Stokes' theorem
Proof via differential forms

The functions R 3 → R 3 {\displaystyle \mathbb {R} ^{3}\to \mathbb {R} ^{3}} can be identified with the differential 1-forms on R 3 {\displaystyle \mathbb {R} ^{3}} via the map F x e 1 + F y e 2 + F z e 3 ↦ F x d x + F y d y + F z d z. {\displaystyle F_{x}\mathbf {e} _{1}+F_{y}\mathbf {e} _{2}+F_{z}\mathbf {e} _{3}\mapsto F_{x}\,\mathrm {d} x+F_{y}\,\mathrm {d} y+F_{z}\,\mathrm {d} z.}

Write the differential 1-form associated to a function F as ω. Then one can calculate that ⋆ ω ∇ × F = d ω F, {\displaystyle \star \omega _{\nabla \times \mathbf {F} }=\mathrm {d} \omega _{\mathbf {F} },} where ★ is the Hodge star and d {\displaystyle \mathrm {d} } is the exterior derivative. Thus, by generalized Stokes' theorem, ∮ ∂ Σ F ⋅ d γ = ∮ ∂ Σ ω F = ∫ Σ d ω F = ∫ Σ ⋆ ω ∇ × F = ∬ Σ ∇ × F ⋅ d Σ {\displaystyle \oint _{\partial \Sigma }{\mathbf {F} \cdot \,\mathrm {d} \mathbf {\gamma } }=\oint _{\partial \Sigma }{\omega _{\mathbf {F} }}=\int _{\Sigma }{\mathrm {d} \omega _{\mathbf {F} }}=\int _{\Sigma }{\star \omega _{\nabla \times \mathbf {F} }}=\iint _{\Sigma }{\nabla \times \mathbf {F} \cdot \,\mathrm {d} \mathbf {\Sigma } }}
combination:two_responses