Calculating integral of absolute integrand

I want to calculate the integral $$\int_0^1\int_0^1\int_0^1 {\rm d}r_1{\rm d}r_2{\rm d}r_3 \, r_1 r_2 r_3\int_0^\pi \int_0^\pi \int_0^\pi {\rm d}\phi_1{\rm d}\phi_2{\rm d}\phi_3 \\ \left| r_1r_2\sin(\phi_2-\phi_1) + r_2r_3\sin(\phi_3-\phi_2) + r_3r_1\sin(\phi_1-\phi_3)\right|$$ The problem obviously is the absolute value. I have tried for ages but don't get anywhere :-(

I tried to split the integral, but that leads to non-integrable terms, but maybe there is a trick using the symmetry?

I'm thinking to somehow transform the $$\phi$$ integrals and do them first.

The absolute value actually is two times the area of a triangle with vertices $$P_1,P_2,P_3$$ inside a semi-circle.

The transform mentioned by Yuriy seems a bit complicated:

• If I'm not mistaken (it's hard to check numerically), we can make a substitution $x=r_1 r_2$, $y=r_2 r_3$, $z=r_3 r_1$, which gives us the same ranges as before, and simplifies the integrand: $$|a r_1 r_2+br_2 r_3 +c r_3 r_1| r_1 r_2 r_3 dr_1 dr_2 dr_3 = \frac{1}{2} |a x+b y +c z| dx dy dz$$ If I found the Jacobian correctly, of course – Yuriy S Oct 14 '18 at 1:13
• @YuriyS Interesting observation. I would like to point out that the range of $x, y, z$ is a bit complicated since we must have $xy \leq z$, $yz \leq x$ and $zx \leq y$. – Sangchul Lee Oct 14 '18 at 1:40
• @SangchulLee, oh, that explains why my numerical check failed – Yuriy S Oct 14 '18 at 8:13