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

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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Let $\mathcal{S}$ denote the value of the following double integral:

$$\begin{align} \mathcal{S} &:=2\int_{0}^{\pi}\mathrm{d}\alpha\int_{0}^{\pi}\mathrm{d}\beta\,\sqrt{\sin^{2}{\left(\alpha\right)}\sin^{2}{\left(\beta\right)}+\sin^{2}{\left(\alpha\right)}\sin^{2}{\left(\alpha-\beta\right)}+\sin^{2}{\left(\beta\right)}\sin^{2}{\left(\alpha-\beta\right)}}.\\ \end{align}$$

This double integral does indeed have the exact value $\mathcal{S}=5\pi$, as we shall demonstrate below.


We begin by applying the tangent half-angle substitutions $\tan{\left(\frac{\alpha}{2}\right)}=x$ and $\tan{\left(\frac{\beta}{2}\right)}=y$, which transforms the integral into

$$\begin{align} \mathcal{S} &=2\int_{0}^{\pi}\mathrm{d}\alpha\int_{0}^{\pi}\mathrm{d}\beta\,\sqrt{\sin^{2}{\left(\alpha\right)}\sin^{2}{\left(\beta\right)}+\sin^{2}{\left(\alpha\right)}\sin^{2}{\left(\alpha-\beta\right)}+\sin^{2}{\left(\beta\right)}\sin^{2}{\left(\alpha-\beta\right)}}\\ &=2\int_{0}^{\pi}\mathrm{d}\alpha\int_{0}^{\pi}\mathrm{d}\beta\,\sqrt{\sin^{2}{\left(\alpha\right)}\sin^{2}{\left(\beta\right)}+\left[\sin^{2}{\left(\alpha\right)}+\sin^{2}{\left(\beta\right)}\right]\sin^{2}{\left(\alpha-\beta\right)}}\\ &=\small{\int_{0}^{\infty}\mathrm{d}x\int_{0}^{\infty}\mathrm{d}y\,\frac{2^{5}\sqrt{x^{2}y^{2}\left(1+x^{2}\right)^{2}\left(1+y^{2}\right)^{2}+\left(x-y\right)^{2}\left(1+xy\right)^{2}\left[x^{2}\left(1+y^{2}\right)^{2}+y^{2}\left(1+x^{2}\right)^{2}\right]}}{\left(1+x^{2}\right)^{3}\left(1+y^{2}\right)^{3}}}.\\ \end{align}$$

In this form, taking the single-variable integral with respect to either of the variables results in a hyperelliptic integral. In fact, the hyperelliptic integral form for $\mathcal{S}$ can be reduced to a double elliptic integral with just a few more simple substitutions:

$$\begin{align} \mathcal{S} &=\small{\int_{0}^{\infty}\mathrm{d}x\int_{0}^{\infty}\mathrm{d}y\,\frac{2^{5}\sqrt{x^{2}y^{2}\left(1+x^{2}\right)^{2}\left(1+y^{2}\right)^{2}+\left(x-y\right)^{2}\left(1+xy\right)^{2}\left[x^{2}\left(1+y^{2}\right)^{2}+y^{2}\left(1+x^{2}\right)^{2}\right]}}{\left(1+x^{2}\right)^{3}\left(1+y^{2}\right)^{3}}}\\ &=\int_{0}^{\infty}\mathrm{d}x\int_{0}^{\infty}\mathrm{d}t\,\frac{2^{5}x^{3}}{\left(1+x^{2}\right)^{3}\left(1+x^{2}t^{2}\right)^{3}}\\ &~~~~~\times\sqrt{t^{2}\left(1+x^{2}\right)^{2}\left(1+x^{2}t^{2}\right)^{2}+\left(1-t\right)^{2}\left(1+x^{2}t\right)^{2}\left[\left(1+x^{2}t^{2}\right)^{2}+t^{2}\left(1+x^{2}\right)^{2}\right]};~~~\small{\left[y=xt\right]}\\ &=\int_{0}^{\infty}\mathrm{d}u\int_{0}^{\infty}\mathrm{d}t\,\frac{2^{4}u}{\left(1+u\right)^{3}\left(1+ut^{2}\right)^{3}}\\ &~~~~~\times\sqrt{t^{2}\left(1+u\right)^{2}\left(1+ut^{2}\right)^{2}+\left(1-t\right)^{2}\left(1+ut\right)^{2}\left[\left(1+ut^{2}\right)^{2}+t^{2}\left(1+u\right)^{2}\right]};~~~\small{\left[x=\sqrt{u}\right]}\\ &=\int_{0}^{\infty}\mathrm{d}t\int_{0}^{\infty}\mathrm{d}u\,\frac{2^{4}t^{3}u}{\left(1+t^{2}u\right)^{3}\left(t+tu\right)^{3}}\\ &~~~~~\times\sqrt{\left(1+t^{2}u\right)^{2}\left(t+tu\right)^{2}+\left(1-t\right)^{2}\left(1+tu\right)^{2}\left[\left(1+t^{2}u\right)^{2}+\left(t+tu\right)^{2}\right]}\\ &=\int_{0}^{\infty}\mathrm{d}t\int_{0}^{\infty}\mathrm{d}v\,\frac{2^{4}tv}{\left(1+tv\right)^{3}\left(t+v\right)^{3}}\\ &~~~~~\times\sqrt{\left(1+tv\right)^{2}\left(t+v\right)^{2}+\left(1-t\right)^{2}\left(1+v\right)^{2}\left[\left(1+tv\right)^{2}+\left(t+v\right)^{2}\right]};~~~\small{\left[u=\frac{v}{t}\right]}\\ &=4\int_{-1}^{1}\mathrm{d}x\int_{-1}^{1}\mathrm{d}y\,\frac{\left(1-x^{2}\right)\left(1-y^{2}\right)\sqrt{\left(1-x^{2}y^{2}\right)^{2}+8x^{2}\left(1+x^{2}y^{2}\right)}}{\left(1-x^{2}y^{2}\right)^{3}};~~~\small{\left[\left(t,u\right)=\left(\frac{1-x}{1+x},\frac{1-y}{1+y}\right)\right]}\\ &=16\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}y\,\frac{\left(1-x^{2}\right)\left(1-y^{2}\right)\sqrt{\left(1-x^{2}y^{2}\right)^{2}+8x^{2}\left(1+x^{2}y^{2}\right)}}{\left(1-x^{2}y^{2}\right)^{3}}.\\ \end{align}$$

