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Green's theorem

From Wikipedia, the free encyclopedia
Theorem in calculus relating line and double integrals
This article is about the theorem in the plane relating double integrals and line integrals. For Green's theorems relating volume integrals involving the Laplacian to surface integrals, seeGreen's identities.
Not to be confused withGreen's law for waves approaching a shoreline.
Part of a series of articles about
Calculus
∫abf′(t)dt=f(b)−f(a){\displaystyle \int _{a}^{b}f'(t)\,dt=f(b)-f(a)}

Invector calculus,Green's theorem relates aline integral around asimple closed curveC to adouble integral over theplane regionD (surface inR2{\displaystyle \mathbb {R} ^{2}}) bounded byC. It is the two-dimensional special case ofStokes' theorem (surface inR3{\displaystyle \mathbb {R} ^{3}}). In one dimension, it is equivalent to thefundamental theorem of calculus. In two dimensions, it isequivalent to the divergence theorem.

Theorem

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LetC be a positivelyoriented,piecewisesmooth,simple closed curve in aplane, and letD be the region bounded byC. IfL andM are functions of(x,y) defined on anopen region containingD and havecontinuouspartial derivatives there, then

∮C(Ldx+Mdy)=∬D(∂M∂x−∂L∂y)dA{\displaystyle \oint _{C}(L\,dx+M\,dy)=\iint _{D}\left({\frac {\partial M}{\partial x}}-{\frac {\partial L}{\partial y}}\right)dA}

where the path of integration alongC iscounterclockwise.[1][2]

Application

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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. Inplane geometry, and in particular, areasurveying, Green's theorem can be used to determine the area and centroid of plane figures solely by integrating over the perimeter.

Proof whenD is a simple region

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IfD is a simple type of region with its boundary consisting of the curvesC1,C2,C3,C4, half of Green's theorem can be demonstrated.

The following is a proof of half of the theorem for the simplified areaD, a type I region whereC1 andC3 are curves connected by vertical lines (possibly of zero length). A similar proof exists for the other half of the theorem whenD is a type II region whereC2 andC4 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 decomposingD into a set of type III regions.

If it can be shown that

∮CLdx=∬D(−∂L∂y)dA{\displaystyle \oint _{C}L\,dx=\iint _{D}\left(-{\frac {\partial L}{\partial y}}\right)dA}1

and

∮C Mdy=∬D(∂M∂x)dA{\displaystyle \oint _{C}\ M\,dy=\iint _{D}\left({\frac {\partial M}{\partial x}}\right)dA}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 regionD is a type I region and can thus be characterized, as pictured on the right, byD={(x,y)∣a≤x≤b,g1(x)≤y≤g2(x)}{\displaystyle D=\{(x,y)\mid a\leq x\leq b,g_{1}(x)\leq y\leq g_{2}(x)\}}whereg1 andg2 arecontinuous functions on[a,b]. Compute the double integral in (1):

∬D∂L∂ydA=∫ab∫g1(x)g2(x)∂L∂y(x,y)dydx=∫ab[L(x,g2(x))−L(x,g1(x))]dx.{\displaystyle {\begin{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}\left[L(x,g_{2}(x))-L(x,g_{1}(x))\right]\,dx.\end{aligned}}}3

Now compute the line integral in (1).C can be rewritten as the union of four curves:C1,C2,C3,C4.

WithC1, use theparametric equations:x =x,y =g1(x),a ≤x ≤b. Then∫C1L(x,y)dx=∫abL(x,g1(x))dx.{\displaystyle \int _{C_{1}}L(x,y)\,dx=\int _{a}^{b}L(x,g_{1}(x))\,dx.}

WithC3, use the parametric equations:x =x,y =g2(x),a ≤x ≤b. Then∫C3L(x,y)dx=∫baL(x,y)dx=−∫abL(x,g2(x))dx.{\displaystyle \int _{C_{3}}L(x,y)\,dx=\int _{b}^{a}L(x,y)\,dx=-\int _{a}^{b}L(x,g_{2}(x))\,dx.}

