Find how the volume density of the elastic deformation energy is distributed in a steel rod depending on the distance r from its axis. The length of the rod is equal to l, the torsion angle to `varphi`.

Explanation

The energy between radii r and `r+dr` is, by differentiation, `(pir^3dr)/(l)Gvarphi^2`
Its density is `(pir^3dr)/(2pirdrl)(Gvarphi^2)/(l)=1/2(Gvarphi^2r^2)/(l^2)`