Then, the integral expression for $\mathcal{S}$ given in the last line above can be further reduced to a double integral in which at least the first integral is elementary as follows:

$$\begin{align} \mathcal{S} &=16\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}y\,\frac{\left(1-x^{2}\right)\left(1-y^{2}\right)\sqrt{\left(1-x^{2}y^{2}\right)^{2}+8x^{2}\left(1+x^{2}y^{2}\right)}}{\left(1-x^{2}y^{2}\right)^{3}}\\ &=16\int_{0}^{1}\mathrm{d}x\int_{0}^{x}\mathrm{d}t\,\frac{\left(1-x^{2}\right)\left(x^{2}-t^{2}\right)\sqrt{\left(1-t^{2}\right)^{2}+8x^{2}\left(1+t^{2}\right)}}{x^{3}\left(1-t^{2}\right)^{3}};~~~\small{\left[y=\frac{t}{x}\right]}\\ &=16\int_{0}^{1}\mathrm{d}t\int_{t}^{1}\mathrm{d}x\,\frac{\left(1-x^{2}\right)\left(x^{2}-t^{2}\right)\sqrt{\left(1-t^{2}\right)^{2}+8x^{2}\left(1+t^{2}\right)}}{x^{3}\left(1-t^{2}\right)^{3}}\\ &=8\int_{0}^{1}\mathrm{d}t\int_{t^{2}}^{1}\mathrm{d}u\,\frac{\left(1-u\right)\left(u-t^{2}\right)\sqrt{\left(1-t^{2}\right)^{2}+8\left(1+t^{2}\right)u}}{\left(1-t^{2}\right)^{3}u^{2}};~~~\small{\left[x=\sqrt{u}\right]}\\ &=\int_{0}^{1}\mathrm{d}w\,\frac{4}{\sqrt{w}}\int_{w}^{1}\mathrm{d}u\,\frac{\left(1-u\right)\left(u-w\right)\sqrt{\left(1-w\right)^{2}+8\left(1+w\right)u}}{\left(1-w\right)^{3}u^{2}};~~~\small{\left[t=\sqrt{w}\right]}\\ &=\int_{0}^{1}\mathrm{d}x\,\frac{2}{\sqrt{1-x^{2}}}\int_{\frac{1-x}{1+x}}^{1}\mathrm{d}u\,\frac{\left(1-u\right)\left(u-\frac{1-x}{1+x}\right)\left(1+x\right)\sqrt{x^{2}+4\left(1+x\right)u}}{x^{3}u^{2}};~~~\small{\left[w=\frac{1-x}{1+x}\right]}\\ &=\int_{0}^{1}\mathrm{d}x\,\frac{1}{\sqrt{1-x^{2}}}\int_{\frac{1-x}{1+x}}^{1}\mathrm{d}u\,\frac{2\left(1-u\right)\left[\left(1+x\right)u-\left(1-x\right)\right]\sqrt{x^{2}+4\left(1+x\right)u}}{x^{3}u^{2}}.\\ \end{align}$$


For convenience let's introduce an auxiliary function $F:\left(0,1\right)\rightarrow\mathbb{R}$ via the definite integral

$$\begin{align} F{\left(x\right)} &:=\int_{\frac{1-x}{1+x}}^{1}\mathrm{d}u\,\frac{2\left(1-u\right)\left[\left(1+x\right)u-\left(1-x\right)\right]\sqrt{x^{2}+4\left(1+x\right)u}}{x^{3}u^{2}}.\\ \end{align}$$

