View source for Green's theorem

{{Short description|Theorem in calculus relating line and double integrals}}
{{Calculus |Vector}}

In vector calculus, '''Green's theorem''' relates a [[Line integral|line integral]] around a [[Curve#Definition|simple closed curve]] {{mvar|C}} to a double integral over the [[Plane (geometry)|plane]] region {{mvar|D}} bounded by {{mvar|C}}. It is the two-dimensional special case of [[Physics:Kelvin–Stokes theorem|Stokes' theorem]].

==Theorem==
Let {{mvar|C}} be a positively [[Curve orientation|oriented]], [[Piecewise|piecewise]] [[Smoothness|smooth]], simple closed curve in a [[Plane (mathematics)|plane]], and let {{mvar|D}} be the region bounded by {{mvar|C}}. If {{mvar|L}} and {{mvar|M}} are functions of {{math|(''x'', ''y'')}} defined on an [[Open set|open region]] containing {{mvar|D}} and have [[Continuous function|continuous]] partial derivatives there, then

<math display="block">\oint_C (L\, dx + M\, dy) = \iint_{D} \left(\frac{\partial M}{\partial x} - \frac{\partial L}{\partial y}\right) dx\, dy</math>

where the path of integration along {{mvar|C}} is anticlockwise.<ref>{{cite book |title=Mathematical Methods for Physics and Engineering |url=https://archive.org/details/mathematicalmeth00rile |url-access=registration |first1=K. F. |last1=Riley |first2=M. P. |last2=Hobson |first3=S. J. |last3=Bence |publisher=Cambridge University Press |year=2010 |isbn=978-0-521-86153-3 }}</ref><ref>{{cite book |title=Vector Analysis |edition=2nd |first1=M. R. |last1=Spiegel |first2=S. |last2=Lipschutz |first3=D. |last3=Spellman |series=Schaum’s Outlines |publisher=McGraw Hill |year=2009 |isbn=978-0-07-161545-7 }}</ref>

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 [[Engineering:Surveying|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==
[[Image:Green's-theorem-simple-region.svg|thumb|300px|right|If ''D'' is a simple type of region with its boundary consisting of the curves ''C''<sub>1</sub>, ''C''<sub>2</sub>, ''C''<sub>3</sub>, ''C''<sub>4</sub>, 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 ''C''<sub>1</sub> and ''C''<sub>3</sub> 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 ''C''<sub>2</sub> and ''C''<sub>4</sub> 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

{{NumBlk|:|<math>\oint_C L\, dx = \iint_{D} \left(- \frac{\partial L}{\partial y}\right) dA</math>|{{EquationRef|1}}}}

and

{{NumBlk|:|<math>\oint_C\ M\, dy = \iint_{D} \left(\frac{\partial M}{\partial x}\right) dA</math>|{{EquationRef|2}}}}

are true, then Green's theorem follows immediately for the region D. We can prove ({{EquationNote|1}}) easily for regions of type I, and ({{EquationNote|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
<math display="block">D = \{(x,y)\mid a\le x\le b, g_1(x) \le y \le g_2(x)\}</math>
where ''g''<sub>1</sub> and ''g''<sub>2</sub> are [[Continuous function|continuous function]]s on {{closed-closed|''a'', ''b''}}. Compute the double integral in ({{EquationNote|1}}):

{{NumBlk|:|<math>
\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 \left[L(x,g_2(x)) - L(x,g_1(x)) \right] \, dx.
\end{align}
</math>|{{EquationRef|3}}}}

Now compute the line integral in ({{EquationNote|1}}). ''C'' can be rewritten as the union of four curves: ''C''<sub>1</sub>, ''C''<sub>2</sub>, ''C''<sub>3</sub>, ''C''<sub>4</sub>.

