Consider the n-dimensional superellipsoid

$∑ k=1n |xk ak |p =1$

where p is an arbitrary real number that allows smooth interpolation between a variety of shapes. Here is how it looks in two dimensions:

For $p=2$ this figure is a standard ellipse. As p increases from that value the figure begins to acquire rounded corners, and when $p\to \infty$ it becomes a rectangle with sharp corners and straight edges. As p decreases from two the edges are less curved, becoming straight for $p=1$ . Further decreasing p below this value causes the edges to curve inward, and when $p\to 0$ the figure collapses onto the two coordinate axes.

The same behavior mutatis mutandis occurs in three dimensions,

and by mental extension in higher dimensions. The purpose of this presentation is to attempt to evaluate the volume and surface area of the superellipsoid in as many dimensions as feasible. The attempt will emphasize what was already seen in the case of ellipsoids: volumes are much easier to determine than surface areas.

It will be interesting in this context to compare the result of the evaluation to the two extreme cases of rectangular cuboid for $p\to \infty$ and collapsed coordinate axes for $p\to 0$ . In the case of collapse onto the axes, the volume is clearly identically zero:

$V(n, p→0) =0$

The ‘surface area’ in two dimensions, being linear itself, becomes a sum of extensions along the axes:

$S(2, p→0) =4(a1 +a2)$

In three and higher dimensions the surface area is no longer linear, so that when it collapses onto the coordinate axes one can expect

$S(n, p→0) =0 , n>2$

For the case of the cuboid, the volume is given by products of extensions along the axes:

$V(n, p→∞) =∏k=1 n 2ak =2n ∏k=1 n ak$

The surface area is a bit trickier. An n-dimensional cuboid is bounded by $2n$ facets of dimension $n-1$ . This statement can be motivated by considering that a line is bounded by two points, a square by four lines, acube by six squares and so on. The ‘area’ of each of these facets is a product of combinations of $n-1$ edges. Summing over the $(n n-1) =n$ combinations gives

$S(n, p→∞) =2∑ Cn-1 n( 2ak) =2n∑ Cn-1 n( ak)$

These will presumably match the suitable limits of the full evaluation.

Begin with the volume of the superellipsoid. The same scaling trick that worked for ellipsoids holds for superellipsoids as well,

$V(n,p) =∏k=1 n ∫dxk =∏k=1 nak ∫d(xk ak) =Vunit (n,p) ∏k=1 nak$

so that one needs to evaluate the volume of a unit supersphere, i.e, a superellipsoid with all denominators equal to unity. This will be implemented first using with the integral of a binomial

$∫dx(1 -axm )r =x F12 (−r, 1m; 1m+1; axm)$

and the Gauss summation formula

$F12 (a,b; c;1) =Γ(c) Γ(c-a -b) Γ(c-a) Γ(c-b)$

In two dimensions the ‘volume’ of the unit supersphere is

$Vunit (2,p) =2∫−1 1dx1 (1-|x1 |p )1/p =4∫01 dx1 (1-x1p )1/p =4F1 2( −1p, 1p; 1p+1; 1) =4Γ( 1p+1 )2 Γ(2p +1)$

where the initial factor of two accounts for the two halves of the figure.

The general evaluation extends this result by integrating over each dimension in turn, working down from ${x}_{n}$ . At each stage the absolute values can be removed by starting from the origin and including a factor of two. With the abbreviation $sm=∑ k=1m |xk |p$ , an intermediate stage of the integration is

$∫0 (1-sm )1/p du (1-sm -up )i/p =(1-sm )i/p ∫0 (1-sm )1/p du (1-up 1-sm )i/p =(1-sm )(i+1) /p F1 2( −ip, 1p; 1p+1; 1)$

Each stage of the integration increases the power of the binomial by one more inverse unit, and adds an overall factor of

$2F1 2( −ip, 1p; 1p+1; 1) =2Γ( 1p+1) Γ(ip +1) Γ(i+1 p+1)$

The final integration, for $i=n-1$ , acquires an extra factor of two just as for the two-dimensional supersphere. The result for n dimensions is thus

