Verify Stokes' theorem, conical surface

Given: H = (3r²/sinθ) a_θ + 54 r cosθ a_φ A/m on the cone θ = 20°, 0 ≤ φ ≤ 2π, 0 ≤ r ≤ 5, with dS along a_θ.

Stokes' theorem: ∮ H · dl = ∫ (∇ × H) · dS. We evaluate both sides.

        z
        │     cone θ = 20°
        │    ╱│╲
        │   ╱ │ ╲   edge: circle r = 5 at θ = 20°
        │  ╱  │  ╲  radius = 5 sin20° = 1.710
        ● ─────────

Right side: surface integral

With H_r = 0, H_θ = 3r²/sinθ, H_φ = 54 r cosθ, the a_θ component of the curl is

(∇ × H)_θ = (1/r) [ (1/sinθ) ∂H_r/∂φ − ∂(r H_φ)/∂r ]
          = −(1/r) ∂(54 r² cosθ)/∂r = −108 cosθ

On the cone dS = r sinθ dr dφ a_θ:

∫ (∇ × H) · dS = ∫₀^(2π) ∫₀⁵ (−108 cosθ)(r sinθ) dr dφ
               = −108 cos20° sin20° (25/2)(2π)
               = −108 (0.32140)(12.5)(6.2832) = −2726.2 A

(Exact form: −1350π sin40° = −2726.16 A.)

Left side: line integral

The surface is bounded by the circle r = 5, θ = 20°. The edge r = 0 is a single point and contributes nothing.

By the right-hand rule, with dS along a_θ, the direction of travel along the rim (where the outward in-surface direction is a_r) is a_θ × a_r = −a_φ. So on the rim dl = −r sinθ dφ a_φ, and H_φ = 54 r cosθ:

∮ H · dl = −∫₀^(2π) 54 (5) cos20° · (5 sin20°) dφ
         = −54 (25) cos20° sin20° (2π)
         = −54 (25)(0.32140)(6.2832) = −2726.2 A

Answer: Both sides = −2726 A (= −1350π sin40°). Stokes' theorem is verified. (The negative sign arises because, with dS along a_θ, the rim must be traversed in the −φ sense.)