Green'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 named after George Green ^[1] and is the two-dimensional special case of the more general Kelvin–Stokes theorem.

Theorem

Let C be a positively oriented, piecewise smooth, simple closed curve in a plane, 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^[2][3]

$\oint_{C} (L\, dx + M\, dy) = \iint_{D} \left(\frac{\partial M}{\partial x} - \frac{\partial L}{\partial y}\right)\, dx\, dy$

where the path of integration along C is counterclockwise.

In physics, Green's theorem is mostly used to solve two-dimensional flow integrals, stating that the sum of fluid outflows from 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, half of Green's theorem can be demonstrated.

The following is a proof of half of the theorem for the simplified area D, a type I region where C₁ and C₃ are curves connected by vertical lines (possibly of zero length). A similar proof exists for the other half of the theorem when D is a type II region where C₂ and C₄ are curves connected by horizontal lines (again, possibly of zero length). Putting these two parts together, the theorem is thus proven for regions of type III (defined as regions which are both type I and type II). The general case can then be deduced from this special case by decomposing D into a set of type III regions.

If it can be shown that if

$\oint_{C} L\, dx = \iint_{D} \left(- \frac{\partial L}{\partial y}\right)\, dA\qquad\mathrm{(1)}$

and

$\oint_{C} M\, dy = \iint_{D} \left(\frac{\partial M}{\partial x}\right)\, dA\qquad\mathrm{(2)}$

are true, then Green's theorem follows immediately for the region D. We can prove (1) easily for regions of type I, and (2) for regions of type II. Green's theorem then follows for regions of type III.

Assume region D is a type I region and can thus be characterized, 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,

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

Combining (3) with (4), we get (1) for regions of type I. A similar treatment yields (2) for regions of type II. Putting the two together, we get the result for regions of type III.

Relationship to the Stokes theorem

Green's theorem is a special case of the Kelvin–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:

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

Kelvin–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

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

Thus we get the right side of Green's theorem

$\iint_S \nabla \times \mathbf{F} \cdot \mathbf{\hat n} \, dS = \iint_D \left(\frac{\partial M}{\partial x} - \frac{\partial L}{\partial y}\right) \, dA.$

Green's theorem is also a straightforward result of the general Stokes' theorem using differential forms and exterior derivatives:

$\oint_C L \,dx + M \,dy = \oint_{\partial D} \omega = \int_D \,d\omega = \int_D \frac{\partial L}{\partial y} \,dy \wedge \,dx + \frac{\partial M}{\partial x} \,dx \wedge \,dy = \iint_D \left(\frac{\partial M}{\partial x} - \frac{\partial L}{\partial y}\right) \,dx \,dy.$

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:

$\iint_D\left(\nabla\cdot\mathbf{F}\right)dA=\oint_C \mathbf{F} \cdot \mathbf{\hat n} \, ds,$

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

To see this, consider the unit normal $\mathbf{\hat n}$ in the right side of the equation. Since in Green's theorem $d\mathbf{r} = (dx, dy)$ is a vector pointing tangential along the 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 of this; one choice would be $(dy, -dx)$. The length of this vector is $\sqrt{dx^2 + dy^2} = ds.$ So $(dy, -dx) = \mathbf{\hat n}\,ds.$

Start with the left side of Green's theorem:

$\oint_{C} (L\, dx + M\, dy) = \oint_{C} (M, -L) \cdot (dy, -dx) = \oint_{C} (M, -L) \cdot \mathbf{\hat n}\,ds.$

Applying the two-dimensional divergence theorem with $\mathbf{F} = (M, -L)$, we get the right side of Green's theorem:

$\oint_{C} (M, -L) \cdot \mathbf{\hat n}\,ds = \iint_D\left(\nabla \cdot (M, -L)\right)dA = \iint_D \left(\frac{\partial M}{\partial x} - \frac{\partial L}{\partial y}\right) \, dA.$

Area calculation

Green's theorem can be used to compute area by line integral.^[4] The area of D is given by $A = \iint_{D}dA$. Then if we choose L and M such that $\frac{\partial M}{\partial x} - \frac{\partial L}{\partial y} = 1$, the area is given by$A = \oint_{C} (L\, dx + M\, dy)$.

Possible formulas for the area of D include:^[4] $A=\oint_{C} x\, dy = -\oint_{C} y\, dx = \tfrac 12 \oint_{C} (-y\, dx + x\, dy).$

See also

• Planimeter
• Method of image charges – A method used in electrostatics that takes advantage of the uniqueness theorem (derived from Green's theorem)
• Shoelace formula - A special case of Green's theorem for simple polygons

References

1. George Green, An Essay on the Application of Mathematical Analysis to the Theories of Electricity and Magnetism (Nottingham, England: T. Wheelhouse, 1828). Green did not actually derive the form of "Green's theorem" which appears in this article; rather, he derived a form of the "divergence theorem", which appears on pages 10-12 of his Essay.
   In 1846, the form of "Green's theorem" which appears in this article was first published, without proof, in an article by Augustin Cauchy: A. Cauchy (1846) "Sur les intégrales qui s'étendent à tous les points d'une courbe fermée" (On integrals that extend over all of the points of a closed curve), Comptes rendus, 23: 251-255. (The equation appears at the bottom of page 254, where (S) denotes the line integral of a function k along the curve s that encloses the area S.)
   A proof of the theorem was finally provided in 1851 by Bernhard Riemann in his inaugural dissertation: Bernhard Riemann (1851) Grundlagen für eine allgemeine Theorie der Functionen einer veränderlichen complexen Grösse (Basis for a general theory of functions of a variable complex quantity), (Göttingen, (Germany): Adalbert Rente, 1867); see pages 8 - 9.
2. Mathematical methods for physics and engineering, K.F. Riley, M.P. Hobson, S.J. Bence, Cambridge University Press, 2010, ISBN 978-0-521-86153-3
3. Vector Analysis (2nd Edition), M.R. Spiegel, S. Lipschutz, D. Spellman, Schaum’s Outlines, McGraw Hill (USA), 2009, ISBN 978-0-07-161545-7
4. Stewart, James. Calculus (6th ed.). Thomson, Brooks/Cole.

Further reading

• Ayres, F.; Mendelson, E. (2009). Calculus. Schaum's Outline (5th ed.). ISBN 978-0-07-150861-2.
• Wrede, R.; Spiegel, M. R. (2010). Advanced Calculus. Schaum's Outline (3rd ed.). ISBN 978-0-07-162366-7.

External links

• Green's Theorem on MathWorld