$Vunit (n,p) =2nΓ( 1p+1) Γ(1p +1) Γ(2p +1) Γ(1p +1)Γ( 2p+1) Γ(3p +1) ⋯Γ( 1p+1) Γ(n−1 p+1) Γ(np +1) =2nΓ( 1p+1 )nΓ( np+1) =2nn pn−1 Γ(1p )nΓ( np)$

and the volume of the superellipsoid is

$V(n,p) =2nΓ( 1p+1 )nΓ( np+1) ∏k=1 nak =2nn pn−1 Γ(1p )nΓ( np) ∏k=1 nak$

For $p=2$ this is

$V(n,2) =2π n/2 nΓ(n2 ) ∏k=1 nak$

which is identical to the result found previously for an n-dimensional ellipsoid. The case of $p=1$ , which separates convex from concave figures, has the simple form

$V(n,1) =2n n! ∏k=1 nak$

using the first form of the general result. Letting $p\to \infty$ in that same form gives

$V(n,∞) =2n ∏k=1 nak$

which is precisely the limit described above. This result can also be derived by expanding the gamma functions in the second form of the general result about zero, where each has a leading singular term.

The other limit described above, for $p\to 0$ , can be evaluated using Stirling’s approximation

$Γ(x+1) ≈2πx (xe )x$

for large argument. Ignoring factors not dependent upon p, the ratio of the two gamma functions in the general formula becomes

$Γ(1p +1)n Γ(np +1) ≈1 pn (1pe )n/p np (npe )n/p ≈[1p (n2n n-1 )1p ]n-12$

For positive nonzero n the combination in the denominator is greater than unity. The ratio inside square brackets is schematically

$limp→0 1p N1p =lim q→∞ qNq =0$

since an exponential function with base greater than unity increases faster than any polynomial. Thus one has $V\left(n,0\right)=0$ as expected as long as $n>1$ .

The two limiting cases are easily visualized by plotting the general result without the product of semimajor axes as a function of the power:

The gray line here is the asymptotic value ${2}^{n}$ .

The technique used here to evaluate the volume could clearly have been used in the presentation on ellipsoids. In the latter case, forming the metric tensor in a spherically symmetric space is a simpler process, leading to somewhat simpler integrals. Forming a metric tensor appropriate for superellipsoids is significantly more complicated. In order to replicate the process used for ellipsoids, it is more expedient to focus on the Jacobian matrix rather than the metric due to a feature of determinants.

As an aside, the metric tensor is formally the product of the transpose of the Jacobian matrix and its direct form. This article makes this clear for two dimensions, and the process extends naturally to higher dimensions.

Consider scaled Cartesian coordinates of the form

$[x1 2/p , x2 2/p , …, xn-1 2/p , xn 2/p ]$

where the individual ${x}_{k}$ have their usual meaning in spherical coordinates. One convenient way to order their values is

$x1 =rcosθ2 xi =rcosθ i+1 ∏k=2 isinθk , 2

which differs a bit from standard Cartesian ordering but is more easily tracked in higher dimensions. All angular variables have a domain of $\left[0,\pi \right]$ apart from the last, which has the domain $\left[0,2\pi \right]$ because is parametrizes two Cartesian variables along with the radial variable.

When forming the Jacobian matrix for the scaled coordinates, each row will be multiplied by a derivative of each of these quantities with respect to itself in addition to the usual spherical coordinate combinations. Since a factor multiplying a row of a matrix also multiplies its determinant, the Jacobian will acquire an extra factor of

$2p x1 2/p-1 2p x2 2/p-1 ⋯ 2p xn-1 2/p-1 2p xn 2/p-1 =2n pn r(2/p -1)n [∏ k=2n cos 2/p-1 θk] [∏ k=2n sin (2/p-1) (n+1-k) θk]$

in addition to the remaining spherically symmetric Jacobian

$rn-1 ∏k=2 n-1 sinn-k θk$

as determined in the previous presentation. The entire integrand is thus

$v(n,p) =2n pn r2n/p -1 [∏ k=2n cos 2/p-1 θk] [∏ k=2n sin 2(n+1 -k)/p-1 θk]$

The angular integrals can be easily evaluated with a trigonometric representation of the beta function:

$B(x,y) =Γ(x) Γ(y) Γ(x+y) =2∫0 π/2 dθ sin2x-1θ cos2y-1θ$

Since the Jacobian has an implicit absolute value omitted here for clarity, integration can be take over the first quadrant, keeping factors of two for most integrals and adding an additional factor of two for the last integral. The volume of the unit supersphere is then

$Vunit (n,p) =∫01 dr [∏k=2 n-1 ∫0π dθk] ∫02π dθn v(n,p) Vunit (n,p) =2n pn× p2n× Γ(1p) Γ(n -1p) Γ(np )× Γ(1p) Γ(n -2p) Γ(n -1p) ×⋯× Γ(1p) Γ(3p) Γ(4p )× Γ(1p) Γ(2p) Γ(3p )× 2Γ(1p )2 Γ(2p) Vunit (n,p) =2n 1pn Γ(1p )n np Γ(np) =2n Γ(1p +1)n Γ(np +1)$

which matches the previous implementation. It is perhaps a bit surprising that this implementation works out as nicely as it does, but the result is clearly correct.