The integral overC3 is negated because it goes in the negative direction fromb toa, asC is oriented positively (anticlockwise). OnC2 andC4,x remains constant, meaning∫C4L(x,y)dx=∫C2L(x,y)dx=0.{\displaystyle \int _{C_{4}}L(x,y)\,dx=\int _{C_{2}}L(x,y)\,dx=0.}

Therefore,

∮CLdx=∫C1L(x,y)dx+∫C2L(x,y)dx+∫C3L(x,y)dx+∫C4L(x,y)dx=∫abL(x,g1(x))dx−∫abL(x,g2(x))dx.{\displaystyle {\begin{aligned}\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{aligned}}}4

Combining (3) with (4), we get (1) for regions of type I. A similar treatment using the same endpoints 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

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We are going to prove the following

Theorem—LetΓ{\displaystyle \Gamma } be a rectifiable, positively orientedJordan curve inR2{\displaystyle \mathbb {R} ^{2}} and letR{\displaystyle R} denote its inner region. Suppose thatA,B:R¯→R{\displaystyle A,B:{\overline {R}}\to \mathbb {R} } are continuous functions with the property thatA{\displaystyle A} has second partial derivative at every point ofR{\displaystyle R},B{\displaystyle B} has first partial derivative at every point ofR{\displaystyle R} and that the functionsD1B,D2A:R→R{\displaystyle D_{1}B,D_{2}A:R\to \mathbb {R} } are Riemann-integrable overR{\displaystyle R}. Then∫Γ(Adx+Bdy)=∫R(D1B(x,y)−D2A(x,y))d(x,y).{\displaystyle \int _{\Gamma }(A\,dx+B\,dy)=\int _{R}\left(D_{1}B(x,y)-D_{2}A(x,y)\right)\,d(x,y).}

We need the following lemmas whose proofs can be found in:[3]

Lemma 1 (Decomposition Lemma)—AssumeΓ{\displaystyle \Gamma } is a rectifiable, positively oriented Jordan curve in the plane and letR{\displaystyle R} be its inner region. For every positive realδ{\displaystyle \delta }, letF(δ){\displaystyle {\mathcal {F}}(\delta )} denote the collection of squares in the plane bounded by the linesx=mδ,y=mδ{\displaystyle x=m\delta ,y=m\delta }, wherem{\displaystyle m} runs through the set of integers. Then, for thisδ{\displaystyle \delta }, there exists a decomposition ofR¯{\displaystyle {\overline {R}}} into a finite number of non-overlapping subregions in such a manner that

  1. Each one of the subregions contained inR{\displaystyle R}, sayR1,R2,…,Rk{\displaystyle R_{1},R_{2},\ldots ,R_{k}}, is a square fromF(δ){\displaystyle {\mathcal {F}}(\delta )}.
  2. Each one of the remaining subregions, sayRk+1,…,Rs{\displaystyle R_{k+1},\ldots ,R_{s}}, has as boundary a rectifiable Jordan curve formed by a finite number of arcs ofΓ{\displaystyle \Gamma } and parts of the sides of some square fromF(δ){\displaystyle {\mathcal {F}}(\delta )}.
  3. Each one of the border regionsRk+1,…,Rs{\displaystyle R_{k+1},\ldots ,R_{s}} can be enclosed in a square of edge-length2δ{\displaystyle 2\delta }.
  4. IfΓi{\displaystyle \Gamma _{i}} is the positively oriented boundary curve ofRi{\displaystyle R_{i}}, thenΓ=Γ1+Γ2+⋯+Γs.{\displaystyle \Gamma =\Gamma _{1}+\Gamma _{2}+\cdots +\Gamma _{s}.}
  5. The numbers−k{\displaystyle s-k} of border regions is no greater than4(Λδ+1){\textstyle 4\!\left({\frac {\Lambda }{\delta }}+1\right)}, whereΛ{\displaystyle \Lambda } is the length ofΓ{\displaystyle \Gamma }.

