# Green's theorem

In mathematicsGreen's theorem gives the relationship between a line integral around a simple closed curve C and a double integralover 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 orientedpiecewise smoothsimple closed curve in a plane, and let D be the region bounded by C. If L and M are functions of (xy) defined on an open regioncontaining D and have continuous partial derivatives there, then

${displaystyle {scriptstyle C}}$ ${displaystyle (L,dx+M,dy)=iint _{D}left({frac {partial M}{partial x}}-{frac {partial L}{partial y}} ight),dx,dy}$

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

In physics, Green's theorem finds many applications. One of which 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

If D is a simple type I region with its boundary consisting of the curves C1C2C3C4, 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

${displaystyle oint _{C}L,dx=iint _{D}left(-{frac {partial L}{partial y}} ight),dAqquad mathrm {(1)} }$

and

${displaystyle oint _{C}M,dy=iint _{D}left({frac {partial M}{partial x}} ight),dAqquad mathrm {(2)} }$

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

${displaystyle D={(x,y)mid aleq xleq b,g_{1}(x)leq yleq g_{2}(x)}}$

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

{displaystyle {egin{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: C1C2C3C4.

With C1, use the parametric equationsx = xy = g1(x), a ≤ x ≤ b. Then

${displaystyle int _{C_{1}}L(x,y),dx=int _{a}^{b}L(x,g_{1}(x)),dx.}$

With C3, use the parametric equations: x = xy = g2(x), a ≤ x ≤ b. Then

${displaystyle int _{C_{3}}L(x,y),dx=-int _{-C_{3}}L(x,y),dx=-int _{a}^{b}L(x,g_{2}(x)),dx.}$

The integral over C3 is negated because it goes in the negative direction from b to a, as C is oriented positively (anticlockwise). On C2 and C4x remains constant, meaning

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

Therefore,

{displaystyle {egin{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_{1}(x)),dx-int _{a}^{b}L(x,g_{2}(x)),dx.qquad mathrm {(4)} end{aligned}}}

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. Let ${displaystyle Gamma }$ be a rectifiable, positively oriented Jordan curve in ${displaystyle mathbf {R} ^{2}}$ and let ${displaystyle R}$ denote its inner region. Suppose that ${displaystyle A,B:{overline {R}}longrightarrow mathbf {R} }$ are continuous functions with the property that ${displaystyle A}$ has second partial derivative at every point of ${displaystyle R}$${displaystyle B}$ has first partial derivative at every point of ${displaystyle R}$ and that the functions ${displaystyle D_{1}B}$${displaystyle D_{2}A:Rlongrightarrow mathbf {R} }$ are Riemann-integrable over ${displaystyle R}$. Then

${displaystyle int _{Gamma }A,dx+B,dy=int _{R},,left(D_{1}B(x,y)-D_{2}A(x,y) ight),d(x,y).}$

We need the following lemmas:

Lemma 1 (Decomposition Lemma). Assume ${displaystyle Gamma }$ is a rectifiable, positively oriented Jordan curve in the plane and let ${displaystyle R}$ be its inner region. For every positive real ${displaystyle delta }$, let ${displaystyle {mathcal {F}}(delta )}$ denote the collection of squares in the plane bounded by the lines ${displaystyle x=mdelta ,y=mdelta }$, where ${displaystyle m}$ runs through the set of integers. Then, for this ${displaystyle delta }$, there exists a decomposition of ${displaystyle {overline {R}}}$ into a finite number of non-overlapping subregions in such a manner that

(i) Each one of the subregions contained in ${displaystyle R}$, say ${displaystyle R_{1},R_{2},ldots ,R_{k}}$, is a square from ${displaystyle {mathcal {F}}(delta )}$.

(ii) Each one of the remaining subregions, say ${displaystyle R_{k+1},ldots ,R_{s}}$, has as boundary a rectifiable Jordan curve formed by a finite number of arcs of ${displaystyle Gamma }$ and parts of the sides of some square from ${displaystyle {mathcal {F}}(delta )}$.

(iii) Each one of the border regions ${displaystyle R_{k+1},ldots ,R_{s}}$ can be enclosed in a square of edge-length ${displaystyle 2delta }$.