And now on to surface areas! Begin with a two-dimensional superellipse parametrized with the Cartesian coordinates

$[ a1 cos2/pφ , a2 sin2/pφ ]$

The space-like interval on this surface contains a single term:

$ds2 =dx12 +dx22 =4p2 (a12 sin2φ cos4/p -2φ +a22 sin4/p -2φ cos2φ) dφ2$

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. The integral to be evaluated is

$S(2,p) =2p ∫0 2π dφ a12 sin2φ cos4/p -2φ +a22 sin4/p -2φ cos2φ S(2,p) =8p ∫0 π/2 dφ sinφcosφ a12 cos2Pφ +a22 sin2Pφ$

where the integration can be reduced to the first quadrant due to the implied square of the Jacobian. The abbreviation $P=2p-2$ has been chosen to simplify the appearance of the integral. This exponent is divided into two regions,

$p<1 ↔ P>0 p>1 ↔ P<0$

which as noted above is the separatrix between convex and concave figures. Under the substitution $φ→π2 -φ$ sines and cosines are interchanged, so that the integral can be written equivalently as

$S(2,p) =8p ∫0 π/2 dφ sinφcosφ a12 sin2Pφ +a22 cos2Pφ$

which simply interchanges the superellipse parameters. One case, $p=1$ , is particularly simple,

$S(2,1) =8a12 +a22 ∫0 π/2 dφsinφcosφ =4a12 +a22$

which is the perimeter of a rhombus. The case $p=2$

$S(2,2) =4∫0 π/2 dφ a12 sin2φ +a22 cos2φ$

is proportional to an elliptic integral of the second kind, a known special function. For equal superellipse parameters, the result is $2\pi$ times the parameter, the circumference of a circle.

In the general case one does not expect a closed form in terms of a known function, but must resort to expansions using the generalized binomial expansion

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

One normally expects the convergence of such expansions to be controlled by the relative values of the two superellipse parameters ${a}_{1}$ and ${a}_{2}$ . In the integral at hand one must factor out either the first term or the second term of the integrand, leading to expressions singular with respect to the angular variable at one endpoint or the other. This means one must expand the integrand with respect to both terms, switching over at the point where they equal one another. For the first form of the integral this occurs when

$a1cosPφ =a2sinPφ → φ12 =tan−1 (a1 a2 )1P$

and for the equivalent form when

$a1sinPφ =a2cosPφ → φ21 =tan−1 (a2 a1 )1P$

where the first subscript indicates which parameter is in the numerator. When the two superellipse parameters are equal, the dividing angle is simply $π4$ , exactly halfway between endpoints.

The two expansions of the first form of the integrand are

$sinφcosφ a12 cos2Pφ +a22 sin2Pφ =a1 ∑k=0 ∞Ck (a22 a12 )k sin2Pk +1φ cos−2Pk +P+1φ =a2 ∑k=0 ∞Ck (a12 a22 )k sin−2Pk +P+1φ cos2Pk +1φ$

with the abbreviation

$Ck =(−1 )kk! Γ(−12 +k) Γ(−12 )$

for a combination that will appear frequently. The corresponding expansions for the equivalent form of the integrand are

$sinφcosφ a12 sin2Pφ +a22 cos2Pφ =a2 ∑k=0 ∞Ck (a12 a22 )k sin2Pk +1φ cos−2Pk +P+1φ =a1 ∑k=0 ∞Ck (a22 a12 )k sin−2Pk +P+1φ cos2Pk +1φ$

with the superellipse parameters interchanged as noted above.

