This presentation will evaluate the volume and surface area of an n-dimensional ellipsoid. The volume is particularly simple to determine, since variables of integration can be scaled to produce integrations over spheres. That is, given the ellipsoid

$∑ k=1n xk2 ak2 =1$

its volume is immediately proportional to that of an n-dimensional unit sphere,

$Ve(n) =∏k=1 n ∫dxk =∏k=1 nak ∫d(xk ak) =Vs(n), a=1 ∏k=1 nak$

where the subscripts designate an ellipsoid or sphere of the given dimension and a will be the single parameter describing a sphere.

Evaluating the volume of an n-sphere requires the Jacobian for a spherically symmetric space, which is the square root of the determinant of its metric tensor. The appropriate metric for two dimensions is easily determined:

$ds(2)2 =dx12 +dx22 =[d(rcosθ) ]2 +[d(rsinθ) ]2 =dr (2)2 +r(2)2 dθ2$

The subscripts in parentheses indicate the dimension of the space in which the quantity is defined.

Rather than manually calculating the result for three or more dimensions, construct a recursion relation. An additional Cartesian variable will be the product of a cosine with the higher-dimensional radius, while the original radial variable will be the product of a sine with the new radial variable:

$ds (n+1)2 =dr (n)2 +r(n)2 dΩ (n)2 +dx n+12 ds (n+1)2 =[d( r(n+1) sinθ)]2 +r(n+1) 2 sin2θ dΩ (n)2 +[d( r(n+1) cosθ)]2 ds (n+1)2 =dr (n+1)2 +r(n+1) 2 [dθ2 +sin2θ dΩ (n)]$

By induction from the metric in two dimensions, the n-dimensional metric is

$ds (n)2 =dr (n)2 +r(n)2 dΩ (n)2 dΩ (n)2 =∑k=2 ndθk2 ∏m=2 k−1 sin2θm$

The first angular variable is labeled with a two because it is the second variable of parametrization. The determinant of this metric consists of products of sines with the radial variable. Its square root can be read off easily because the metric is diagonal:

$detg(n) =rn-1 ∏k=2 n-1 sinn-k θk$

The the last angular variable does not appear in this expression. The domain of this variable is twice that of the other angular variables, since along with the radial variable it parametrizes two Cartesian variables. Integration over the other angular variables is done with a trigonometric integral,

$∫0π dθ sinpθ =Γ(p +12) Γ(12) Γ(p+2 2) =π Γ(p+1 2) Γ(p+2 2)$

where symmetry of the integrand gives a factor of two. The volume of the n-sphere is

$Vs(n) =∫0a dr rn-1 ×∏k=2 n-1 ∫0π dθk sinn-k θk ×∫0 2π dθn Vs(n) =2π ann ×π(n -2)/2 Γ(n -12) Γ(n2 ) Γ(n -22) Γ(n -12) ⋯ Γ(32) Γ(42) Γ(22) Γ(32) Vs(n) =2π n/2 nΓ(n2 ) an$

As a check

$Vs(2) =πa2 Vs(3) =43π a3$

which are the expected values for the area of a circle and the volume of a sphere in three dimensions. The volume of the ellipsoid is simply

$Ve(n) =2π n/2 nΓ(n2 ) ∏k=1 nak$

which holds for an arbitrary number of dimensions two or greater. The volume of the ellipsoid differs from that of the sphere merely in having the single parameter replaced by a product of semiaxes.

The surface area of an ellipsoid is much more involved to evaluate. 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 tensor. Unfortunately the semiaxis parameters cannot be simply scaled away but remain throughout the evaluation.

As a point of reference, first consider the surface area of an n-sphere. The relevant metric is found by setting the radial variable in the n-dimensional spherically symmetric metric equal to the constant radius, leaving merely the angular part of the metric. The surface area of an n-dimensional sphere is

$Ss(n) =an-1 ×∏k=2 n-1 ∫0π dθk sinn-k θk ×∫0 2π dθn =2π n/2 Γ(n2) an-1$

The differs from its volume in having one less radial factor and not having the dimension of the space in the denominator. As a check

$Ss(2) =2πa Ss(3) =4πa2$

which are the expected values for the circumference of a circle and the area of a sphere in three dimensions. Evaluating them via metric determinants is somewhat simpler than derivations appearing in elementary calculus textbooks.

To begin to see how different the surface area of a general ellipsoid will be from this result, first consider a two-dimensional ellipse. A typical parametrization in Cartesian coordinates is

$r e(2) =[acosφ, bsinφ]$

The metric corresponding to this parametrization is

$ds e(2)2 =dx12 +dx22 =(a2 sin2φ +b2 cos2φ) dφ2$

Integrating the square root of the single tensor component results in a known but nontrivial function, a complete elliptic integral of the second kind,

$Se(2) =∫0 2π dφ a2sin2φ +b2cos2φ Se(2) =4b∫0 π/2 dφ 1-b2 -a2b2 sin2φ Se(2) =4bE(1 -a2 b2) , a

where the elliptic parameter is used in preference to the traditional elliptic modulus and the condition on the semiaxes keeps it positive. Since complete elliptic integrals can be defined with either sines or cosines, one also has

$Se(2) =4a∫0 π/2 dφ 1-a2 -b2a2 cos2φ =4aE(1 -b2 a2) , a>b$

which is the result normally seen. This is equivalent to applying an identity for elliptic functions of negative parameter.

