Total current through z = 4 surface

The total current through a surface is I = ∫ J·dS, with J = ∇×H (Ampere's law, static or conduction form).

H = [(x + 2y)/z²] a_y + (2/z) a_z, so H_x = 0, H_y = (x + 2y)/z², H_z = 2/z.

Step 1: a_z component of the curl (the surface z = 4 has normal a_z).

(∇×H)_z = ∂H_y/∂x − ∂H_x/∂y
        = ∂/∂x [(x + 2y)/z²] − 0
        = 1/z²

(For information, the other components are (∇×H)_x = ∂H_z/∂y − ∂H_y/∂z = 2(x+2y)/z³ and (∇×H)_y = 0, but they do not cross the plane z = 4.)

Step 2: integrate over the surface z = 4, 1 < x < 5, 3 < y < 5, dS = dx dy a_z.

I = ∫_{y=3}^{5} ∫_{x=1}^{5} (1/z²) dx dy   at z = 4
  = (1/16) × (5 − 1) × (5 − 3)
  = 8/16
  = 0.5 A

Check with Stokes' theorem: go round the rectangle at z = 4 (counter-clockwise seen from +z). H_x = 0, so only the edges x = 5 (going +y) and x = 1 (going −y) contribute.

x = 5 edge: ∫₃⁵ (5 + 2y)/16 dy = 26/16 = 1.625
x = 1 edge: ∫₃⁵ (1 + 2y)/16 dy = 18/16 = 1.125
I = 1.625 − 1.125 = 0.5 A

This matches.

Answer: I = 0.5 A in the a_z direction.