With ''C''<sub>1</sub>, use the [[Parametric equation|parametric equation]]s: ''x'' = ''x'', ''y'' = ''g''<sub>1</sub>(''x''), ''a'' ≤ ''x'' ≤ ''b''. Then
<math display="block">\int_{C_1} L(x,y)\, dx = \int_a^b L(x,g_1(x))\, dx.</math>

With ''C''<sub>3</sub>, use the parametric equations: ''x'' = ''x'', ''y'' = ''g''<sub>2</sub>(''x''), ''a'' ≤ ''x'' ≤ ''b''. Then
<math display="block"> \int_{C_3} L(x,y)\, dx = -\int_{-C_3} L(x,y)\, dx = - \int_a^b L(x,g_2(x))\, dx.</math>

The integral over ''C''<sub>3</sub> is negated because it goes in the negative direction from ''b'' to ''a'', as ''C'' is oriented positively (anticlockwise). On ''C''<sub>2</sub> and ''C''<sub>4</sub>, ''x'' remains constant, meaning
<math display="block"> \int_{C_4} L(x,y)\, dx = \int_{C_2} L(x,y)\, dx = 0.</math>

Therefore,

{{NumBlk|:|<math>
\begin{align}
\oint_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_1(x))\, dx - \int_a^b L(x,g_2(x))\, dx .
\end{align}
</math>|{{EquationRef|4}}}}

Combining ({{EquationNote|3}}) with ({{EquationNote|4}}), we get ({{EquationNote|1}}) for regions of type I. A similar treatment yields ({{EquationNote|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

{{math theorem|math_statement= Let <math>\Gamma</math> be a rectifiable, positively oriented [[Jordan curve theorem|Jordan curve]] in <math>\R^2</math> and let <math>R</math> denote its inner region. Suppose that <math>A, B:\overline{R} \to \R</math> are continuous functions with the property that <math>A</math> has second partial derivative at every point of <math>R</math>, <math>B</math> has first partial derivative at every point of <math>R</math> and that the functions <math>D_1 B, D_2 A : R \to \R</math> are Riemann-integrable over <math>R</math>. Then
<math display="block">\int_{\Gamma}(A\,dx + B\,dy) = \int_R \left(D_1 B(x,y)-D_2 A(x,y)\right)\,d(x,y).</math>}}

We need the following lemmas whose proofs can be found in:<ref>{{cite book |last1=Apostol |first1=Tom |title=Mathematical Analysis |date=1960 |publisher=Addison-Wesley Publishing Company, INC |location=Reading, Massachusetts, U.S.A. |edition=1}}</ref>

{{math theorem|name=Lemma 1 (Decomposition Lemma)|math_statement= Assume <math>\Gamma</math> is a rectifiable, positively oriented Jordan curve in the plane and let <math>R</math> be its inner region. For every positive real <math>\delta</math>, let <math>\mathcal{F}(\delta)</math> denote the collection of squares in the plane bounded by the lines <math>x=m\delta, y=m\delta</math>, where <math>m</math> runs through the set of integers. Then, for this <math>\delta</math>, there exists a decomposition of <math>\overline{R}</math> into a finite number of non-overlapping subregions in such a manner that
<ol type="i">
<li>Each one of the subregions contained in <math>R</math>, say <math>R_1, R_2,\ldots, R_k</math>, is a square from <math>\mathcal{F}(\delta)</math>.</li>
<li>Each one of the remaining subregions, say <math>R_{k+1},\ldots,R_s</math>, has as boundary a rectifiable Jordan curve formed by a finite number of arcs of <math>\Gamma</math> and parts of the sides of some square from <math>\mathcal{F}(\delta)</math>.</li>
<li>Each one of the border regions <math>R_{k+1},\ldots,R_s</math> can be enclosed in a square of edge-length <math>2\delta</math>.</li>
<li>If <math>\Gamma_i</math> is the positively oriented boundary curve of <math>R_i</math>, then <math>\Gamma=\Gamma_1+\Gamma_2+\cdots+\Gamma_s.</math></li>
<li>The number <math>s-k</math> of border regions is no greater than <math display="inline">4\!\left(\frac{\Lambda}{\delta} + 1\right)</math>, where <math>\Lambda</math> is the length of <math>\Gamma</math>.</li>
</ol>}}
{{math theorem|name=Lemma 2|math_statement= Let <math>\Gamma</math> be a rectifiable curve in the plane and let <math>\Delta_\Gamma(h)</math> be the set of points in the plane whose distance from (the range of) <math>\Gamma</math> is at most <math>h</math>. The outer Jordan content of this set satisfies <math>\overline{c}\,\,\Delta_\Gamma(h)\le2h\Lambda +\pi h^2</math>.}}