Angular integrals can be evaluated with a pair of trigonometric integrals,

$∫dθ sinpθ cosmθ =sin p+1θ p+1 F12 (1-m2 , p+12 ; p+32 ; sin2θ)$

$∫dθ sinpθ cosmθ =−cos m+1θ m+1 F12 (1-p2 , m+12 ; m+32 ; cos2θ)$

one in terms of sines and the other in terms of cosines, depending on on which stays finite at one endpoint or the other.

Consider first $P>0$ , starting with the first form of the integrand. The integral in terms of sines can be used with the first expansion on the first half of the interval, and the integral in terms of cosines with the second expansion on the second half of the interval. With the appropriate angle dividing the interval, one has

$S(2,p) =8a1p ∑k=0 ∞Ck (a22 a12 )k sin 2Pk+2 φ12 2Pk+2 F12 (Pk-P2 , Pk+1; Pk+2; sin2φ12) +8a2p ∑k=0 ∞Ck (a12 a22 )k cos 2Pk+2 φ12 2Pk+2 F12 (Pk-P2 , Pk+1; Pk+2; cos2φ12)$

Given the symmetry in exponents of the two expansions, the hypergeometric functions here have identical parameters.

If instead one starts with the equivalent form of the integrand, the situation is quite similar. The integrals can be used in exactly the same way, but the superellipse parameters interchange and the other dividing angle is necessary. This gives

$S(2,p) =8a2p ∑k=0 ∞Ck (a12 a22 )k sin 2Pk+2 φ21 2Pk+2 F12 (Pk-P2 , Pk+1; Pk+2; sin2φ21) +8a1p ∑k=0 ∞Ck (a22 a12 )k cos 2Pk+2 φ21 2Pk+2 F12 (Pk-P2 , Pk+1; Pk+2; cos2φ21)$

That these two results are identical can be seen by noting that

$sintan−1x =x 1+x2 =coscot−1x sincot−1x =1 1+x2 =costan−1x$

which implies

$sinφ12 =cosφ21 cosφ12 =sinφ21$

and the two expressions are clearly term-by-term equivalent. This also means one can write two additional forms of the surface area for $P>0$ by interchanging sines and cosines and the dividing angle, but this becomes superfluous.

For $P<0$ the application of the integrals is slightly different. The integral in terms of sines can be used with the second expansion on the first half of the interval, and the integral in terms of cosines with the first expansion on the second half of the interval. Again with the appropriate angle dividing the interval, one has

$S(2,p) =8a2p ∑k=0 ∞Ck (a12 a22 )k sin −2Pk +P+2 φ12 −2Pk +P+2 × F12 (−Pk, −Pk+P2 +1; −Pk+P2 +2; sin2φ12) +8a1p ∑k=0 ∞Ck (a22 a12 )k cos −2Pk +P+2 φ12 −2Pk +P+2 × F12 (−Pk, −Pk+P2 +1; −Pk+P2 +2; cos2φ12)$

This can be simplified a bit using $P2+1 =1p$ , but then one risks confusion between capital and small letters under the sum: better to stick to one or the other. Also note that since P never goes below negative two for finite p, the hypergeometric functions remain nonsingular.

Starting from the equivalent form of the integrand interchanges superellipse parameters and dividing angle:

$S(2,p) =8a1p ∑k=0 ∞Ck (a22 a12 )k sin −2Pk +P+2 φ21 −2Pk +P+2 × F12 (−Pk, −Pk+P2 +1; −Pk+P2 +2; sin2φ21) +8a2p ∑k=0 ∞Ck (a12 a22 )k cos −2Pk +P+2 φ21 −2Pk +P+2 × F12 (−Pk, −Pk+P2 +1; −Pk+P2 +2; cos2φ21)$

This expression is again term-by-term equivalent to the preceding one. One can again write two more forms for the surface area for $P<0$ using the equality of trigonometric functions, but this is again superfluous.

These expressions for surface area can be simplified by noting that

$sin2P φ12 =a12 (a1 2/P +a2 2/P )P cos2P φ12 =a22 (a1 2/P +a2 2/P )P$

so that for $P>0$ one has

$S(2,p) =8a1p ∑k=0 ∞Ck sin2 φ12 cos2Pk φ12 2Pk+2 F12 (Pk-P2 , Pk+1; Pk+2; sin2φ12) +8a2p ∑k=0 ∞Ck sin2Pk φ12 cos2 φ12 2Pk+2 F12 (Pk-P2 , Pk+1; Pk+2; cos2φ12)$

while for $P<0$ one has

$S(2,p) =8a2p ∑k=0 ∞Ck sinP+2 φ12 cos−2Pk φ12 −2Pk +P+2 × F12 (−Pk, −Pk+P2 +1; −Pk+P2 +2; sin2φ12) +8a1p ∑k=0 ∞Ck sin−2Pk φ12 cosP+2 φ12 −2Pk +P+2 × F12 (−Pk, −Pk+P2 +1; −Pk+P2 +2; cos2φ12)$

