Relationship between line current & surface current density

[QuantumDuality]

Homework Statement

In a configuration having axial symmetry about the z axis, a line current I
flows in the −z direction along the z axis. This current is returned at the
radii a and b, where there are uniform surface current densities K_za and
K_zb , respectively. The current density is zero in the regions 0 < r < b, b <
r < a and a < r.

(a) Given that K_za = 2K_zb , show that K_za= I/π(2a + b).
(b) Show that H is:
-I/2πr for 0<r<b
-Ia/πr(2a+b) for b < r < a

Homework Equations

I = lim_{|J|→∞ A→0}∫_S J ⋅da (1)
2 π r K⋅ i_n≈ ∫_S J⋅da (2)
∫_cH⋅ds = ∫_S J ⋅da (3)
K_za = 2K_zb

The Attempt at a Solution

[/B]
To get K_za, I used ∫_S J ⋅da on the region a < r.
Because there is not current density on this region, the only contribution to the integral comes form the surface current densities, using (2):

∫_S J ⋅da = 2π( b K_zb + aK_za )

Substituting the value of K_za:

∫_S J ⋅da = 2πK_zb( b + 2 a)

As the problem is talking about Line current and surface current densities, I am assuming is a tick wire, but that is length is much bigger than its thickness. Therefore, (I think) I can equate (1) and (2)

2πK_zb( b + 2 a) = I

K_zb = I/2π(2a + b)
K_za = I/π(2a + b)For answer (b), On the region r < b, the only contribution to the magnetic field comes from -I, therefore:
∫_cH⋅ds = - I
H = -I/2πr i for r < b

For the region b < r < a, the contribution to the integral should come from the surface current density for r = b and I, therefore:

H = (bK_zb/r -I/2πr) i_φ for b < r < a
H = (bI/2π(a+b)r -I/2πr) i_φ for b < r < a

I don't understand why for b < r < a K_za should be taken into account, because r doesn't necessarily have to be near a

 

[rude man]

Having a hard time seeing the setup.