Undergraduate Mathematics ← Line integral Green's theorem Stoke's theorem →

In mathematics, Green's theorem gives the relationship between a line integral around a simple closed curve C and a double integral over the plane region D bounded by C. It is the two-dimensional special case of the more general Stokes' theorem, and is named after British mathematician George Green.

Let C be a positively oriented, piecewise smooth, simple closed curve in the plane $\mathbb {R}$ 2, and let D be the region bounded by C. If L and M are functions of (x, y) defined on an open region containing D and have continuous partial derivatives there, then

$\oint _{C}(L\,\mathrm {d} x+M\,\mathrm {d} y)=\iint _{D}\left({\frac {\partial M}{\partial x}}-{\frac {\partial L}{\partial y}}\right)\,\mathrm {d} x\,\mathrm {d} y=\int \int _{D}(\nabla \times )\cdot kdA$ For positive orientation, an arrow pointing in the counterclockwise direction may be drawn in the small circle in the integral symbol.

In physics, Green's theorem is mostly used to solve two-dimensional flow integrals, stating that the sum of fluid outflows at any point inside a volume is equal to the total outflow summed about an enclosing area. In plane geometry, and in particular, area surveying, Green's theorem can be used to determine the area and centroid of plane figures solely by integrating over the perimeter.

Proof when D is a simple region If D is a simple region with its boundary consisting of the curves C1, C2, C3, C4, Green's theorem can be demonstrated.

The following is a proof of the theorem for the simplified area D, a type I region where C2 and C4 are vertical lines. A similar proof exists for when D is a type II region where C1 and C3 are straight lines. The general case can be deduced from this special case by approximating the domain D by a union of simple domains.

If it can be shown that

$\int _{C}L\,dx=\iint _{D}\left(-{\frac {\partial L}{\partial y}}\right)\,dA\qquad \mathrm {(1)}$ and

$\int _{C}M\,dy=\iint _{D}\left({\frac {\partial M}{\partial x}}\right)\,dA\qquad \mathrm {(2)}$ are true, then Green's theorem is proven in the first case.

Define the type I region D as pictured on the right by:

$D=\{(x,y)|a\leq x\leq b,g_{1}(x)\leq y\leq g_{2}(x)\}$ where g1 and g2 are continuous functions on [a, b]. Compute the double integral in (1):

{\begin{aligned}\iint _{D}{\frac {\partial L}{\partial y}}\,dA&=\int _{a}^{b}\,\int _{g_{1}(x)}^{g_{2}(x)}{\frac {\partial L}{\partial y}}(x,y)\,dy\,dx\\&=\int _{a}^{b}{\Big \{}L(x,g_{2}(x))-L(x,g_{1}(x)){\Big \}}\,dx\qquad \mathrm {(3)} \end{aligned}} Now compute the line integral in (1). C can be rewritten as the union of four curves: C1, C2, C3, C4.

With C1, use the parametric equations: x = x, y = g1(x), axb. Then

$\int _{C_{1}}L(x,y)\,dx=\int _{a}^{b}L(x,g_{1}(x))\,dx$ With C3, use the parametric equations: x = x, y = g2(x), axb. Then

$\int _{C_{3}}L(x,y)\,dx=-\int _{-C_{3}}L(x,y)\,dx=-\int _{a}^{b}L(x,g_{2}(x))\,dx$ The integral over C3 is negated because it goes in the negative direction from b to a, as C is oriented positively (counterclockwise). On C2 and C4, x remains constant, meaning

$\int _{C_{4}}L(x,y)\,dx=\int _{C_{2}}L(x,y)\,dx=0$ Therefore,

{\begin{aligned}\int _{C}L\,dx&=\int _{C_{1}}L(x,y)\,dx+\int _{C_{2}}L(x,y)\,dx+\int _{C_{3}}L(x,y)\,dx+\int _{C_{4}}L(x,y)\,dx\\&=-\int _{a}^{b}L(x,g_{2}(x))\,dx+\int _{a}^{b}L(x,g_{1}(x))\,dx\qquad \mathrm {(4)} \end{aligned}} Combining (3) with (4), we get (1). Similar computations give (2).