{{math theorem|name=Lemma 3|math_statement= Let <math>\Gamma</math> be a rectifiable curve in <math>\R^2</math> and let <math>f:\text{range of }\Gamma \to \R</math> be a continuous function. Then
<math display="block">\left\vert\int_\Gamma f(x,y)\,dy\right\vert \le\frac{1}{2} \Lambda\Omega_f,</math> and
<math display="block">\left\vert\int_\Gamma f(x,y)\,dx\right\vert \le\frac{1}{2} \Lambda\Omega_f,</math>
where <math>\Omega_f</math> is the oscillation of <math>f</math> on the range of <math>\Gamma</math>.}}

Now we are in position to prove the theorem:

'''Proof of Theorem.''' Let <math>\varepsilon</math> be an arbitrary positive real number. By continuity of <math>A</math>, <math>B</math> and compactness of <math>\overline{R}</math>, given <math>\varepsilon>0</math>, there exists <math>0<\delta<1</math> such that whenever two points of <math>\overline{R}</math> are less than <math>2\sqrt{2}\,\delta</math> apart, their images under <math>A, B</math> are less than <math>\varepsilon</math> apart. For this <math>\delta</math>, consider the decomposition given by the previous Lemma. We have
<math display="block">\int_\Gamma A\,dx+B\,dy=\sum_{i=1}^k \int_{\Gamma_i} A\,dx+B\,dy\quad +\sum_{i=k+1}^s \int_{\Gamma_i}A\,dx+B\,dy.</math>

Put <math>\varphi := D_1 B - D_2 A</math>.

For each <math>i\in\{1,\ldots,k\}</math>, the curve <math>\Gamma_i</math> is a positively oriented square, for which Green's formula holds. Hence
<math display="block">\sum_{i=1}^k \int_{\Gamma_i}A\,dx + B\,dy =\sum_{i=1}^k \int_{R_i} \varphi = \int_{\bigcup_{i=1}^k R_i}\,\varphi.</math>

Every point of a border region is at a distance no greater than <math>2\sqrt{2}\,\delta</math> from <math>\Gamma</math>. Thus, if <math>K</math> is the union of all border regions, then <math>K\subset \Delta_{\Gamma}(2\sqrt{2}\,\delta)</math>; hence <math>c(K)\le\overline{c}\,\Delta_{\Gamma}(2\sqrt{2}\,\delta)\le 4\sqrt{2}\,\delta+8\pi\delta^2</math>, by Lemma 2. Notice that
<math display="block">\int_R \varphi\,\,-\int_{\bigcup_{i=1}^k R_i} \varphi=\int_K \varphi.</math>
This yields
<math display="block">\left\vert\sum_{i=1}^k \int_{\Gamma_i} A\,dx+B\,dy\quad-\int_R\varphi \right\vert \le M \delta(1+\pi\sqrt{2}\,\delta) \text{ for some } M > 0.</math>

We may as well choose <math>\delta</math> so that the RHS of the last inequality is <math><\varepsilon.</math>

The remark in the beginning of this proof implies that the oscillations of <math>A</math> and <math>B</math> on every border region is at most <math>\varepsilon</math>. We have
<math display="block">\left\vert\sum_{i=k+1}^s \int_{\Gamma_i}A\,dx+B\,dy\right\vert\le\frac{1}{2} \varepsilon\sum_{i=k+1}^s \Lambda_i.</math>

By Lemma 1(iii),
<math display="block">\sum_{i=k+1}^s \Lambda_i \le\Lambda + (4\delta)\,4\!\left(\frac{\Lambda}{\delta}+1\right)\le17\Lambda+16.</math>

