
Arturo O. answered 04/17/17
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Start with Gauss' law. For the Gaussian surface, use a hollow cylinder of radius r (a < r < b) centered at the same axis. (It looks like you tried to use a sphere for the Gaussian surface, which will not work here. You need to use a cylindrical surface.)
EA = Q/ε0
A = surface area of the Gaussian surface cylinder without the end surfaces (radial electric field)
Let L = length of the cylinder. It is actually infinity, but you will find that L drops out of the math later.
A = 2πrL
Q = (the charge on the line) + (the charge enclosed inside the shell)
Q = λL + ρV = λL + ρ[π(r2 - a2)L],
where V = π(r2 - a2)L is the volume of the shell with charge in its interior over a radius range from a to r.
Put this all together in Gauss' law and get
E(r)(2πrL) = (1/ε0){λL + ρ[π(r2 - a2)L]}
Note that L cancels out, so we do not have to worry about an infinity. Then
E(r) = [1/(2πε0r)] [λ + ρπ(r2 - a2)]
Be sure to carefully check my math.