# Electric flux density outside cylindrical surface?

What I did is that I draw a Gaussian contour with radius p>a.

The electric flux outside the cylinder is Ps/2

The electric flux inside the cylinder is PL/(2*pi*p)

Consider a length $L$ of the cylindrical surface and the line. Draw a cylindrical contour with radius $\rho>a$ and length $L$, centered about the charged cylinder and line charge.
The charge contained within the contour is equal to the linear charge density of the line of charge times its length, plus the surface charge density of the cylindrical surface times its surface area, that is: $$Q = (\rho_L)(L)+(\rho_S)(A)$$ The surface area of the charged cylinder is the circumference of the cylinder times its length: $A=2\pi a L$. Thus, the charge contained within the Gaussian contour is: $$Q = (\rho_L)(L)+(\rho_S)(2\pi a L)$$ The flux density is equal to the flux passing through the Gaussian contour divided by the area of the contour: $$|\vec{D}|=\frac{Q}{A}=\frac{(\rho_L)(L)+(\rho_S)(2\pi a L)}{2\pi\rho L}$$ Note that the length $L$ can be canceled from each term. Due to the symmetry of the problem, the electric flux acts in the radial direction: $$\vec{D}=\frac{\rho_L+2\pi a \rho_S}{2\pi\rho}\vec{a}_\rho$$