Difference between revisions of "Green's Theorem"

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In vector calculus, '''Green's theorem''' relates a line integral around a simple closed curve {{mvar|C}} to a double integral over the plane region {{mvar|D}} bounded by {{mvar|C}}. It is the two-dimensional special case of Stokes' theorem.
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==Theorem==
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Let {{mvar|C}} be a positively oriented, piecewise smooth, simple closed curve in a 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 region containing {{mvar|D}} and having continuous partial derivatives there, then
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<math>\oint_{\scriptstyle C} (L\, dx + M\, dy) = \iint_{D} \left(\frac{\partial M}{\partial x} - \frac{\partial L}{\partial y}\right) dx\, dy</math>
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where the path of integration along {{mvar|C}} is anticlockwise.
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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. 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.
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==Proof when ''D'' is a simple region==
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[[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.]]
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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.
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If it can be shown that if
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<math>\oint_C L\, dx = \iint_{D} \left(- \frac{\partial L}{\partial y}\right) dA \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;</math> (1)
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and
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<math>\oint_C\ M\, dy = \iint_{D} \left(\frac{\partial M}{\partial x}\right) dA \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;</math> (2)
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are true, then Green's theorem follows immediately for the region D. We can prove the first equation easily for regions of type I, and the second equation for regions of type II. Green's theorem then follows for regions of type III.
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Assume region ''D'' is a type I region and can thus be characterized, as pictured on the right, by
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:<math>D = \{(x,y)\mid a\le x\le b, g_1(x) \le y \le g_2(x)\}</math>
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where ''g''<sub>1</sub> and ''g''<sub>2</sub> are continuous functions on [''a'', ''b'']. Compute the double integral in (1):
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<math>
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\begin{align}
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\iint_D \frac{\partial L}{\partial y}\, dA
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& = \int_a^b\,\int_{g_1(x)}^{g_2(x)} \frac{\partial L}{\partial y} (x,y)\,dy\,dx \\
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& = \int_a^b \left[L(x,g_2(x)) - L(x,g_1(x)) \right] \, dx.
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\end{align}
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\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;</math> (3)
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Now compute the line integral in (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>.
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With ''C''<sub>1</sub>, use the parametric equations: ''x'' = ''x'', ''y'' = ''g''<sub>1</sub>(''x''), ''a'' ≤ ''x'' ≤ ''b''. Then
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:<math>\int_{C_1} L(x,y)\, dx = \int_a^b L(x,g_1(x))\, dx.</math>
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With ''C''<sub>3</sub>, use the parametric equations: ''x'' = ''x'', ''y'' = ''g''<sub>2</sub>(''x''), ''a'' ≤ ''x'' ≤ ''b''. Then
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:<math> \int_{C_3} L(x,y)\, dx = -\int_{-C_3} L(x,y)\, dx = - \int_a^b L(x,g_2(x))\, dx.</math>
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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
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:<math> \int_{C_4} L(x,y)\, dx = \int_{C_2} L(x,y)\, dx = 0.</math>
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Therefore,
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<math>
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\begin{align}
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\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 \\
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& = \int_a^b L(x,g_1(x))\, dx - \int_a^b L(x,g_2(x))\, dx .
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\end{align}
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\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;</math> (4)
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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.
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== Proof for rectifiable Jordan curves ==
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We are going to prove the following
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'''Theorem.''' Let <math>\Gamma</math> be a rectifiable, positively oriented 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
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<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>
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We need the following lemmas whose proofs can be found in:
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'''Lemma 1 (Decomposition Lemma).''' 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
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<ol type="i">
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<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>
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<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>
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<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>
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<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>
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<li>The number <math>s-k</math> of border regions is no greater than <math>4\!\left(\frac{\Lambda}{\delta} + 1\right)</math>, where <math>\Lambda</math> is the length of <math>\Gamma</math>.</li>
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</ol>
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'''Lemma 2.''' 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>.
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'''Lemma 3.''' 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
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: <math>\left\vert\int_\Gamma f(x,y)\,dy\right\vert</math> and
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: <math>\left\vert\int_\Gamma f(x,y)\,dx\right\vert</math> are <math>\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>.
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Now we are in position to prove the Theorem:
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'''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
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: <math>\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>
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Put <math>\varphi := D_1 B - D_2 A</math>.
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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
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: <math>\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>
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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
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: <math>\int_R \varphi\,\,-\int_{\bigcup_{i=1}^k R_i} \varphi=\int_K \varphi.</math>
