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Relationship to the divergence theorem

Considering only two-dimensional vector fields, Green's theorem is equivalent to the two-dimensional version of the divergence theorem:

where ∇ ⋅ F {\displaystyle \nabla \cdot \mathbf {F} } is the divergence on the two-dimensional vector field F {\displaystyle \mathbf {F} }, and n ^ {\displaystyle \mathbf {\hat {n}} } is the outward-pointing unit normal vector on the boundary.

To see this, consider the unit normal n ^ {\displaystyle \mathbf {\hat {n}} } in the right side of the equation. Since in Green's theorem d r = (d x, d y) {\displaystyle d\mathbf {r} =(dx,dy)} is a vector pointing tangential along the curve, and the curve C is the positively oriented (i.e. anticlockwise) curve along the boundary, an outward normal would be a vector which points 90° to the right of this; one choice would be (d y, − d x) {\displaystyle (dy,-dx)}. The length of this vector is d x 2 + d y 2 = d s. {\textstyle {\sqrt {dx^{2}+dy^{2}}}=ds.} So (d y, − d x) = n ^ d s. {\displaystyle (dy,-dx)=\mathbf {\hat {n}} \,ds.}

Start with the left side of Green's theorem: ∮ C (L d x + M d y) = ∮ C (M, − L) ⋅ (d y, − d x) = ∮ C (M, − L) ⋅ n ^ d s. {\displaystyle \oint _{C}(L\,dx+M\,dy)=\oint _{C}(M,-L)\cdot (dy,-dx)=\oint _{C}(M,-L)\cdot \mathbf {\hat {n}} \,ds.} Applying the two-dimensional divergence theorem with F = (M, − L) {\displaystyle \mathbf {F} =(M,-L)}, we get the right side of Green's theorem: ∮ C (M, − L) ⋅ n ^ d s = ∬ D (∇ ⋅ (M, − L)) d A = ∬ D (∂ M ∂ x − ∂ L ∂ y) d A. {\displaystyle \oint _{C}(M,-L)\cdot \mathbf {\hat {n}} \,ds=\iint _{D}\left(\nabla \cdot (M,-L)\right)\,dA=\iint _{D}\left({\frac {\partial M}{\partial x}}-{\frac {\partial L}{\partial y}}\right)\,dA.}