Lemma 2—LetΓ{\displaystyle \Gamma } be a rectifiable curve in the plane and letΔΓ(h){\displaystyle \Delta _{\Gamma }(h)} be the set of points in the plane whose distance from (the range of)Γ{\displaystyle \Gamma } is at mosth{\displaystyle h}. The outer Jordan content of this set satisfiesc¯ΔΓ(h)≤2hΛ+πh2{\displaystyle {\overline {c}}\,\,\Delta _{\Gamma }(h)\leq 2h\Lambda +\pi h^{2}}.

Lemma 3—LetΓ{\displaystyle \Gamma } be a rectifiable closed curve inR2{\displaystyle \mathbb {R} ^{2}} and letf:range of Γ→R{\displaystyle f:{\text{range of }}\Gamma \to \mathbb {R} } be a continuous function. Then|∫Γf(x,y)dy|≤12ΛΩf,{\displaystyle \left\vert \int _{\Gamma }f(x,y)\,dy\right\vert \leq {\frac {1}{2}}\Lambda \Omega _{f},} and|∫Γf(x,y)dx|≤12ΛΩf,{\displaystyle \left\vert \int _{\Gamma }f(x,y)\,dx\right\vert \leq {\frac {1}{2}}\Lambda \Omega _{f},}whereΩf{\displaystyle \Omega _{f}} is the oscillation off{\displaystyle f} on the range ofΓ{\displaystyle \Gamma }.

Now we are in position to prove the theorem:

Proof of Theorem. Letε{\displaystyle \varepsilon } be an arbitrary positivereal number. By continuity ofA{\displaystyle A},B{\displaystyle B} and compactness ofR¯{\displaystyle {\overline {R}}}, givenε>0{\displaystyle \varepsilon >0}, there exists0<δ<1{\displaystyle 0<\delta <1} such that whenever two points ofR¯{\displaystyle {\overline {R}}} are less than22δ{\displaystyle 2{\sqrt {2}}\,\delta } apart, their images underA,B{\displaystyle A,B} are less thanε{\displaystyle \varepsilon } apart. For thisδ{\displaystyle \delta }, consider the decomposition given by the previous Lemma. We have∫ΓAdx+Bdy=∑i=1k∫ΓiAdx+Bdy+∑i=k+1s∫ΓiAdx+Bdy.{\displaystyle \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.}

Putφ:=D1B−D2A{\displaystyle \varphi :=D_{1}B-D_{2}A}.

For eachi∈{1,…,k}{\displaystyle i\in \{1,\ldots ,k\}}, the curveΓi{\displaystyle \Gamma _{i}} is a positively oriented square, for which Green's formula holds. Hence∑i=1k∫ΓiAdx+Bdy=∑i=1k∫Riφ=∫⋃i=1kRiφ.{\displaystyle \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 .}

Every point of a border region is at a distance no greater than22δ{\displaystyle 2{\sqrt {2}}\,\delta } fromΓ{\displaystyle \Gamma }. Thus, ifK{\displaystyle K} is the union of all border regions, thenK⊂ΔΓ(22δ){\displaystyle K\subset \Delta _{\Gamma }(2{\sqrt {2}}\,\delta )}; hencec(K)≤c¯ΔΓ(22δ)≤42δ+8πδ2{\displaystyle c(K)\leq {\overline {c}}\,\Delta _{\Gamma }(2{\sqrt {2}}\,\delta )\leq 4{\sqrt {2}}\,\delta +8\pi \delta ^{2}}, by Lemma 2. Notice that∫Rφ−∫⋃i=1kRiφ=∫Kφ.{\displaystyle \int _{R}\varphi \,\,-\int _{\bigcup _{i=1}^{k}R_{i}}\varphi =\int _{K}\varphi .}This yields|∑i=1k∫ΓiAdx+Bdy−∫Rφ|≤Mδ(1+π2δ) for some M>0.{\displaystyle \left\vert \sum _{i=1}^{k}\int _{\Gamma _{i}}A\,dx+B\,dy\quad -\int _{R}\varphi \right\vert \leq M\delta (1+\pi {\sqrt {2}}\,\delta ){\text{ for some }}M>0.}