These expressions are mathematically cleaner than those given previously, since it is clear here that the sums are over quantities dependent only on the dividing angle. The downside is that the trigonometric factors here are numerically unstable as $p\to 1$ , which corresponds to $P\to 0$ , requiring great precision for accurate evaluation in that regime.

In this context it is worth noting that these series expansions, while a bit ungainly, are preferable to direct numerical evaluation of the integrals from which they arise. The latter approach is numerically unstable, particularly as $p\to \infty$ , while the series expansions in general remain stable. These series expansions also give one a sense of what sort of functions the integrals represent, even if only as infinite sums.

In the special case ${a}_{1}={a}_{2}=a$ the two dividing angles are equal: $φ1=φ2 =tan−11 =π4$ . Sine and cosine are equal at that angle, $sinπ4 =cosπ4 =12$ , so that the two sums in all the expressions above are identical. The surface area in this special case is given by a single sum:

$Sequal (2,p) =16ap ∑k=0 ∞Ck (12 )Pk+1 2Pk+2 F12 (Pk-P2 , Pk+1; Pk+2; 12) ,P>0 =16ap ∑k=0 ∞Ck (12 )−Pk +P/2+1 −2Pk +P+2 F12 (−Pk, −Pk+P2 +1; −Pk+P2 +2; 12) ,P<0$

And now to see it in action! As a function of the original exponent p, using the unsimplified forms for stabler numerical evaluation, the surface area in two dimensions looks like this:

The gray line here is the asymptotic value of the area, which for two dimensions is equal in both limits. The red line is the exact value for the separatrix $p=1$ , and clearly the minimum value. The purple line is the exact value for $p=2$ and one other exponent value near one half.

These series converge numerically rather slowly, but do represent exact solutions with the expected asymptotic values. In order not to overburden the browser, the number of terms is restricted to a value that renders as quickly as possible, as well as a reduced tolerance on the hypergeometric functions.

Now consider confirming some additional special cases. For $p=1$ , which corresponds to $P=0$ , the hypergeometric functions are unity in all expressions given. In the first expression for $P>0$ one then has

$S(2,1) =4a1 sin2φ12 ∑k=0 ∞Ck (a22 a12 )k +4a2 cos2φ12 ∑k=0 ∞Ck (a12 a22 )k S(2,1) =4( sin2φ12 +cos2φ12 ) a12 +a22 =4a12 +a22$

since the ${C}_{k}$ are just the expansion coefficients for the square root of one plus the argument. The first expression for $P<0$ interchanges sine and cosine but is otherwise identical. The result is naturally the exact result given above.

As $p\to \infty$ one can initially set $P=-2$ in the appropriate expression. The exponents of trigonometric functions are $4k$ , and the parameters of the hypergeometric functions are $2k$ , $2k$ and $2k+1$ , all finite values. Since there is an overall infinite multiplicative factor, the denominators in the sum must be handled explicitly:

$1p× 1−2Pk +P+2 =1p× 1−2( 2p -2)k +2p =14(p -1)k+2$

This denominator approaches zero for all but the first terms of the sums, so that

$S(2,∞) =4(a1 +a2)$

which is the expected result for two dimensions, where the combinations are simply linear quantities.

As $p\to 0$ then $P\to \infty$  and the dividing angles both approach $π4$ regardless of the relative values of the superellipse parameters. For nonzero k the denominators can be handled similarly,

$1p× 12Pk+2 =1p× 12(2p -2)k+2 =14(1 -p)k+2$

which in this case remains finite. The hypergeometric functions appear not to remain finite, but they are balanced by the values of the trigonometric functions. Approximating the terms of the hypergeometric functions by the leading powers of their polynomial numerators and denominators, the dependence on P in this balance is

$limP→∞ (12 )Pk (Pk -P/2)m (Pk)m (Pk)m ≈lim P→∞ Pm 2Pk =0$

for some finite term m in the expansion of the hypergeometric function. This ratio is zero for nonzero k for the same reason as above with volumes: an exponential function with base greater than unity increases faster than any polynomial. This just leaves the first terms of the sums, where one can use direct integration plus a binomial integral to identify the hypergeometric function in question:

