quality:correction, information:filtering
Relationship to Stokes' theorem

Green's theorem is a special case of the Kelvin–Stokes theorem, when applied to a region in the x y {\displaystyle xy} -plane.

We can augment the two-dimensional field into a three-dimensional field with a z component that is always 0. Write F for the vector -valued function F = (L, M, 0) {\displaystyle \mathbf {F} =(L,M,0)}. Start with the left side of Green's theorem: ∮ C (L d x + M d y) = ∮ C (L, M, 0) ⋅ (d x, d y, d z) = ∮ C F ⋅ d r. {\displaystyle \oint _{C}(L\,dx+M\,dy)=\oint _{C}(L,M,0)\cdot (dx,dy,dz)=\oint _{C}\mathbf {F} \cdot d\mathbf {r}.}

The Kelvin–Stokes theorem: ∮ C F ⋅ d r = ∬ S ∇ × F ⋅ n ^ d S. {\displaystyle \oint _{C}\mathbf {F} \cdot d\mathbf {r} =\iint _{S}\nabla \times \mathbf {F} \cdot \mathbf {\hat {n}} \,dS.}

The surface S {\displaystyle S} is just the region in the plane D {\displaystyle D}, with the unit normal n ^ {\displaystyle \mathbf {\hat {n}} } defined (by convention) to have a positive z component in order to match the "positive orientation" definitions for both theorems.

The expression inside the integral becomes ∇ × F ⋅ n ^ = ⋅ k = (∂ M ∂ x − ∂ L ∂ y). {\displaystyle \nabla \times \mathbf {F} \cdot \mathbf {\hat {n}} =\left\cdot \mathbf {k} =\left({\frac {\partial M}{\partial x}}-{\frac {\partial L}{\partial y}}\right).}

Thus we get the right side of Green's theorem ∬ S ∇ × F ⋅ n ^ d S = ∬ D (∂ M ∂ x − ∂ L ∂ y) d A. {\displaystyle \iint _{S}\nabla \times \mathbf {F} \cdot \mathbf {\hat {n}} \,dS=\iint _{D}\left({\frac {\partial M}{\partial x}}-{\frac {\partial L}{\partial y}}\right)\,dA.}

Green's theorem is also a straightforward result of the general Stokes' theorem using differential forms and exterior derivatives: ∮ C L d x + M d y = ∮ ∂ D ω = ∫ D d ω = ∫ D ∂ L ∂ y d y ∧ d x + ∂ M ∂ x d x ∧ d y = ∬ D (∂ M ∂ x − ∂ L ∂ y) d x d y. {\displaystyle \oint _{C}L\,dx+M\,dy=\oint _{\partial D}\!\omega =\int _{D}d\omega =\int _{D}{\frac {\partial L}{\partial y}}\,dy\wedge \,dx+{\frac {\partial M}{\partial x}}\,dx\wedge \,dy=\iint _{D}\left({\frac {\partial M}{\partial x}}-{\frac {\partial L}{\partial y}}\right)\,dx\,dy.}