We may as well chooseδ{\displaystyle \delta } so that the RHS of the last inequality is<ε.{\displaystyle <\varepsilon .}

The remark in the beginning of this proof implies that the oscillations ofA{\displaystyle A} andB{\displaystyle B} on every border region is at mostε{\displaystyle \varepsilon }. We have|∑i=k+1s∫ΓiAdx+Bdy|≤12ε∑i=k+1sΛi.{\displaystyle \left\vert \sum _{i=k+1}^{s}\int _{\Gamma _{i}}A\,dx+B\,dy\right\vert \leq {\frac {1}{2}}\varepsilon \sum _{i=k+1}^{s}\Lambda _{i}.}

By Lemma 1(iii),∑i=k+1sΛi≤Λ+(4δ)4(Λδ+1)≤17Λ+16.{\displaystyle \sum _{i=k+1}^{s}\Lambda _{i}\leq \Lambda +(4\delta )\,4\!\left({\frac {\Lambda }{\delta }}+1\right)\leq 17\Lambda +16.}

Combining these, we finally get|∫ΓAdx+Bdy−∫Rφ|<Cε,{\displaystyle \left\vert \int _{\Gamma }A\,dx+B\,dy\quad -\int _{R}\varphi \right\vert <C\varepsilon ,}for someC>0{\displaystyle C>0}. Since this is true for everyε>0{\displaystyle \varepsilon >0}, we are done.

Validity under different hypotheses

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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 functionsA,B:R¯→R{\displaystyle A,B:{\overline {R}}\to \mathbb {R} } are still assumed to be continuous. However, we now require them to be Fréchet-differentiable at every point ofR{\displaystyle R}. This implies the existence of all directional derivatives, in particularDeiA=:DiA,DeiB=:DiB,i=1,2{\displaystyle D_{e_{i}}A=:D_{i}A,D_{e_{i}}B=:D_{i}B,\,i=1,2}, where, as usual,(e1,e2){\displaystyle (e_{1},e_{2})} is the canonical ordered basis ofR2{\displaystyle \mathbb {R} ^{2}}. In addition, we require the functionD1B−D2A{\displaystyle D_{1}B-D_{2}A} to be Riemann-integrable overR{\displaystyle R}.

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

Theorem (Cauchy)—IfΓ{\displaystyle \Gamma } is a rectifiable Jordan curve inC{\displaystyle \mathbb {C} } and iff:closure of inner region of Γ→C{\displaystyle f:{\text{closure of inner region of }}\Gamma \to \mathbb {C} } is a continuous mapping holomorphic throughout the inner region ofΓ{\displaystyle \Gamma }, then∫Γf=0,{\displaystyle \int _{\Gamma }f=0,} the integral being a complex contour integral.

Proof

We regard the complex plane asR2{\displaystyle \mathbb {R} ^{2}}. Now, defineu,v:R¯→R{\displaystyle u,v:{\overline {R}}\to \mathbb {R} } to be such thatf(x+iy)=u(x,y)+iv(x,y).{\displaystyle f(x+iy)=u(x,y)+iv(x,y).} These functions are clearly continuous. It is well known thatu{\displaystyle u} andv{\displaystyle v} are Fréchet-differentiable and that they satisfy the Cauchy-Riemann equations:D1v+D2u=D1u−D2v=zero function{\displaystyle D_{1}v+D_{2}u=D_{1}u-D_{2}v={\text{zero function}}}.

Now, analyzing the sums used to define the complex contour integral in question, it is easy to realize that∫Γf=∫Γudx−vdy+i∫Γvdx+udy,{\displaystyle \int _{\Gamma }f=\int _{\Gamma }u\,dx-v\,dy\quad +i\int _{\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.