$∫dx (1-x)r =−(1 -x) r+1 r+1 +C =xF1 2 (−r,1; 2;x) F12 (−r,1; 2;x) =1-(1-x )r+1 (r+1)x$

Remembering to include the initial singular factor when $k=0$ , the surface area is

$S(2,0) =limp→0 8p (a1 +a2)× 12× 12× F12 (−P2, 1;2; 12) S(2,0) =2(a1 +a2) limp→0 1-(12 )1/p 12 =4(a1 +a2)$

which again is the expected result for two dimensions.

The final special case is $p=2$ , which corresponds to $P=-1$ . Expressions above for $P<0$ give the mathematically equivalent forms

$S(2,2) =4a2 ∑k=0 ∞Ck (a12 a22 )k sin2k+1 φ12 2k+1 F12 (k, k+12; k+32; sin2φ12) +4a1 ∑k=0 ∞Ck (a22 a12 )k cos2k+1 φ12 2k+1 F12 (k, k+12; k+32; cos2φ12) S(2,2) =4a2 ∑k=0 ∞Ck sinφ12 cos2k φ12 2k+1 F12 (k, k+12; k+32; sin2φ12) +4a1 ∑k=0 ∞Ck sin2k φ12 cosφ12 2k+1 F12 (k, k+12; k+32; cos2φ12)$

There does not appear to be an easy way to simplify this further due to the different factors multiplying each sum. One does know from the integral itself that

$S(2,2) =4a1E(1 -a22 a12) =4a2E(1 -a12 a22)$

so that the combination of sums must be equal to the two solutions in terms of a complete elliptic integral. This can be verified numerically with a visualization of the two equal expressions, using the unsimplified summations for stabler evaluation:

An interesting side note is that for the case ${a}_{1}={a}_{2}$ of equal superellipse parameters, the elliptic function is half of $\pi$ and one can write

$∑k=0 ∞Ck (12 )k+1/2 2k+1 F12 (k, k+12; k+32; 12) =π4$

a summation that does not appear to be well known, but which can be verified numerically.

Whew! All that work for just two dimensions. As promised above, surface areas are a lot more difficult to evaluate than volumes.

Now consider a three-dimensional superellipsoid parametrized with the Cartesian coordinates

$[a1 (cosφ1 )2/p , a2 (sinφ1 cosφ2 )2/p , a3 (sinφ1 sinφ2 )2/p ]$

where the dependence has been chosen for consistency with the presentation on ellipsoids. As in that presentation, for conciseness use the abbreviations

$sinφi =si cosφi =ci$

The space-like interval on this surface is

$ds2 =dx12 +dx22 +dx32 ds2 =4p2 [a12 s12 c1N-2 +s1N-2 c12( a22 c2N +a32 s2N)] dφ12 +8p2 s1N-1 c1( a32 s2N-1 c2 -a22 s2 c2N-1) dφ1 dφ2 +4p2 s1N( a22 s22 c2N-2 +a32 s2N-2 c22) dφ22$

with $N=4p$ for convenience. After a wee bit of algebra, the determinant of the metric tensor is

$detg =16p4 [a22 a32 s12N-2 c12 s2N-2 c2N-2 +a12 s1N+2 c1N-2 (a22 s22 c2N-2 +a32 s2N-2 c22)]$

which reduces to the determinant of the ellipsoid for $p=2$ . In terms of the previous abbreviation $P=2p-2$ the exponents appearing here are

$2N-2 =4P+6 N+2 =2P+6 N-2 =2P+2$

so that the determinant can be written a bit more simply as

$detg =16p4 s12P+6 c12 s22 c22 [a22 a32 s12P s22P c22P +a12 c12P (a22 c22P +a32 s22P )]$

The surface area is given by the integral

$S(3,p) =4p2 ∫0π dφ1 ∫02π dφ2 s1P+3 c1 s2 c2 a22 a32 s12P s22P c22P +a12 c12P (a22 c22P +a32 s22P ) S(3,p) =32p2 ∫0 π/2 dφ1 ∫0 π/2 dφ2 s1P+3 c1 s2 c2 a22 a32 s12P s22P c22P +a12 c12P (a22 c22P +a32 s22P )$

where the integration can be reduced to the first octant due to the implied square of the Jacobian. Note again that P never goes below negative two for finite p, so the integral is not immediately singular.

