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Stokes' theorem and line integral of H

Short answer

Stokes' theorem equates a closed line integral to the surface integral of the curl. For the given H and rectangular path, the closed line integral evaluates to −76 A.

State and derive the expression for Stokes' theorem. Evaluate the closed line integral of H from P1(5, 4, 1) to P2(5, 6, 1) to P3(0, 6, 1) to P4(0, 4, 1) to P1 using straight line segments, if H = 0.1y³ a_x + 0.4x a_z A/m.

Stokes' theorem (statement)

The line integral of H around a closed path equals the surface integral of ∇ × H over any surface bounded by the path: ∮ H · dl = ∫ (∇ × H) · dS.

Derivation

  1. Divide the surface into small patches ΔS_k. For a small rectangle Δx × Δy in the xy-plane, adding the four sides gives 
    ∮ H · dl ≈ H_x(y)Δx + H_y(x+Δx)Δy − H_x(y+Δy)Δx − H_y(x)Δy
         = (∂H_y/∂x − ∂H_x/∂y) Δx Δy = (∇ × H)_z ΔS
    So for any small patch, ∮ H · dl = (∇ × H) · ΔS (the same holds for the other two components).
  2. Sum over all patches. The common edge between two adjacent patches is traversed in opposite directions, so interior contributions cancel, leaving only the outer boundary.
  3. In the limit of ΔS → 0: ∮_L H · dl = ∫_S (∇ × H) · dS.

Numerical: H = 0.1y³ a_x + 0.4x a_z A/m

Path: P1(5, 4, 1) → P2(5, 6, 1) → P3(0, 6, 1) → P4(0, 4, 1) → P1. All points lie in the plane z = 1, so dz = 0 and the a_z term of H does not contribute.

Segment 1: P1→P2, x = 5, z = 1: dl = dy a_y.  H · dl = H_y dy = 0   →  0
Segment 2: P2→P3, y = 6, z = 1: dl = dx a_x.  H_x = 0.1(6)³ = 21.6
           ∫ H_x dx from x = 5 to 0 = 21.6 (0 − 5) = −108 A
Segment 3: P3→P4, x = 0: dl = dy a_y.  H · dl = 0   →  0
Segment 4: P4→P1, y = 4, z = 1: dl = dx a_x.  H_x = 0.1(4)³ = 6.4
           ∫ H_x dx from x = 0 to 5 = 6.4 (5) = +32 A

∮ H · dl = 0 − 108 + 0 + 32 = −76 A

Check using Stokes' theorem:

∇ × H = (∂H_z/∂y − ∂H_y/∂z) a_x + (∂H_x/∂z − ∂H_z/∂x) a_y + (∂H_y/∂x − ∂H_x/∂y) a_z
      = −0.4 a_y − 0.3y² a_z

The path goes counter-clockwise in the xy-plane, so dS = dx dy a_z:

∫ (∇ × H) · dS = ∫₀⁵ ∫₄⁶ (−0.3 y²) dy dx = −0.3 (5)(6³ − 4³)/3 = −0.3(5)(50.67) = −76 A

Answer: ∮ H · dl = −76 A

✓ Answer checked by Claude Opus 5.5 (high effort)

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