# What is the surface area of the 3-dimensional elliptope?

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?