Multiply-connected regions

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Theorem. LetΓ0,Γ1,…,Γn{\displaystyle \Gamma _{0},\Gamma _{1},\ldots ,\Gamma _{n}} be positively oriented rectifiable Jordan curves inR2{\displaystyle \mathbb {R} ^{2}} satisfyingΓi⊂R0,if 1≤i≤nΓi⊂R2∖R¯j,if 1≤i,j≤n and i≠j,{\displaystyle {\begin{aligned}\Gamma _{i}\subset R_{0},&&{\text{if }}1\leq i\leq n\\\Gamma _{i}\subset \mathbb {R} ^{2}\setminus {\overline {R}}_{j},&&{\text{if }}1\leq i,j\leq n{\text{ and }}i\neq j,\end{aligned}}}whereRi{\displaystyle R_{i}} is the inner region ofΓi{\displaystyle \Gamma _{i}}. LetD=R0∖(R¯1∪R¯2∪⋯∪R¯n).{\displaystyle D=R_{0}\setminus ({\overline {R}}_{1}\cup {\overline {R}}_{2}\cup \cdots \cup {\overline {R}}_{n}).}

Supposep:D¯→R{\displaystyle p:{\overline {D}}\to \mathbb {R} } andq:D¯→R{\displaystyle q:{\overline {D}}\to \mathbb {R} } are continuous functions whose restriction toD{\displaystyle D} is Fréchet-differentiable. If the function(x,y)⟼∂q∂e1(x,y)−∂p∂e2(x,y){\displaystyle (x,y)\longmapsto {\frac {\partial q}{\partial e_{1}}}(x,y)-{\frac {\partial p}{\partial e_{2}}}(x,y)}is Riemann-integrable overD{\displaystyle D}, then∫Γ0p(x,y)dx+q(x,y)dy−∑i=1n∫Γip(x,y)dx+q(x,y)dy=∫D{∂q∂e1(x,y)−∂p∂e2(x,y)}d(x,y).{\displaystyle {\begin{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)\right\}\,d(x,y).\end{aligned}}}

Relationship to Stokes' theorem

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Green's theorem is a special case of theKelvin–Stokes theorem, when applied to a region in thexy{\displaystyle xy}-plane.

We can augment the two-dimensional field into a three-dimensional field with az component that is always 0. WriteF for thevector-valued functionF=(L,M,0){\displaystyle \mathbf {F} =(L,M,0)}. Start with the left side of Green's theorem:∮C(Ldx+Mdy)=∮C(L,M,0)⋅(dx,dy,dz)=∮CF⋅dr.{\displaystyle \oint _{C}(L\,dx+M\,dy)=\oint _{C}(L,M,0)\cdot (dx,dy,dz)=\oint _{C}\mathbf {F} \cdot d\mathbf {r} .}

The Kelvin–Stokes theorem:∮CF⋅dr=∬S∇×F⋅n^dS.{\displaystyle \oint _{C}\mathbf {F} \cdot d\mathbf {r} =\iint _{S}\nabla \times \mathbf {F} \cdot \mathbf {\hat {n}} \,dS.}

The surfaceS{\displaystyle S} is just the region in the planeD{\displaystyle D}, with the unit normaln^{\displaystyle \mathbf {\hat {n}} } 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∇×F⋅n^=[(∂0∂y−∂M∂z)i+(∂L∂z−∂0∂x)j+(∂M∂x−∂L∂y)k]⋅k=(∂M∂x−∂L∂y).{\displaystyle \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∬S∇×F⋅n^dS=∬D(∂M∂x−∂L∂y)dA.{\displaystyle \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 usingdifferential forms andexterior derivatives:∮CLdx+Mdy=∮∂Dω=∫Ddω=∫D∂L∂ydy∧dx+∂M∂xdx∧dy=∬D(∂M∂x−∂L∂y)dxdy.{\displaystyle \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

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Considering only two-dimensional vector fields, Green's theorem is equivalent to the two-dimensional version of thedivergence theorem:

∬D(∇⋅F)dA=∮CF⋅n^ds,{\displaystyle \iint _{D}\left(\nabla \cdot \mathbf {F} \right)dA=\oint _{C}\mathbf {F} \cdot \mathbf {\hat {n}} \,ds,}

where∇⋅F{\displaystyle \nabla \cdot \mathbf {F} } is the divergence on the two-dimensional vector fieldF{\displaystyle \mathbf {F} }, andn^{\displaystyle \mathbf {\hat {n}} } is the outward-pointing unit normal vector on the boundary.

