Expected value of distance between two points chosen at random on the edges of a unit cube Consider the probability space formed by the set $A$ of the edges of the unit cube $[0,1]^3$ in $\mathbb{R}^3$, the corresponding Borel $\sigma$- algebra, and the normalized length measure. Compute the expected value of the distance of $2$ distinct points chosen independently and uniformly in $A$.
My attempt: I know that the expected value can be calculated as 
$$\int \int \int \sqrt{(x_1-y_1)^2+(x_2-y_2)^2+(x_3-y_3)^2}\,dx\,dy\,dz. $$ But since the points are all chosen on the edges, then depending on the edge some of coordinates are fixed value, for example $0$ or $1$. But there are many scenarios for example two points are on the same edge, or two points are chosen on two jointed edges and etc. How do I calculate the total expected value?
Help please.
 A: There are five cases of symmetry.   Relative to the edge on which lies the first point, the second point may lie upon (A) the very same edge, (B) one of the four adjacent edges sharing a vertex (C) one of the two opposite edges sharing a face, (D) the opposite edge of the cube, (E) one of the remaining four edges.
So the expected distance between points $\vec P,\vec Q$ when given that they independently lie on the edges of the unit cube with uniform distribution, is: 
$$\newcommand{\vector}[3]{\left[\begin{smallmatrix}#1\\#2\\#3\end{smallmatrix}\right]}
\begin{align}
\mathsf E(\delta(\vec P,\vec Q)) ~&=~ {\quad\tfrac 1{12}\int_0^1\!\int_0^1 \delta\Bigl(\vector u 0 0,\vector v 0 0 \Bigr)\operatorname d u\operatorname d v \\ + \tfrac 4{12}\int_0^1\!\int_0^1 \delta\Bigl(\vector u 0 0 ,\vector  0 v 0 \Bigr)\operatorname d u\operatorname d v \\ +\tfrac 2{12}\int_0^1\!\int_0^1 \delta\Bigl(\vector u 0 0 ,\vector v 1 0 \Bigr)\operatorname d u\operatorname d v \\ +\tfrac 1{12}\int_0^1\!\int_0^1 \delta\Bigl(\vector u 0 0 ,\vector v 1 1 \Bigr)\operatorname d u\operatorname d v\\ +\tfrac 4{12}\int_0^1\!\int_0^1 \delta\Bigl(\vector u 0 0 ,\vector 0 v 1\Bigr)\operatorname d u\operatorname d v }\end{align}$$

Where $\delta\Bigl(\vector{x_1}{y_1}{c_1}, \vector{x_2}{y_2}{z_2}\Bigr) = \sqrt{~{(x_1-x_2)}^2+{(y_1-y_2)}^2+{(z_1-z_2)}^2~}$