(iv) If ${displaystyle Gamma _{i}}$ is the positively oriented boundary curve of ${displaystyle R_{i}}$, then ${displaystyle Gamma =Gamma _{1}+Gamma _{2}+cdots +Gamma _{s}.}$

(v) The number ${displaystyle s-k}$ of border regions is no greater than ${displaystyle 4!left({frac {Lambda }{delta }}+1 ight)}$, where ${displaystyle Lambda }$ is the length of ${displaystyle Gamma }$.

Lemma 2. Let ${displaystyle Gamma }$ be a rectifiable curve in the plane and let ${displaystyle Delta _{Gamma }(h)}$ be the set of points in the plane whose distance from (the range of) ${displaystyle Gamma }$ is at most ${displaystyle h}$. The outer Jordan content of this set satisfies ${displaystyle {overline {c}},,Delta _{Gamma }(h)leq 2hLambda +pi h^{2}}$.

Lemma 3. Let ${displaystyle Gamma }$ be a rectifiable curve in ${displaystyle mathbf {R} ^{2}}$ and let ${displaystyle f:{ ext{range of }}Gamma longrightarrow mathbf {R} }$ be a continuous function. Then

${displaystyle leftvert int _{Gamma }f(x,y),dy ightvert }$ and
${displaystyle leftvert int _{Gamma }f(x,y),dx ightvert }$ are ${displaystyle leq {frac {1}{2}}Lambda Omega _{f},}$where ${displaystyle Omega _{f}}$ is the oscillation of ${displaystyle f}$ on the range of ${displaystyle Gamma }$.

Now we are in position to prove the Theorem:

Proof of Theorem. Let ${displaystyle varepsilon }$ be an arbitrary positive real number. By continuity of ${displaystyle A}$${displaystyle B}$ and compactness of ${displaystyle {overline {R}}}$, given ${displaystyle varepsilon >0}$, there exists ${displaystyle 0 such that whenever two points of ${displaystyle {overline {R}}}$ are less than ${displaystyle 2{sqrt {2}},delta }$ apart, their images under ${displaystyle A,B}$ are less than ${displaystyle varepsilon }$ apart. For this ${displaystyle delta }$, consider the decomposition given by the previous Lemma. We have

${displaystyle int _{Gamma }A,dx+B,dy=sum _{i=1}^{k}int _{Gamma _{i}}A,dx+B,dyquad +sum _{i=k+1}^{s}int _{Gamma _{i}}A,dx+B,dy.}$

Put ${displaystyle varphi :=D_{1}B-D_{2}A}$.

For each ${displaystyle iin {1,ldots ,k}}$, the curve ${displaystyle Gamma _{i}}$ is a positively oriented square, for which Green's formula holds. Hence

${displaystyle sum _{i=1}^{k}int _{Gamma _{i}}A,dx+B,dy=sum _{i=1}^{k}int _{R_{i}},,varphi =int _{igcup _{i=1}^{k}R_{i}},varphi .}$

Every point of a border region is at a distance no greater than ${displaystyle 2{sqrt {2}},delta }$ from ${displaystyle Gamma }$. Thus, if ${displaystyle K}$ is the union of all border regions, then ${displaystyle Ksubset Delta _{Gamma }(2{sqrt {2}},delta )}$; hence ${displaystyle c(K)leq {overline {c}},Delta _{Gamma }(2{sqrt {2}},delta )leq 4{sqrt {2}},delta +8pi delta ^{2}}$, by Lemma 2. Notice that

${displaystyle int _{R}varphi ,,-int _{igcup _{i=1}^{k}R_{i}}varphi =int _{K}varphi .}$ This yields
${displaystyle leftvert sum _{i=1}^{k}int _{Gamma _{i}}A,dx+B,dyquad -int _{R}varphi ightvert leq Mdelta (1+pi {sqrt {2}},delta ){ ext{ for some }}M>0.}$

We may as well choose ${displaystyle delta }$ so that the RHS of the last inequality is ${displaystyle

The remark in the beginning of this proof implies that the oscillations of ${displaystyle A}$ and ${displaystyle B}$ on every border region is at most ${displaystyle varepsilon }$. We have

${displaystyle leftvert sum _{i=k+1}^{s}int _{Gamma _{i}}A,dx+B,dy ightvert leq {frac {1}{2}}varepsilon sum _{i=k+1}^{s}Lambda _{i}.}$

