Editing Green's theorem (section)

==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]]:

:<math>\iint_D\left(\nabla\cdot\mathbf{F}\right)dA=\oint_C \mathbf{F} \cdot \mathbf{\hat n} \, ds,</math>

where <math>\nabla\cdot\mathbf{F}</math> is the divergence on the two-dimensional vector field <math>\mathbf{F}</math>, and <math>\mathbf{\hat n}</math> is the outward-pointing unit normal vector on the boundary.

To see this, consider the unit normal <math>\mathbf{\hat n}</math> in the right side of the equation. Since in Green's theorem <math>d\mathbf{r} = (dx, dy)</math> 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 <math>(dy, -dx)</math>. The length of this vector is <math display="inline">\sqrt{dx^2 + dy^2} = ds.</math> So <math>(dy, -dx) = \mathbf{\hat n}\,ds.</math>

Start with the left side of Green's theorem:
<math display="block">\oint_C (L\, dx + M\, dy) = \oint_C (M, -L) \cdot (dy, -dx) = \oint_C (M, -L) \cdot \mathbf{\hat n}\,ds.</math>
Applying the two-dimensional divergence theorem with <math>\mathbf{F} = (M, -L)</math>, we get the right side of Green's theorem:
<math display="block">\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.</math>