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 $$\Gamma = \{(x,y,z)\in[-1,1]^3 : x^2+y^2+z^2\leq 1+2xyz\}.$$

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 $z=f_{\pm}(x,y)=x y\pm \sqrt{1-x^2}\sqrt{1-y^2}$, 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 $$S=2\int_{-1}^1\int_{-1}^1 \sqrt{1+\left(\frac{\partial f_+}{\partial x}\right)^2+\left(\frac{\partial f_+}{\partial y}\right)^2}\,dx\,dy,$$ where $$\frac{\partial f_+}{\partial x} = y-x\sqrt{\frac{1-y^2}{1-x^2}},\quad \frac{\partial f_+}{\partial y}=x-y\sqrt{\frac{1-x^2}{1-y^2}}.$$ If one substitutes $x=\cos\alpha,y=\cos\beta$ over the ranges $0\leq \alpha,\beta\leq \pi$, then the result may be placed in the form

$$S = 2\int_{0}^\pi\int_0^\pi \sqrt{\sin^2(\alpha) \sin^2(\alpha -\beta )+\sin^2(\beta ) \sin^2(\alpha -\beta )+\sin^2(\alpha) \sin^2(\beta) }\;d\alpha \,d\beta.$$ 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 $5\pi$. Can anyone show that this result is correct?


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