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This yields
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: <math>\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>
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We may as well choose <math>\delta</math> so that the RHS of the last inequality is <math><\varepsilon.</math>
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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
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: <math>\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>
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By Lemma 1(iii),
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: <math>\sum_{i=k+1}^s \Lambda_i \le\Lambda + (4\delta)\,4\!\left(\frac{\Lambda}{\delta}+1\right)\le17\Lambda+16.</math>
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Combining these, we finally get
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: <math>\left\vert\int_\Gamma A\,dx+B\,dy\quad-\int_R \varphi\right\vert< C \varepsilon,</math>
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for some <math>C > 0</math>. Since this is true for every <math>\varepsilon > 0</math>, we are done.
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== 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:
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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>.
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As a corollary of this, we get the Cauchy Integral Theorem for rectifiable Jordan curves:
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'''Theorem (Cauchy).''' 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
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: <math>\int_\Gamma f=0,</math>
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the integral being a complex contour integral.
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'''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>.
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Now, analyzing the sums used to define the complex contour integral in question, it is easy to realize that
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: <math>\int_\Gamma f=\int_\Gamma u\,dx-v\,dy\quad+i\int_\Gamma v\,dx+u\,dy,</math>
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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.
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== Multiply-connected regions ==
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'''Theorem.''' Let <math>\Gamma_0,\Gamma_1,\ldots,\Gamma_n</math> be positively oriented rectifiable Jordan curves in <math>\R^{2}</math> satisfying
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: <math>\begin{aligned}
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\Gamma_i \subset R_0,&&\text{if } 1\le i\le n\\
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\Gamma_i \subset \R^2 \setminus \overline{R}_j,&&\text{if }1\le i,j \le n\text{ and }i\ne j,
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\end{aligned}</math>
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where <math>R_i</math> is the inner region of <math>\Gamma_i</math>. Let
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: <math>D = R_0 \setminus (\overline{R}_1 \cup \overline{R}_2 \cup \cdots \cup \overline{R}_n).</math>
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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
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: <math>(x,y)\longmapsto\frac{\partial q}{\partial e_1}(x,y)-\frac{\partial p}{\partial e_2}(x,y)</math>
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is Riemann-integrable over <math>D</math>, then
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: <math>
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\begin{align}
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& \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]
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= {} & \int_D\left\{\frac{\partial q}{\partial e_1}(x,y)-\frac{\partial p}{\partial e_2}(x,y)\right\} \, d(x,y).
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\end{align}
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</math>
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==Relationship to Stokes' theorem==
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Green's theorem is a special case of the Kelvin–Stokes theorem, when applied to a region in the <math>xy</math>-plane.
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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 <math>\mathbf{F}=(L,M,0)</math>. Start with the left side of Green's theorem:
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:<math>\oint_C (L\, dx + M\, dy) = \oint_C (L, M, 0) \cdot (dx, dy, dz) = \oint_C \mathbf{F} \cdot d\mathbf{r}. </math>
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The Kelvin–Stokes theorem:
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:<math>\oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_S \nabla \times \mathbf{F} \cdot \mathbf{\hat n} \, dS. </math>
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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.
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The expression inside the integral becomes
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: <math>\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>
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Thus we get the right side of Green's theorem
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:<math>\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>
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Green's theorem is also a straightforward result of the general Stokes' theorem using differential forms and exterior derivatives:
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: <math>
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\begin{align}
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& \oint_C L \,dx + M \,dy = \oint_{\partial D} \omega = \int_D \,d\omega \\[5pt]
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= {} & \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.
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\end{align}
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</math>
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==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 the divergence theorem:
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<math>\iiint_V\left(\mathbf{\nabla}\cdot\mathbf{F}\right)\,dV= \oiint_{\partial \scriptstyle V} (\mathbf{F}\cdot\mathbf{\hat n})\,dS .</math>
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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.
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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>
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Start with the left side of Green's theorem:
 +
:<math>\oint_C (L\, dx + M\, dy) = \oint_C (M, -L) \cdot (dy, -dx) = \oint_C (M, -L) \cdot \mathbf{\hat n}\,ds.</math>
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Applying the two-dimensional divergence theorem with <math>\mathbf{F} = (M, -L)</math>, we get the right side of Green's theorem:
 +
:<math>\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>
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==Area calculation==
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Green's theorem can be used to compute area by line integral. The area of a planar region <math>D</math> is given by
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:<math>A = \iint_D dA.</math>
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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
 +
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:<math>A = \oint_{C} (L\, dx + M\, dy).</math>
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 +
Possible formulas for the area of <math>D</math> include
 +
 +
:<math>A=\oint_C x\, dy = -\oint_C y\, dx = \tfrac 12 \oint_C (-y\, dx + x\, dy).</math>
 +
 +
 +
 +
==Resources==
 
