This is to be handled by the same trick as in 3.96, We have effectively a two dinensional situation. For a uniform cylinder full or charge with chagre density `rho_(0)` (charge per unit volume), the electric field `vec(E)` at an inside point is along the (cyclinder) radius `vec(r)` and equal to,
`vec(E) = (1)/(2epsilon_(0)) rho vec(r)`
`(di v vec(E) = (1)/(r) (r E_(r)) = (rho)/(epsilon_(0)), hence E_(r) = (rho)/(2epsilon_(0)) r)`
Therefore the plarized cylinder can be thought of as two equal and opposite charge distributions displaced with respect with respect to each other
`vec(E) = (1)/(2epsilon_(0)) rho vec(r) = (1)/(2epsilon_(0)) rho (vec(r) - b vec(r)) = (1)/(2epsilon_(0)) rho b vec(r) = (vec(P))/(2 epsilon_(0))`
Since `vec(P) = -rho sigma vec(r)`. (direction of electric dipole moment vector being from the negative cahrge to positive charge.)