By Lemma 1(iii),

${displaystyle sum _{i=k+1}^{s}Lambda _{i}leq Lambda +(4delta ),4!left({frac {Lambda }{delta }}+1 ight)leq 17Lambda +16.}$

Combining these, we finally get

${displaystyle leftvert int _{Gamma }A,dx+B,dyquad -int _{R}varphi ightvert

for some ${displaystyle C>0}$. Since this is true for every ${displaystyle varepsilon >0}$, 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 ${displaystyle A,B:{overline {R}}longrightarrow mathbf {R} }$ are still assumed to be continuous. However, we now require them to be Fréchet-differentiable at every point of ${displaystyle R}$. This implies the existence of all directional derivatives, in particular ${displaystyle D_{e_{i}}A=:D_{i}A,D_{e_{i}}B=:D_{i}B,,i=1,2}$, where, as usual, ${displaystyle (e_{1},e_{2})}$ is the canonical ordered basis of ${displaystyle mathbf {R} ^{2}}$. In addition, we require the function ${displaystyle D_{1}B-D_{2}A}$ to be Riemann-integrable over ${displaystyle R}$.

It suffices to prove this for squares which are contained in ${displaystyle R}$ and have sides parallel to the axes. The proof then follows the lines of the method employed to prove the Cauchy-Goursat Theorem for triangles.

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

Theorem (Cauchy). If ${displaystyle Gamma }$ is a rectifiable Jordan curve in ${displaystyle mathbf {C} }$ and if ${displaystyle f:{ ext{closure of inner region of }}Gamma longrightarrow mathbf {C} }$ is a continuous mapping holomorphic throughout the inner region of ${displaystyle Gamma }$, then

${displaystyle int _{Gamma }f=0,}$

the integral being a complex contour integral.

Proof. We regard the complex plane as ${displaystyle mathbf {R} ^{2}}$. Now, define ${displaystyle u,v:{overline {R}}longrightarrow mathbf {R} }$ to be such that ${displaystyle f(x+iy)=u(x,y)+iv(x,y).}$ These functions are clearly continuous. It is well-known that ${displaystyle u}$ and ${displaystyle v}$ are Fréchet-differentiable and that they satisfy the Cauchy-Riemann equations: ${displaystyle D_{1}v+D_{2}u=D_{1}u-D_{2}v={ ext{zero function}}}$.

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

${displaystyle int _{Gamma }f=int _{Gamma }u,dx-v,dyquad +iint _{Gamma }v,dx+u,dy,}$

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 ${displaystyle D_{i}A,D_{i}B,,i=1,2}$, are defined at every point of ${displaystyle R}$, with the exception of a countable subset.

(ii) The function ${displaystyle D_{1}B-D_{2}A}$ is Lebesgue-integrable over ${displaystyle R}$.

## Multiply-connected regions

Theorem. Let ${displaystyle Gamma _{0},Gamma _{1},ldots ,Gamma _{n}}$ be positively oriented rectifiable Jordan curves in ${displaystyle mathbf {R} ^{2}}$ satisfying

{displaystyle {egin{aligned}Gamma _{i}subset R_{0},&&{ ext{if }}1leq ileq n\Gamma _{i}subset mathbf {R} ^{2}smallsetminus {overline {R}}_{j},&&{ ext{if }}1leq i,jleq n{ ext{ and }}i eq j,end{aligned}}}

where ${displaystyle R_{i}}$ is the inner region of ${displaystyle Gamma _{i}}$. Let

${displaystyle D=R_{0}smallsetminus ({overline {R}}_{1}cup {overline {R}}_{2}cup cdots cup {overline {R}}_{n}).}$

Suppose ${displaystyle p:{overline {D}}longrightarrow mathbf {R} }$ and ${displaystyle q:{overline {D}}longrightarrow mathbf {R} }$ are continuous functions whose restriction to ${displaystyle D}$ is Fréchet-differentiable. If the function

${displaystyle (x,y)longmapsto {frac {partial q}{partial e_{1}}}(x,y)-{frac {partial p}{partial e_{2}}}(x,y)}$

is Riemann-integrable over ${displaystyle D}$, then

{displaystyle {egin{aligned}&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) ight},d(x,y).end{aligned}}}