To see this, consider the unit normaln^{\displaystyle \mathbf {\hat {n}} } in the right side of the equation. Since in Green's theoremdr=(dx,dy){\displaystyle d\mathbf {r} =(dx,dy)} is a vector pointing tangential along the curve, and the curveC 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(dy,−dx){\displaystyle (dy,-dx)}. The length of this vector isdx2+dy2=ds.{\textstyle {\sqrt {dx^{2}+dy^{2}}}=ds.} So(dy,−dx)=n^ds.{\displaystyle (dy,-dx)=\mathbf {\hat {n}} \,ds.}

Start with the left side of Green's theorem:∮C(Ldx+Mdy)=∮C(M,−L)⋅(dy,−dx)=∮C(M,−L)⋅n^ds.{\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 withF=(M,−L){\displaystyle \mathbf {F} =(M,-L)}, we get the right side of Green's theorem:∮C(M,−L)⋅n^ds=∬D(∇⋅(M,−L))dA=∬D(∂M∂x−∂L∂y)dA.{\displaystyle \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

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Green's theorem can be used to compute area by line integral.[4] The area of a planar regionD{\displaystyle D} is given byA=∬DdA.{\displaystyle A=\iint _{D}dA.}

ChooseL{\displaystyle L} andM{\displaystyle M} such that∂M∂x−∂L∂y=1{\displaystyle {\frac {\partial M}{\partial x}}-{\frac {\partial L}{\partial y}}=1}, the area is given byA=∮C(Ldx+Mdy).{\displaystyle A=\oint _{C}(L\,dx+M\,dy).}

Possible formulas for the area ofD{\displaystyle D} include[4]A=∮Cxdy=−∮Cydx=12∮C(−ydx+xdy).{\displaystyle A=\oint _{C}x\,dy=-\oint _{C}y\,dx={\tfrac {1}{2}}\oint _{C}(-y\,dx+x\,dy).}

History

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It is named afterGeorge Green, who stated a similar result in an 1828 paper titledAn Essay on the Application of Mathematical Analysis to the Theories of Electricity and Magnetism. In 1846,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. 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 onpages 10–12 of hisEssay.
In 1846, the form of "Green's theorem" which appears in this article was first published, without proof, in an article byAugustin 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 functionk along the curves that encloses the areaS.)
A proof of the theorem was finally provided in 1851 byBernhard 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.[5]

See also

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  • Planimeter – Tool for measuring area
  • 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

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  1. ^Riley, Kenneth F.; Hobson, Michael P.; Bence, Stephen J. (2010).Mathematical methods for physics and engineering (3rd ed.). Cambridge:Cambridge University Press.ISBN 978-0-521-86153-3.
  2. ^Lipschutz, Seymour;Spiegel, Murray R. (2009).Vector analysis and an introduction to tensor analysis. Schaum's outline series (2nd ed.). New York:McGraw Hill Education.ISBN 978-0-07-161545-7.OCLC 244060713.
  3. ^Apostol, Tom (1960).Mathematical Analysis. Reading, Massachusetts, U.S.A.:Addison-Wesley.OCLC 6699164.
  4. ^abStewart, James (1999).Calculus. GWO - A Gary W. Ostedt book (4. ed.). Pacific Grove, Calif. London:Brooks/Cole.ISBN 978-0-534-35949-2.
  5. ^Katz, Victor J. (2009). "22.3.3: Complex Functions and Line Integrals".A history of mathematics: an introduction(PDF) (3. ed.). Boston, Mass. Munich:Addison-Wesley. pp. 801–5.ISBN 978-0-321-38700-4.

Further reading

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External links

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