)^(1/2)] EllipticE[(c^2 (b^2 - ... a^2))/(a^2 (b^2 - c^2))] + π/IEllipticE[ArcCos[b/c], (c^2 (b^2 - a^2) )/(a^2 (b^2 - c^2))]//N

The second integral is very similar to the first :

$NIntegrate[1/(x (1 - x)^(1/2)) EllipticF[ArcSin[(-c^2/b^2 + x)/(c^2/a^2 - c^2/b^2)^(1/2)], (c^2 (b^2 - a^2))/(a^2 (b^2 - c^2))], {x, c^2/a^2, c^2/b^2}]$

$(2 (a^2 - b^2) b)/(a^2 (b^2 - c^2)^(1/2) ) NIntegrate[(EllipticF[z, (c^2 (b^2 - a^2))/(a^2 (b^ ... - b^2/c^2) Sin[z]^2) (1 - (c^2 (b^2 - a^2))/(a^2 (b^2 - c^2)) Sin[z]^2)^(1/2)), {z, 0, π/2}]$

This is – with Gradshteyn Ryzhik (6.113-2) – :

2ArcTanh[ ( 1 - c^2/a^2)^(1/2)] EllipticK[(c^2 (b^2 - a^2))/(a^2 (b^2 - c^2))] + π/IEllipticF[ArcCos[b/c], (c^2 (b^2 - a^2))/(a^2 (b^2 - c^2))]//N

Put them into the formula for the surface area above :

N[2π b^2 - (2π a b^2)/(I (b^2 - c^2)^(1/2)) EllipticF[ArcCos[b/c], (c^2 (b^2 - a^2)) ... + (2π a (b^2 - c^2)^(1/2))/IEllipticE[ArcCos[b/c], (c^2 (b^2 - a^2))/(a^2 (b^2 - c^2))], 17]

This is already the correct expression. Now rename the axes, that a > b > c :

a=. ; b=. ; c=. ; aa = b ; bb = c ; cc = a ; 2π bb^2 - (2π aa bb^2)/(I (bb^2 - cc^2 ... 960; aa (bb^2 - cc^2)^(1/2))/IEllipticE[ArcCos[bb/cc], (cc^2 (bb^2 - aa^2))/(aa^2 (bb^2 - cc^2))]

2 c^2 π - 2  b (-a^2 + c^2)^(1/2) π EllipticE[ArcCos[c/a], (a^2 (-b^2 + c^2) ... 10; b c^2 π EllipticF[ArcCos[c/a], (a^2 (-b^2 + c^2))/(b^2 (-a^2 + c^2))])/(-a^2 + c^2)^(1/2)

The final result is ( incidentally valid for any set of semi-axes – if a ≠ b ≠ c) :

$RowBox[{RowBox[{a, =, 4.7}], ;, RowBox[{b, =, 1.2}], ;, RowBox[{c, =, 1.1}], ;, t = ArcCos[c/a ... , ;, 2π c^2 + (2π a b)/Sin[t] (Cos[t]^2 EllipticF[t, k^2] + Sin[t]^2EllipticE[t, k^2])}]$

c→b ; surface of revolution (radius b)

b→a ; surface of sphere

As a 'pocket calculator' approximation to the surface area of an ellipsoid, that is better than 2.1% (if a > b ≥ c), we may use:

r = ArcCos[c/a]/(1 - c^2/a^2)^(1/2) ; 2 π (c^2 + a b r + (b^2 - c^2)/(3a b) r^3 (c^2 - a^2/2 + (a^4 b^2 + 3 a^4c^2 - 12 a^2c^4 + 8 b^2c^4)/(40 a^2 b^2) r^2))

Note that the first two terms correspond to the surface of revolution.

References

1. Wolfram: The Mathematica Book, Wolfram Media, Inc., Fourth Edition, 1999
2. Gradshteyn/Ryzhik: Table of Integrals, Series and Products, Academic Press, Second Printing, 1981

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