## Relationship to Stokes' theorem

Green's theorem is a special case of the Kelvin–Stokes theorem, when applied to a region in the ${displaystyle xy}$-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 ${displaystyle mathbf {F} =(L,M,0)}$. Start with the left side of Green's theorem:

${displaystyle oint _{C}(L,dx+M,dy)=oint _{C}(L,M,0)cdot (dx,dy,dz)=oint _{C}mathbf {F} cdot dmathbf {r} .}$

The Kelvin–Stokes theorem:

${displaystyle oint _{C}mathbf {F} cdot dmathbf {r} =iint _{S} abla imes mathbf {F} cdot mathbf {hat {n}} ,dS.}$

The surface ${displaystyle S}$ is just the region in the plane ${displaystyle D}$, with the unit normal ${displaystyle mathbf {hat {n}} }$ pointing up (in the positive ${displaystyle z}$ direction) to match the "positive orientation" definitions for both theorems.

The expression inside the integral becomes

${displaystyle abla imes mathbf {F} cdot mathbf {hat {n}} =left[left({frac {partial 0}{partial y}}-{frac {partial M}{partial z}} ight)mathbf {i} +left({frac {partial L}{partial z}}-{frac {partial 0}{partial x}} ight)mathbf {j} +left({frac {partial M}{partial x}}-{frac {partial L}{partial y}} ight)mathbf {k} ight]cdot mathbf {k} =left({frac {partial M}{partial x}}-{frac {partial L}{partial y}} ight).}$

Thus we get the right side of Green's theorem

${displaystyle iint _{S} abla imes mathbf {F} cdot mathbf {hat {n}} ,dS=iint _{D}left({frac {partial M}{partial x}}-{frac {partial L}{partial y}} ight),dA.}$

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

{displaystyle {egin{aligned}&oint _{C}L,dx+M,dy=oint _{partial D}omega =int _{D},domega \[5pt]={}&int _{D}{frac {partial L}{partial y}},dywedge ,dx+{frac {partial M}{partial x}},dxwedge ,dy=iint _{D}left({frac {partial M}{partial x}}-{frac {partial L}{partial y}} ight),dx,dy.end{aligned}}}

## 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:

${displaystyle iiint _{V}left(mathbf { abla } cdot mathbf {F} ight),dV=}$ ${displaystyle partial scriptstyle V}$ ${displaystyle (mathbf {F} cdot mathbf {hat {n}} ),dS.}$

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

To see this, consider the unit normal ${displaystyle mathbf {hat {n}} }$ in the right side of the equation. Since in Green's theorem ${displaystyle dmathbf {r} =(dx,dy)}$ 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 ${displaystyle (dy,-dx)}$. The length of this vector is ${displaystyle {sqrt {dx^{2}+dy^{2}}}=ds.}$ So ${displaystyle (dy,-dx)=mathbf {hat {n}} ,ds.}$

Start with the left side of Green's theorem:

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

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

${displaystyle oint _{C}(M,-L)cdot mathbf {hat {n}} ,ds=iint _{D}left( abla cdot (M,-L) ight),dA=iint _{D}left({frac {partial M}{partial x}}-{frac {partial L}{partial y}} ight),dA.}$

## Area calculation

Green's theorem can be used to compute area by line integral.[4] The area of a planar region ${displaystyle D}$ is given by

${displaystyle A=iint _{D}dA.}$

Choose ${displaystyle L}$ and ${displaystyle M}$ such that ${displaystyle {frac {partial M}{partial x}}-{frac {partial L}{partial y}}=1}$, the area is given by

${displaystyle A=oint _{C}(L,dx+M,dy).}$

Possible formulas for the area of ${displaystyle D}$ include[4]

${displaystyle A=oint _{C}x,dy=-oint _{C}y,dx={ frac {1}{2}}oint _{C}(-y,dx+x,dy).}$

## 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 rendus23: 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. ^ Riley, K. F.; Hobson, M. P.; Bence, S. J. (2010). Mathematical Methods for Physics and Engineering. Cambridge University Press. ISBN 978-0-521-86153-3.
3. ^ Spiegel, M. R.; Lipschutz, S.; Spellman, D. (2009). Vector Analysis. Schaum’s Outlines (2nd ed.). McGraw Hill. ISBN 978-0-07-161545-7.
4.  Stewart, James. Calculus (6th ed.). Thomson, Brooks/Cole.

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