Relationship to the Stokes theorem

Green's theorem is a special case of Stokes' theorem, when applied to a region on the xy-plane:

We can augment the two-dimensional field into a three-dimensional field with a z-component that is always 0. Start with the left side of Green's theorem.

$\mathbf {F} =\oint _{C}(L\,dx+M\,dy)=\oint _{C}(L,M,0)\cdot (dx,dy,dz)=\oint _{C}\mathbf {F} \cdot d\mathbf {r}$ Then by Stokes' Theorem:

$\oint _{C}\mathbf {F} \cdot d\mathbf {r} =\iint _{S}\nabla \times \mathbf {F} \cdot \mathbf {\hat {n}} \,dS$ The surface $S$ is just the region in the plane $D$ , with the unit normals pointing up (in +z direction) to match the "positive orientation" definitions for both theorems.

The expression inside the integral becomes

$\nabla \times \mathbf {F} \cdot \mathbf {\hat {n}} =\left[\left({\frac {\partial 0}{\partial y}}-{\frac {\partial M}{\partial z}}\right)\mathbf {i} +\left({\frac {\partial L}{\partial z}}-{\frac {\partial 0}{\partial x}}\right)\mathbf {j} +\left({\frac {\partial M}{\partial x}}-{\frac {\partial L}{\partial y}}\right)\mathbf {k} \right]\cdot \mathbf {k} =\left({\frac {\partial M}{\partial x}}-{\frac {\partial L}{\partial y}}\right)$ Thus we get the right side of Green's theorem

$\iint _{S}\nabla \times \mathbf {F} \cdot \mathbf {\hat {n}} \,dS=\iint _{D}\left({\frac {\partial M}{\partial x}}-{\frac {\partial L}{\partial y}}\right)\,dA$ Relationship to the divergence theorem

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

$\iint _{D}\left(\nabla \cdot \mathbf {F} \right)dA=\oint _{C}\mathbf {F} \cdot \mathbf {\hat {n}} \,ds,$ where $\mathbf {\hat {n}}$ is the outward-pointing unit normal vector on the boundary.

To see this, consider the unit normal in the right side of the equation. Since $d\mathbf {r} =(dx,dy)$ is a vector pointing tangential along a curve, and the curve C is the positively-oriented (i.e. counterclockwise) curve along the boundary, an outward normal would be a vector which points 90° to the right, which would be $(dy,-dx)$ . The length of this vector is ${\sqrt {dx^{2}+dy^{2}}}=ds$ . So $\mathbf {\hat {n}} \,ds=(dy,-dx).$ Now let the components of $\mathbf {F} =(P,Q)$ . Then the right hand side becomes

$\int _{C}\mathbf {F} \cdot \mathbf {\hat {n}} \,ds=\int _{C}Pdy-Qdx$ which by Green's theorem becomes

$\int _{C}-Qdx+Pdy=\iint _{D}\left({\frac {\partial P}{\partial x}}+{\frac {\partial Q}{\partial y}}\right)\,dA=\iint _{D}\left(\nabla \cdot \mathbf {F} \right)dA.$ Area Calculation

Green's theorem can be used to compute area by line integral. The area of D is given by:

$A=\iint _{D}\mathrm {d} A.$ Provided we choose L and M such that:

${\frac {\partial M}{\partial x}}-{\frac {\partial L}{\partial y}}=1$ Then the area is given by:

$A=\oint _{C}(L\,\mathrm {d} x+M\,\mathrm {d} y)$ Possible formulas for the area of D include:

$A=\oint _{C}x\,\mathrm {d} y=-\oint _{C}y\,\mathrm {d} x={\tfrac {1}{2}}\oint _{C}(x\,\mathrm {d} y-y\,\mathrm {d} x)$ Reference

1. a b Stewart, James. Calculus (6th ed.). Thomson, Brooks/Cole.