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# 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, though its first proof is due to Bernhard Riemann[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 having continuous partial derivatives there, then

where the path of integration along C is anticlockwise.[2][3]

In physics, Green's theorem finds many applications. One is solving two-dimensional flow integrals, stating that the sum of fluid outflowing 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

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

and

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

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

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

With C3, use the parametric equations: x = x, y = g2(x), axb. Then

The integral over C3 is negated because it goes in the negative direction from b to a, as C is oriented positively (anticlockwise). On C2 and C4, x remains constant, meaning

Therefore,

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.

## Proof for rectifiable Jordan curves

We are going to prove the following

Theorem. Letbe a rectifiable, positively orientedJordan curveinand letdenote its inner region. Suppose thatare continuous functions with the property thathas second partial derivative at every point of,has first partial derivative at every point ofand that the functions,are Riemann-integrable over. Then

We need the following lemmas:

Lemma 1 (Decomposition Lemma). Assumeis a rectifiable, positively oriented Jordan curve in the plane and letbe its inner region. For every positive real, letdenote the collection of squares in the plane bounded by the lines, whereruns through the set of integers. Then, for this, there exists a decomposition ofinto a finite number of non-overlapping subregions in such a manner that
(i) Each one of the subregions contained in, say, is a square from.
(ii) Each one of the remaining subregions, say, has as boundary a rectifiable Jordan curve formed by a finite number of arcs ofand parts of the sides of some square from.
(iii) Each one of the border regionscan be enclosed in a square of edge-length.
(iv) Ifis the positively oriented boundary curve of, then
(v) The numberof border regions is no greater than, whereis the length of.
Lemma 2. Letbe a rectifiable curve in the plane and letbe the set of points in the plane whose distance from (the range of)is at most. The outer Jordan content of this set satisfies.
Lemma 3. Letbe a rectifiable curve inand letbe a continuous function. Then
andarewhereis the oscillation ofon the range of.

Now we are in position to prove the Theorem:

Proof of Theorem. Letbe an arbitrary positive real number. By continuity of,and compactness of, given, there existssuch that whenever two points ofare less thanapart, their images underare less thanapart. For this, consider the decomposition given by the previous Lemma. We have
Put.
For each, the curveis a positively oriented square, for which Green's formula holds. Hence
Every point of a border region is at a distance no greater thanfrom. Thus, ifis the union of all border regions, then; hence, by Lemma 2. Notice that
This yields
We may as well chooseso that the RHS of the last inequality is
The remark in the beginning of this proof implies that the oscillations ofandon every border region is at most. We have

By Lemma 1(iii),

Combining these, we finally get

for some. Since this is true for every, we are done.

## Validity under different hypotheses

The hypothesis of the last theorem are not the only ones under which Green's formula is true. Another common set of conditions is the following:

The functionsare still assumed to be continuous. However, we now require them to be Fréchet-differentiable at every point of. This implies the existence of all directional derivatives, in particular, where, as usual,is the canonical ordered basis of. In addition, we require the functionto be Riemann-integrable over.

As a corollary of this, we get the Cauchy Integral Theorem for rectifiable Jordan curves:

Theorem (Cauchy). Ifis a rectifiable Jordan curve inand ifis a continuous mapping holomorphic throughout the inner region of, then

the integral being a complex contour integral.

Proof. We regard the complex plane as. Now, defineto be such thatThese functions are clearly continuous. It is well known thatandare Fréchet-differentiable and that they satisfy the Cauchy-Riemann equations:.

Now, analysing the sums used to define the complex contour integral in question, it is easy to realize that

the integrals on the RHS being usual line integrals. These remarks allow us to apply Green's Theorem to each one of these line integrals, finishing the proof.

### Measure-theoretic assumptions

Green's formula also holds when, besides continuity assumptions,

(i) The functions, are defined at every point of, with the exception of a countable subset.
(ii) The functionis Lebesgue-integrable over.

## Multiply-connected regions

Theorem. Letbe positively oriented rectifiable Jordan curves insatisfying
whereis the inner region of. Let
Supposeandare continuous functions whose restriction tois Fréchet-differentiable. If the function
is Riemann-integrable over, then

## Relationship to Stokes' theorem

Green's theorem is a special case of theKelvin–Stokes theorem, when applied to a region in the-plane.
We can augment the two-dimensional field into a three-dimensional field with a z component that is always 0. Write F for thevector-valued function. Start with the left side of Green's theorem:

The Kelvin–Stokes theorem:

The surfaceis just the region in the plane, with the unit normalpointing up (in the positivedirection) to match the "positive orientation" definitions for both theorems.

The expression inside the integral becomes

Thus we get the right side of Green's theorem

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

## 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:

whereis the divergence on the two-dimensional vector field, andis the outward-pointing unit normal vector on the boundary.
To see this, consider the unit normalin the right side of the equation. Since in Green's theoremis 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. The length of this vector isSo

Applying the two-dimensional divergence theorem with, we get the right side of Green's theorem:

## Area calculation

Green's theorem can be used to compute area by line integral.[4] The area of a planar regionis given by
Chooseandsuch that, the area is given by
Possible formulas for the area ofinclude[4]

• 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]
Citation Linkbooks.google.comGeorge 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.
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[2]
Citation Linkopenlibrary.orgRiley, K. F.; Hobson, M. P.; Bence, S. J. (2010). Mathematical Methods for Physics and Engineering. Cambridge University Press. ISBN 978-0-521-86153-3.
Sep 22, 2019, 12:24 AM
[3]
Citation Linkopenlibrary.orgSpiegel, M. R.; Lipschutz, S.; Spellman, D. (2009). Vector Analysis. Schaum’s Outlines (2nd ed.). McGraw Hill. ISBN 978-0-07-161545-7.
Sep 22, 2019, 12:24 AM
[4]
Citation Linkopenlibrary.orgStewart, James. Calculus (6th ed.). Thomson, Brooks/Cole.
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[5]
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[6]
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[8]
Citation Linkarchive.org"Sur les intégrales qui s'étendent à tous les points d'une courbe fermée"
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[9]
Citation Linkbooks.google.comGrundlagen für eine allgemeine Theorie der Functionen einer veränderlichen complexen Grösse
Sep 22, 2019, 12:24 AM
[10]