Elliptic integrals were of course introduced to “rectify the ellipse”, as the old phrase goes. What is critical in the current context is that the ellipse parameters cannot simply be factored out of the result as was the case for volume.

It is instructive at this point to consider the same calculation in elliptic coordinates. The parametrization for these coordinates is

$r e(2) =[ccoshucosv, csinhusinv]$

with the standard relation ${c}^{2}={a}^{2}-{b}^{2}$ from squaring the identifications $a=ccoshu$ and $b=csinhu$ . The metric corresponding to this parametrization is

$ds e(2)2 =c2 (sinh2u +sin2v) (du2 +dv2)$

Using standard simple integrals, the volume from this metric is

$Ve(2) =c2 ∫0u du ∫02π dv (sinh2u +sin2v) =2πc2 sinh2u4 =πab$

which is the expected value for the area of an ellipse. The surface area from this metric, which will be the perimeter of the ellipse, is found by setting the parameter u to a constant value. The first term of the metric drops out, taking its multiplicative factor with it, and the integral to be evaluated is no longer simple:

$Se(2) =c∫0 2π dv sinh2u +sin2v S e(2) =c∫0 2π dv cosh2u -cos2v Se(2) =4ccoshu ∫0 π/2 dv 1-c2 c2 cosh2u cos2v =4aE(1 -b2 a2)$

The metric for spherically symmetric spaces has a coefficient of one multiplying the differential of the radial variable, so when it drops out for surface area evaluations the change to the process is minor. This is no longer the case for elliptic coordinates, and in a sense is the origin of elliptic integrals.

Determining the metric for an ellipsoidal surface rapidly becomes complicated as the number of dimensions increases. Fortunately evaluating the surface area from the metric determinant for the first few dimensions indicates the general result for an arbitrary number of dimensions. This will be done using the generalized binomial expansion

$(1-x )r =1-rx +r (r-1) 2! x2 -r(r-1) (r-2) 3! x3 +⋯ (1-x )r =∑k=0 ∞ Γ(−r +k) Γ(−r) xkk!$

as well as the trigonometric integral

$∫0 π/2dθ sinpθ cosmθ =Γ(p +12) Γ(m +12) 2Γ(p +m2 +1)$

The determinants of metric tensors are evaluated in Mathematica and available here. Since the trigonometric quantities appearing in them are all squared, angular regions of integration can always be reduced to that of the trigonometric integral.

First repeat the two-dimensional case, starting with the parametrization

$r e(2) =[a1 cosφ1, a2 sinφ1]$

For conciseness introduce the abbreviations

$si=sinφi ci=cosφi αi =1-an2 ai2$

where n is the dimension of space. Note that the last angle will have a domain equal to twice that of other angles. The evaluation then proceeds as

$Se(2) =∫0 2π dφ1 a12 s12 +a22 c12 =4a1∫0 π/2 dφ1 1-α1 c12 Se(2) =4a1 ∑k=0 ∞ Γ(−12 +k)Γ( −12) α1k k! ∫0 π/2 dφ1 c12k Se(2) =2a1 ∑k=0 ∞ Γ(−12 +k)Γ( −12) α1k k!· Γ(12) Γ(k +12) Γ(k+1) Se(2) =2πa1 ∑k=0 ∞ Γ(−12 +k)Γ( −12) Γ(12 +k) Γ(12) Γ(1) Γ(1+k) α1k k!$

This is the series for a Gauss hypergeometric function. While it is already known that this is an complete elliptical integral of the second kind, it will be left in this form for comparison with other cases.

For the three-dimensional case, the parametrization is

$r e(3) =[a1 cosφ1, a2 sinφ1 cosφ2, a3 sinφ1 sinφ2]$

and the evaluation proceeds as

$Se(3) =∫0π dφ1 ∫0 2π dφ2 s1 a22 a32 c12 +a12 s12( a32 c22 +a22 s22) Se(3) =8a1a2 ∫0 π/2 dφ1 ∫0 π/2 dφ2 s1 1-α1 c12 1-α2 s12 1-α1 c12 c22 Se(3) =8a1a2 ∑k=0 ∞ Γ(−12 +k)Γ( −12) α2k k! ∫0 π/2 dφ1 s12k+1 (1-α1 c12 )12-k ∫0 π/2 dφ2 c22k Se(3) =8a1a2 ∑k,l=0 ∞ Γ(−12 +k+l)Γ( −12) α1l l! α2k k! ∫0 π/2 dφ1 s12k+1 c12l ∫0 π/2 dφ2 c22k Se(3) =2a1a2 ∑k,l=0 ∞ Γ( −12 +k+l)Γ( −12) α1l l! α2k k!· Γ(k+1) Γ(l +12) Γ(k+l +32)· Γ(12) Γ(k +12) Γ(k+1) Se(3) =4πa1 a2 ∑k,l=0 ∞ Γ(−12 +k+l) Γ( −12) Γ(12 +k) Γ(12) Γ(12 +l) Γ(12) Γ(32) Γ(32 +k+l) α1l l! α2k k!$

This is the series for an Appell double hypergeometric function of the first kind, consistent with the results of a previous presentation.