Then, for $x\in\left(0,1\right)$ we obtain

$$\begin{align} F{\left(x\right)} &=\int_{\frac{1-x}{1+x}}^{1}\mathrm{d}u\,\frac{2\left(1-u\right)\left[\left(1+x\right)u-\left(1-x\right)\right]\sqrt{x^{2}+4\left(1+x\right)u}}{x^{3}u^{2}}\\ &=\int_{\left(2-x\right)^{2}}^{\left(2+x\right)^{2}}\mathrm{d}v\,\frac{\left[4\left(1+x\right)-\left(v-x^{2}\right)\right]\left[\left(v-x^{2}\right)-4\left(1-x\right)\right]\sqrt{v}}{2x^{3}\left(v-x^{2}\right)^{2}};~~~\small{\left[x^{2}+4\left(1+x\right)u=v\right]}\\ &=\int_{\left(2-x\right)^{2}}^{\left(2+x\right)^{2}}\mathrm{d}v\,\frac{\left[\left(2+x\right)^{2}-v\right]\left[v-\left(2-x\right)^{2}\right]\sqrt{v}}{2x^{3}\left(v-x^{2}\right)^{2}}\\ &=\int_{2-x}^{2+x}\mathrm{d}w\,\frac{\left[\left(2+x\right)^{2}-w^{2}\right]\left[w^{2}-\left(2-x\right)^{2}\right]w^{2}}{x^{3}\left(w^{2}-x^{2}\right)^{2}};~~~\small{\left[\sqrt{v}=w\right]}\\ &=\frac{1}{x^{3}}\int_{2-x}^{2+x}\mathrm{d}w\,\left[8-w^{2}+\frac{8\left(3x^{2}-2\right)}{\left(w^{2}-x^{2}\right)}-\frac{16x^{2}\left(1-x^{2}\right)}{\left(w^{2}-x^{2}\right)^{2}}\right]\\ &=\frac{1}{x^{3}}\int_{2-x}^{2+x}\mathrm{d}w\,\frac{\mathrm{d}}{\mathrm{d}w}\bigg{[}8w-\frac13w^{3}+\frac{8\left(1-x^{2}\right)w}{\left(w^{2}-x^{2}\right)}+\frac{4\left(2x^{2}-1\right)}{x}\ln{\left(\frac{w-x}{w+x}\right)}\bigg{]}\\ &=\frac{1}{x^{3}}\left[4x-\frac{2x^{3}}{3}-\frac{4\left(2x^{2}-1\right)}{x}\ln{\left(1-x^{2}\right)}\right]\\ &=\frac23\left[\frac{6}{x^{2}}-1-\frac{6\left(2x^{2}-1\right)}{x^{4}}\ln{\left(1-x^{2}\right)}\right].\\ \end{align}$$

Finally, $\mathcal{S}$ can be reduced to the single-variable (elementary) integral

$$\begin{align} \mathcal{S} &=\int_{0}^{1}\mathrm{d}x\,\frac{1}{\sqrt{1-x^{2}}}\int_{\frac{1-x}{1+x}}^{1}\mathrm{d}u\,\frac{2\left(1-u\right)\left[\left(1+x\right)u-\left(1-x\right)\right]\sqrt{x^{2}+4\left(1+x\right)u}}{x^{3}u^{2}}\\ &=\int_{0}^{1}\mathrm{d}x\,\frac{F{\left(x\right)}}{\sqrt{1-x^{2}}}\\ &=\frac23\int_{0}^{1}\mathrm{d}x\,\frac{1}{\sqrt{1-x^{2}}}\left[\frac{6}{x^{2}}-1-\frac{6\left(2x^{2}-1\right)}{x^{4}}\ln{\left(1-x^{2}\right)}\right]\\ &=\int_{0}^{1}\mathrm{d}x\,\frac{\mathrm{d}}{\mathrm{d}x}\bigg{[}10\arcsin{\left(x\right)}+\frac{4\sqrt{1-x^{2}}\left[\left(4x^{2}-1\right)\ln{\left(1-x^{2}\right)}-x^{2}\right]}{3x^{3}}\bigg{]}\\ &=\lim_{x\to1}\bigg{[}10\arcsin{\left(x\right)}+\frac{4\sqrt{1-x^{2}}\left[\left(4x^{2}-1\right)\ln{\left(1-x^{2}\right)}-x^{2}\right]}{3x^{3}}\bigg{]}\\ &~~~~~-\lim_{x\to0}\bigg{[}10\arcsin{\left(x\right)}+\frac{4\sqrt{1-x^{2}}\left[\left(4x^{2}-1\right)\ln{\left(1-x^{2}\right)}-x^{2}\right]}{3x^{3}}\bigg{]}\\ &=10\arcsin{\left(1\right)}\\ &=5\pi.\blacksquare\\ \end{align}$$


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