Combining these, we finally get
<math display="block">\left\vert\int_\Gamma A\,dx+B\,dy\quad-\int_R \varphi\right\vert< C \varepsilon,</math>
for some <math>C > 0</math>. Since this is true for every <math>\varepsilon > 0</math>, 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 functions <math>A, B:\overline{R} \to \R</math> are still assumed to be continuous. However, we now require them to be Fréchet-differentiable at every point of <math>R</math>. This implies the existence of all directional derivatives, in particular <math>D_{e_i}A=:D_i A, D_{e_i}B=:D_i B, \,i=1,2</math>, where, as usual, <math>(e_1,e_2)</math> is the canonical ordered basis of <math>\R^2</math>. In addition, we require the function <math>D_1 B-D_2 A</math> to be Riemann-integrable over <math>R</math>.

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

{{math theorem|name=Theorem (Cauchy)|math_statement=If <math>\Gamma</math> is a rectifiable Jordan curve in <math>\Complex</math> and if <math>f : \text{closure of inner region of } \Gamma \to \Complex</math> is a continuous mapping holomorphic throughout the inner region of <math>\Gamma</math>, then <math display="block">\int_\Gamma f=0,</math> the integral being a complex contour integral.}}

{{math proof|proof=We regard the complex plane as <math>\R^2</math>. Now, define <math>u,v:\overline{R} \to \R</math> to be such that <math>f(x + iy) = u(x, y) + iv(x, y).</math> These functions are clearly continuous. It is well known that <math>u</math> and <math>v</math> are Fréchet-differentiable and that they satisfy the Cauchy-Riemann equations: <math>D_1 v + D_2 u = D_1 u - D_2 v = \text{zero function}</math>.

Now, analyzing the sums used to define the complex contour integral in question, it is easy to realize that
<math display="block">\int_\Gamma f=\int_\Gamma u\,dx-v\,dy\quad+i\int_\Gamma v\,dx+u\,dy,</math>
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.}}

== Multiply-connected regions ==
'''Theorem.''' Let <math>\Gamma_0,\Gamma_1,\ldots,\Gamma_n</math> be positively oriented rectifiable Jordan curves in <math>\R^{2}</math> satisfying
<math display="block">\begin{aligned}
\Gamma_i \subset R_0,&&\text{if } 1\le i\le n\\
\Gamma_i \subset \R^2 \setminus \overline{R}_j,&&\text{if }1\le i,j \le n\text{ and }i\ne j,
\end{aligned}</math>
where <math>R_i</math> is the inner region of <math>\Gamma_i</math>. Let
<math display="block">D = R_0 \setminus (\overline{R}_1 \cup \overline{R}_2 \cup \cdots \cup \overline{R}_n).</math>

Suppose <math>p: \overline{D} \to \R</math> and <math>q: \overline{D} \to \R</math> are continuous functions whose restriction to <math>D</math> is Fréchet-differentiable. If the function
<math display="block">(x,y)\longmapsto\frac{\partial q}{\partial e_1}(x,y)-\frac{\partial p}{\partial e_2}(x,y)</math>
is Riemann-integrable over <math>D</math>, then
<math display="block">\begin{align}
& \int_{\Gamma_0} p(x,y)\,dx+q(x,y)\,dy-\sum_{i=1}^n \int_{\Gamma_i} p(x,y)\,dx + q(x,y)\,dy \\[5pt]
= {} & \int_D\left\{\frac{\partial q}{\partial e_1}(x,y)-\frac{\partial p}{\partial e_2}(x,y)\right\} \, d(x,y).
\end{align}</math>

==Relationship to Stokes' theorem==
Green's theorem is a special case of the [[Physics:Stokes' theorem#Theorem|Kelvin–Stokes theorem]], when applied to a region in the <math>xy</math>-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 [[Euclidean vector|vector]]-valued function <math>\mathbf{F}=(L,M,0)</math>. Start with the left side of Green's theorem:
<math display="block">\oint_C (L\, dx + M\, dy) = \oint_C (L, M, 0) \cdot (dx, dy, dz) = \oint_C \mathbf{F} \cdot d\mathbf{r}. </math>

The Kelvin–Stokes theorem:
<math display="block">\oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_S \nabla \times \mathbf{F} \cdot \mathbf{\hat n} \, dS. </math>

The surface <math>S</math> is just the region in the plane <math>D</math>, with the unit normal <math>\mathbf{\hat n}</math> defined (by convention) to have a positive z component in order to match the "positive orientation" definitions for both theorems.

