State and prove Stokes' theorem

Statement: The line integral of a vector field A around a closed path L equals the surface integral of the curl of A over any open surface S bounded by L.

∮_L A · dl = ∫_S (∇ × A) · dS

The direction of dS (normal to S) is related to the direction of travel along L by the right-hand rule.

Proof

        ┌────┬────┬────┐   S is divided into many
        │ ΔS │ ΔS │ ΔS │   small patches ΔS_k.
        ├────┼────┼────┤   Each patch has boundary
        │ ΔS │ ΔS │ ΔS │   ΔL_k, traversed in the
        └────┴────┴────┘   same sense as L.

1. Divide the open surface S into a very large number of small patches, each of area ΔS_k, bounded by a small closed path ΔL_k, all traversed in the same (counter-clockwise) sense.
2. For one small rectangular patch Δx × Δy in the xy-plane, the circulation of A around it (the four sides) is

   ∮ A · dl ≈ A_x(y)Δx + A_y(x+Δx)Δy − A_x(y+Δy)Δx − A_y(x)Δy
            = (∂A_y/∂x − ∂A_x/∂y) Δx Δy
            = (∇ × A)_z ΔS
   

   So for any small patch: ∮ A · dl = (∇ × A) · ΔS_k.
3. Add this over all patches. On every interior common edge, the two neighbouring patches traverse the edge in opposite directions, so the contributions cancel. Only the edges lying on the outer boundary L survive:

   Σ_k ∮_ΔL_k A · dl = ∮_L A · dl
   

4. In the limit ΔS_k → 0 the sum on the right-hand side becomes an integral:

   ∮_L A · dl = Σ_k (∇ × A) · ΔS_k  →  ∫_S (∇ × A) · dS
   

Hence ∮ A · dl = ∫ (∇ × A) · dS. Proved.

Note: The theorem converts a line integral into a surface integral (and vice versa). It holds for any surface bounded by L, so the result does not depend on the shape of S. It is used to derive Maxwell's equation ∮ H · dl = I from ∇ × H = J.