For the four-dimensional case, the parametrization is

$r e(4) =[a1 cosφ1, a2 sinφ1 cosφ2, a3 sinφ1 sinφ2 cosφ3, a4 sinφ1 sinφ2 sinφ3]$

and (deep breath!) the evaluation proceeds as

$Se(4) =∫0π dφ1 ∫0π dφ2 ∫02π dφ3 s12s2 a22 a32a42 c12 +a12 s12( a32 a42 c22 +a22 s22( a42 c32 +a32 s32)) Se(3) =16a1 a2a3 ∫0 π/2 dφ1 ∫0 π/2 dφ2 ∫0 π/2 dφ3 s12s2 1-α1 c12 -α2 s12 c22 1-α3 s12 s22 1-α1 c12 -α2 s12 c22 c32 Se(4) =16a1 a2a3 ∑k=0 ∞ Γ(−12 +k)Γ( −12) α3k k! ∫0 π/2 dφ1 s12k +2 ×∫0 π/2 dφ2 s22k +1 (1-α1 c12 -α2 s12 c22 )12-k ∫0 π/2 dφ3 c32k Se(4) =16a1 a2a3 ∑k=0 ∞ Γ(−12 +k)Γ( −12) α3k k! ∫0 π/2 dφ1 s12k +2 (1-α1 c12 )12 -k ×∫0 π/2 dφ2 s22k +1 (1-α2 s12 1-α1 c12 c22 )12 -k ∫0 π/2 dφ3 c32k Se(4) =16a1 a2a3 ∑k,l=0 ∞ Γ(−12 +k+l)Γ( −12) α2l l! α3k k! ∫0 π/2 dφ1 s12k+2l +2 (1-α1 c12 )12 -k-l ×∫0 π/2 dφ2 s22k +1 c22l ∫0 π/2 dφ3 c32k Se(4) =16a1 a2a3 ∑ k,l,m=0 ∞ Γ(−12 +k+l+m) Γ(−12 ) α1m m! α2l l! α3k k! ×∫0 π/2 dφ1 s12k+2l +2 c12m ∫0 π/2 dφ2 s22k +1 c22l ∫0 π/2 dφ3 c32k Se(4) =2a1 a2a3 ∑ k,l,m=0 ∞ Γ(−12 +k+l+m) Γ(−12 ) α1m m! α2l l! α3k k! ×Γ(k+l +32) Γ(m+12) Γ(k+l+m +2) ·Γ(k+1) Γ(l+12) Γ(k+l +32) ·Γ(12) Γ(k+12) Γ(k+1) Se(4) =2π2a1 a2a3 ∑ k,l,m=0 ∞ Γ(−12 +k+l+m) Γ(−12 ) Γ(12 +k) Γ(12) Γ(12 +l) Γ(12) Γ(12 +m) Γ(12) ×Γ(2) Γ(k+l +m+2) α1m m! α2l l! α3k k!$

This is the series for a Lauricella hypergeometric function of type D of three variables. This function reduces to the Appell double hypergeometric function for two variables and to the Gauss hypergeometric function for a single variable. This is enough evidence to establish the general result that the surface area of an n-dimensional ellipsoid is proportional to a Lauricella hypergeometric function of type D of n−1 variables. This is clear from how the function builds up during the explicit evaluation, a process that will naturally extend to higher dimensions.

All that remains is to determine the constant of proportionality, apart from the product of semiaxes. There are n−1 angular integrals, each of which contributes a factor of two from symmetry of the integrand, apart from the last one which contributes twice that value. Evaluation of each angular integration contributes a factor of one-half. The first binomial expansion contributes a direct factor of gamma of one-half, and each of the n−1 binomial expansions must be offset by a factor of gamma of one-half for the series. The series also requires a numerator of gamma of half the dimension of space, so that the total constant of proportionality is

$2n-2 ·22 ·12 n-1 ·Γ( 12) ·Γ(12 )n-1 ·1Γ( n2) =2Γ(12 )n Γ(n2) =2π n/2 Γ(n2)$

which is the same constant of proportionality as for an n-dimensional sphere, as one would expect. The general expression for the surface area of an n-dimensional ellipsoid is thus

$Se(n) =2π n/2 Γ(n2) (∏k=1 n-1 ak) FD (n-1)( −12; 12,…, 12; n2; α1,…, αn-1) , αi=1 -an2 ai2$

When all of the ${\alpha }_{i}$ are equal the hypergeometric function is unity, and the expression matches that above for the sphere.

What is perhaps most interesting about this result is that, while the volume of the ellipsoid is an analytic function of dimension, the surface area is decidedly not. Unless of course there is a sense in which one can interpolate between discrete summations. Something to ponder...