* [https://www.youtube.com/watch?v=sDby9VzPoj4 Green's Theorem Part 1] by James Sousa, Math is Power 4U
 
* [https://www.youtube.com/watch?v=sDby9VzPoj4 Green's Theorem Part 1] by James Sousa, Math is Power 4U
 
* [https://www.youtube.com/watch?v=oXISOEqd0X4 Green's Theorem Part 2] by James Sousa, Math is Power 4U
 
* [https://www.youtube.com/watch?v=oXISOEqd0X4 Green's Theorem Part 2] by James Sousa, Math is Power 4U
Line 20: Line 249:
 
* [https://www.youtube.com/watch?v=gGXnILbrhsM Green's Theorem Example 1] by Khan Academy
 
* [https://www.youtube.com/watch?v=gGXnILbrhsM Green's Theorem Example 1] by Khan Academy
 
* [https://www.youtube.com/watch?v=sSyPAAyL8nQ Green's Theorem Example 2] by Khan Academy
 
* [https://www.youtube.com/watch?v=sSyPAAyL8nQ Green's Theorem Example 2] by Khan Academy
 +
 +
==Licensing==
 +
Content obtained and/or adapted from:
 +
* [https://en.wikipedia.org/wiki/Green%27s_theorem Green's Theorem, Wikipedia] under a CC BY-SA license

Latest revision as of 17:14, 2 November 2021

In vector calculus, Green's theorem relates a line integral around a simple closed curve C to a double integral over the plane region D bounded by C. It is the two-dimensional special case of 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.

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

If D is a simple type of 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

(1)

and

(2)

are true, then Green's theorem follows immediately for the region D. We can prove the first equation easily for regions of type I, and the second equation 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):

(3)

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,

(4)

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 be a rectifiable, positively oriented Jordan curve in and let denote its inner region. Suppose that are continuous functions with the property that has second partial derivative at every point of , has first partial derivative at every point of and that the functions are Riemann-integrable over . Then

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

Lemma 1 (Decomposition Lemma). Assume is a rectifiable, positively oriented Jordan curve in the plane and let be its inner region. For every positive real , let denote the collection of squares in the plane bounded by the lines , where runs through the set of integers. Then, for this , there exists a decomposition of into a finite number of non-overlapping subregions in such a manner that

  1. Each one of the subregions contained in , say , is a square from .
  2. Each one of the remaining subregions, say , has as boundary a rectifiable Jordan curve formed by a finite number of arcs of and parts of the sides of some square from .
  3. Each one of the border regions can be enclosed in a square of edge-length .
  4. If is the positively oriented boundary curve of , then
  5. The number of border regions is no greater than , where is the length of .

Lemma 2. Let be a rectifiable curve in the plane and let be 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. Let be a rectifiable curve in and let be a continuous function. Then

and
are where is the oscillation of on the range of .

Now we are in position to prove the Theorem:

Proof of Theorem. Let be an arbitrary positive real number. By continuity of , and compactness of , given , there exists such that whenever two points of are less than apart, their images under are less than apart. For this , consider the decomposition given by the previous Lemma. We have

Put .

For each , the curve is a positively oriented square, for which Green's formula holds. Hence

Every point of a border region is at a distance no greater than from . Thus, if is the union of all border regions, then ; hence , by Lemma 2. Notice that

This yields

We may as well choose so that the RHS of the last inequality is

The remark in the beginning of this proof implies that the oscillations of and on 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 functions are 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 function to be Riemann-integrable over .

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

Theorem (Cauchy). If is a rectifiable Jordan curve in and if is 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, define to be such that These functions are clearly continuous. It is well known that and are Fréchet-differentiable and that they satisfy the Cauchy-Riemann equations: .

Now, analyzing 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.

Multiply-connected regions

Theorem. Let be positively oriented rectifiable Jordan curves in satisfying

where is the inner region of . Let

Suppose and are continuous functions whose restriction to is Fréchet-differentiable. If the function

is Riemann-integrable over , then

Relationship to Stokes' theorem

Green's theorem is a special case of the Kelvin–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 the vector-valued function . Start with the left side of Green's theorem:

The Kelvin–Stokes theorem:

The surface is just the region in the plane , with the unit normal 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

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:



where is the divergence on the two-dimensional vector field , and is the outward-pointing unit normal vector on the boundary.

To see this, consider the unit normal in the right side of the equation. Since in Green's theorem 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 . The length of this vector is So

Start with the left side of Green's theorem:

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. The area of a planar region is given by

Choose and such that , the area is given by

Possible formulas for the area of include


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