As in two dimensions, the case $p=1$ is particularly simple,

$S(3,1) =32a12 a22 +a12 a32 +a22 a32 ∫0 π/2 dφ1 ∫0 π/2 dφ2 s13c1 s2c2 =4a12 a22 +a12 a32 +a22 a32$

which is the surface area of an octahedron with semiaxes given by the superellipsoid parameters. The case $p=2$

$S(3,2) =8∫0 π/2 dφ1 ∫0 π/2 dφ2 s1 a22 a32 c12 +a12 s12 (a22 s22 +a32 c22)$

was shown to be proportional to an Appell double hypergeometric function of the first kind in the previous presentation. For equal superellipsoid parameters the result is $4\pi$ times the parameter squared, the surface area of a sphere.

The general integral is naturally more complicated than that for two dimensions, but can be treated in the same general manner. One again needs to pay attention to endpoint singularities, dividing the two intervals of integration at some middle point. To find the dividing angle for the first angular variable, set the two terms of the integrand equal as before:

$a2a3 s1P s2P c2P =a1 c1P a22 c22P +a32 s22P Φ(φ2) =tan −1( a1 a2 a3R )1P R=a22 s2−2P +a32 c2−2P$

where the square root is separated to simplify the next step. The two expansions of the integrand with respect to the first angular variable are then

$s1P+3 c1 s2 c2 a22 a32 s12P s22P c22P +a12 c12P (a22 c22P +a32 s22P ) =s1P+3 c1 s2P+1 c2P+1 a22 a32 s12P +a12 c12P R2 =a1 ∑k=0 ∞Ck (a22 a32 a12 )k s12Pk +P+3 c1−2Pk +P+1 s2P+1 c2P+1 R−2k +1 =a2a3 ∑k=0 ∞Ck (a12 a22 a32 )k s1−2Pk +2P+3 c12Pk +1 s2P+1 c2P+1 R2k$

Now consider integration over this first angular variable, using the dividing angle above. For $P>0$ the trigonometric integral in terms of sines can be used with the first expansion on the first half of the interval, while that in terms of cosines with the second expansion on the second half of the interval:

$S(3,p) =32a1 p2 ∑k=0 ∞Ck (a22 a32 a12 )k × ∫0 π/2 dφ2 s2P+1 c2P+1 R−2k +1 sin 2Pk +P+4Φ 2Pk +P+4 × F12 (Pk -P2, Pk+P2 +2; Pk+P2 +3; sin2Φ) +32a2 a3p2 ∑k=0 ∞Ck (a12 a22 a32 )k × ∫0 π/2 dφ2 s2P+1 c2P+1 R2k cos 2Pk+2Φ 2Pk+2 × F12 (Pk-P -1, Pk+1; Pk+2; cos2Φ)$

For $P<0$ the trigonometric integral in terms of sines can be used with the second expansion on the first half of the interval, while that in terms of cosines with the first expansion on the second half of the interval:

$S(3,p) =32a2 a3p2 ∑k=0 ∞Ck (a12 a22 a32 )k × ∫0 π/2 dφ2 s2P+1 c2P+1 R2k sin −2Pk +2P+4Φ −2Pk +2P+4 × F12 (−Pk, −Pk+P +2; −Pk+P +3; sin2Φ) +32a1 p2 ∑k=0 ∞Ck (a22 a32 a12 )k × ∫0 π/2 dφ2 s2P+1 c2P+1 R−2k +1 cos −2Pk +P+2Φ −2Pk +P+2 × F12 (−Pk -P2 -1, −Pk+P2 +1; −Pk+P2 +2; cos2Φ)$

Since the exponents of trigonometric functions are no longer symmetric as for the two-dimensional case, all hypergeometric functions here have different parameters. And since P never goes below negative two for finite p, the hypergeometric functions remain nonsingular.

