Green's theorem

Fundamental theorem
Limits of functions
Continuity
Mean value theorem

Differential calculus
Derivative Change of variables Implicit differentiation Taylor's theorem Related rates Rules and identities: Power rule, Product rule, Quotient rule, Chain rule

Integral calculus
Integral Lists of integrals Improper integrals Integration by: parts, disks, cylindrical shells, substitution, trigonometric substitution, partial fractions, changing order

Vector calculus
Gradient Divergence Curl Laplacian Gradient theorem Green's theorem Stokes' theorem Divergence theorem

Multivariable calculus
Matrix calculus Partial derivative Multiple integral Line integral Surface integral Volume integral Jacobian

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

$\oint_{C} (L\, \mathrm{d}x %2B M\, \mathrm{d}y) = \iint_{D} \left(\frac{\partial M}{\partial x} - \frac{\partial L}{\partial y}\right)\, \mathrm{d}x\, \mathrm{d}y.$

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.

1 Proof when D is a simple region
2 Relationship to the Stokes theorem
3 Relationship to the divergence theorem
4 Area Calculation
5 See also
6 Reference
7 External links

Proof when D is a simple region

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:

$D = \{(x,y)|a\le x\le b, g_1(x) \le y \le g_2(x)\}$

where g₁ and g₂ are continuous functions on [a, b]. Compute the double integral in (1):

$\begin{align} \iint_D \frac{\partial L}{\partial y}\, dA & =\int_a^b\,\int_{g_1(x)}^{g_2(x)} \frac{\partial L}{\partial y} (x,y)\,dy\,dx \\ & = \int_a^b \Big\{L(x,g_2(x)) - L(x,g_1(x)) \Big\} \, dx\qquad\mathrm{(3)} \end{align}$

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), a ≤ x ≤ b. 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), a ≤ x ≤ b. 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,

$\begin{align} \int_{C} L\, dx & = \int_{C_1} L(x,y)\, dx %2B \int_{C_2} L(x,y)\, dx %2B \int_{C_3} L(x,y)\, dx %2B \int_{C_4} L(x,y)\, dx \\ & = -\int_a^b L(x,g_2(x))\, dx %2B \int_a^b L(x,g_1(x))\, dx\qquad\mathrm{(4)} \end{align}$

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 in the 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 $\mathbf{F}=(L,M,0)$ . Start with the left side of Green's theorem:

$\oint_{C} (L\, dx %2B M\, dy) = \oint_{C} (L, M, 0) \cdot (dx, dy, dz) = \oint_{C} \mathbf{F} \cdot d\mathbf{r}$

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 $\mathbf{\hat n}$ pointing up (in the positive z direction) to match the "positive orientation" definitions for both theorems.

The expression inside the integral becomes

$\nabla \times \mathbf{F} \cdot \mathbf{\hat n} = \left[ \left(\frac{\partial 0}{\partial y} - \frac{\partial M}{\partial z}\right) \mathbf{i} %2B \left(\frac{\partial L}{\partial z} - \frac{\partial 0}{\partial x}\right) \mathbf{j} %2B \left(\frac{\partial M}{\partial x} - \frac{\partial L}{\partial y}\right) \mathbf{k} \right] \cdot \mathbf{k} = \left(\frac{\partial M}{\partial x} - \frac{\partial L}{\partial y}\right)$

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 $\sqrt{dx^2 %2B dy^2} = ds$ . So $\mathbf{\hat n}\,ds = ( dy, -dx).$

Now let the components of $\mathbf{F} = ( P, Q)$ . Then the right hand side becomes

$\int_C \mathbf{F} \cdot \mathbf{\hat n} \, ds = \int_C P dy - Q dx$

which by Green's theorem becomes

$\int_C -Q dx %2B P dy = \iint_{D} \left(\frac{\partial P}{\partial x} %2B \frac{\partial Q}{\partial y}\right)\, dA = \iint_D\left(\nabla\cdot\mathbf{F}\right)dA.$