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The $n$-elliptope is defined as the set of $n$-by-$n$ correlation matrices; that is, the set of $n$-by-$n$ symmetric positive-definite matrices with ones on the diagonal. Such matrices are parametrized by their $n(n-1)/2$ upper off-diagonal elements. In the case of $n=3$, this yields the 3-elliptope $$Gamma = (x,y,z)in[-1,1]^3 : x^2+y^2+z^2leq 1+2xyz.$$

The volume of $Gamma$ was considered in an earlier question (What is the volume of the $3$-dimensional elliptope?) and shown to be $pi^2/2$. However, what I'm interested in presently is the subset of singular 3-by-3 correlation matrices. This corresponds precisely to the boundary of the above set: $$partial Gamma = (x,y,z)in [-1,1]^3: x^2+y^2+z^2=1+2xyz.$$

With that in mind, what I want to know is the the surface area of $partial Gamma.$ Formally, this is not so hard: The surface can be expressed as the union of the surfaces $z=f_pm(x,y)=x ypm sqrt1-x^2sqrt1-y^2$, and the bottom surface is the mirror of the top. Hence their areas are the same, so the total area is given the double integral $$S=2int_-1^1int_-1^1 sqrt1+left(fracpartial f_+partial xright)^2+left(fracpartial f_+partial yright)^2,dx,dy,$$ where
$$fracpartial f_+partial x = y-xsqrtfrac1-y^21-x^2,quad fracpartial f_+partial y=x-ysqrtfrac1-x^21-y^2.$$ If one substitutes $x=cosalpha,y=cosbeta$ over the ranges $0leq alpha,betaleq pi$, then the result may be placed in the form

$$S = 2int_0^piint_0^pi sqrtsin^2(alpha) sin^2(alpha -beta )+sin^2(beta ) sin^2(alpha -beta )+sin^2(alpha) sin^2(beta) ;dalpha ,dbeta.$$
Alas, while this integral is intriguing it has defied my attempts at analytical solution (as well as those of Mathematica). Numerically, however, the integral seems to be exactly $5pi$. Can anyone show that this result is correct?