These expressions can be simplified by noting that

$R=a2 a3a1 tanPΦ$

so that for $P>0$ one has

$S(3,p) =32a2 a3 p2 ∑k=0 ∞Ck ∫0 π/2 dφ2 s2P+1 c2P+1 sin2P +4Φ cos2Pk -PΦ 2Pk +P+4 × F12 (Pk -P2, Pk+P2 +2; Pk+P2 +3; sin2Φ) +32a2 a3p2 ∑k=0 ∞Ck ∫0 π/2 dφ2 s2P+1 c2P+1 sin 2PkΦ cos2Φ 2Pk+2 × F12 (Pk-P -1, Pk+1; Pk+2; cos2Φ)$

while for $P<0$ one has

$S(3,p) =32a2 a3p2 ∑k=0 ∞Ck ∫0 π/2 dφ2 s2P+1 c2P+1 sin2P +4Φ cos−2Pk Φ −2Pk +2P+4 × F12 (−Pk, −Pk+P +2; −Pk+P +3; sin2Φ) +32a2 a3 p2 ∑k=0 ∞Ck ∫0 π/2 dφ2 s2P+1 c2P+1 sin−2Pk +PΦ cos2Φ −2Pk +P+2 × F12 (−Pk -P2 -1, −Pk+P2 +1; −Pk+P2 +2; cos2Φ)$

These expressions are somewhat mathematically cleaner, since now only trigonometric functions appear. One should keep in mind that the expressions in terms of $\Phi$ are still complicated functions of the second angular variable and the superellipsoid parameters.

After expansion of the hypergeometric functions,

$F12 (a,b;c; x) =∑l=0 ∞ Flxl Fl (a,b,c) =1l! Γ(a+l) Γ(a) Γ(b+l) Γ(b) Γ(c) Γ(c+l)$

the expressions for surface area are summations over integrals of two forms: either in terms of products of R with trigonometric functions,

$∫0 π/2 dφ2 s2P+1 c2P+1 RMsinNΦ ∫0 π/2 dφ2 s2P+1 c2P+1 RMcosNΦ$

or simply in terms of trigonometric functions:

$I(P,M,N) =∫0 π/2 dφ2 s2P+1 c2P+1 sinMΦ cosNΦ$

Given that

$sinΦ=( a1R a2a3 )1P 1+( a1R a2a3 )2P cosΦ=1 1+( a1R a2a3 )2P$

one might look for expansions of the common denominator radical using the generalized binomial expansion. Unfortunately there do not appear to be any good options. The second term under the radical is typically quite different from unity, so a dividing angle is not possible here. The quantity R is also not guaranteed to be small for either positive or negative P and cannot be used an an expansion parameter. Indeed for $P>0$ this quantity blows up severely at both endpoints of the interval of integration.

Any possible series expansions of either two integral forms are likely to be so complicated as to be unuseful in practice. It is simpler at this point to evaluate the remaining angular integrals numerically. The numerical behavior of the quantity R makes the first integral form impractical in an interactive environment. Fortunately the second form is stable enough numerically for evaluation here, as anticipated by the notation above.

For $P>0$ the surface area in three dimensions is thus

$S(3,p) =32a2 a3 p2 ∑k,l=0 ∞ Ck 2Pk +P+4 ×Fl (Pk -P2, Pk+P2 +2, Pk+P2 +3) ×I(P, 2P+4+2l, 2Pk-P) +32a2 a3p2 ∑k,l=0 ∞ Ck 2Pk+2 ×Fl (Pk-P -1, Pk+1, Pk+2) ×I(P, 2Pk, 2+2l)$

while for $P<0$ the surface area is

$S(3,p) =32a2 a3p2 ∑k,l=0 ∞ Ck −2Pk +2P+4 ×Fl (−Pk, −Pk+P +2, −Pk+P +3) ×I(P, 2P+4+2l, −2Pk) +32a2 a3 p2 ∑k,l=0 ∞ Ck −2Pk +P+2 ×Fl (−Pk -P2 -1, −Pk+P2 +1, −Pk+P2 +2) ×I(P, −2Pk+P, 2+2l)$

As a function of the original exponent p the surface area in three dimensions looks like this:

This graphic takes some time to update due to the amount of calculation required for a decent result. Patience is required!

The presence of integrals in this final result complicates evaluation of special cases. From the graphic one can at least see that $S\left(3,0\right)=0$ as anticipated. It is currently unclear how to proceed with confirmation of other special cases. To be considered at some other time...

The second half of the presentation on ellipsoids allowed an inference on the form of surface area for an arbitrary number of dimensions in terms of a Lauricella hypergeometric function. Unfortunately no such inference appears possible at this stage of this presentation. Since this one has gone on rather long, that will be left as a possible future addition. Schade!