The expression inside the integral becomes
<math display="block">\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). </math>

Thus we get the right side of Green's theorem
<math display="block">\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. </math>

Green's theorem is also a straightforward result of the general Stokes' theorem using [[Differential form|differential form]]s and [[Exterior derivative|exterior derivative]]s:
<math display="block">\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.</math>

==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|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>

==Area calculation==
Green's theorem can be used to compute area by line integral.<ref name="stuart">{{cite book|last=Stewart|first=James |title=Calculus |year=1999|url=https://archive.org/details/calculus00stew|url-access=registration|publisher=Thomson, Brooks/Cole |isbn=9780534359492 | edition=6th}}</ref> The area of a planar region <math>D</math> is given by
<math display="block">A = \iint_D dA.</math>

Choose <math>L</math> and <math>M</math> such that <math>\frac{\partial M}{\partial x} - \frac{\partial L}{\partial y} = 1</math>, the area is given by
<math display="block">A = \oint_{C} (L\, dx + M\, dy).</math>

Possible formulas for the area of <math>D</math> include<ref name="stuart" />
<math display="block">A=\oint_C x\, dy = -\oint_C y\, dx = \tfrac 12 \oint_C (-y\, dx + x\, dy).</math>

== History ==
It is named after [[Biography:George Green (mathematician)|George Green]], who stated a similar result in an 1828 paper titled ''An Essay on the Application of Mathematical Analysis to the Theories of Electricity and Magnetism''. In 1846, [[Biography:Augustin-Louis Cauchy|Augustin-Louis Cauchy]] published a paper stating Green's theorem as the penultimate sentence. This is in fact the first printed version of Green's theorem in the form appearing in modern textbooks. [[Biography:Bernhard Riemann|Bernhard Riemann]] gave the first proof of Green's theorem in his doctoral dissertation on the theory of functions of a complex variable.<ref>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 [https://books.google.com/books?id=GwYXAAAAYAAJ&pg=PA10 pages 10–12] of his ''Essay''.<br />
In 1846, the form of "Green's theorem" which appears in this article was first published, without proof, in an article by [[Biography:Augustin-Louis Cauchy|Augustin Cauchy]]: A. Cauchy (1846) [https://archive.org/stream/ComptesRendusAcademieDesSciences0023/ComptesRendusAcadmieDesSciences-Tome023-Juillet-dcembre1846#page/n254/mode/1up "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''.)<br />
A proof of the theorem was finally provided in 1851 by [[Biography:Bernhard Riemann|Bernhard Riemann]] in his inaugural dissertation: Bernhard Riemann (1851) [https://books.google.com/books?id=PpALAAAAYAAJ&pg=PP5 ''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.</ref><ref>{{Cite book|last=Katz|first=Victor|title=A History of Mathematics: An Introduction|publisher=Addison-Wesley|year=2009|isbn=978-0-321-38700-4|pages=801–5|chapter=22.3.3: Complex Functions and Line Integrals}}</ref>

==See also==
*{{Annotated link|Planimeter}}
*[[Physics:Method of image charges|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==
{{Reflist}}

==Further reading==
* {{cite book |first1=Jerrold E. |last1=Marsden |first2=Anthony J. |last2=Tromba |chapter=The Integral Theorems of Vector Analysis |title=Vector Calculus |location=New York |publisher=Freeman |year=2003 |edition=Fifth |isbn=0-7167-4992-0 |pages=518–608 |chapter-url=https://books.google.com/books?id=LiRLJf2m_dwC&pg=PA518 }}

==External links==
*[http://mathworld.wolfram.com/GreensTheorem.html Green's Theorem on MathWorld]

{{Calculus topics}}

[[Category:Theorems in calculus]]
[[Category:Articles containing proofs]]

{{Sourceattribution|Green's theorem}}