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Green's Theorem

Green's theorem relates a line integral around a positively oriented boundary curve to a double integral over the region enclosed by that curve.
GREEN'S THEOREM
Let CC be a positively oriented, piecewise-smooth, simple closed curve in the plane, and let DD be the region bounded by CC. If PP and QQ have continuous partial derivatives on an open region that contains DD, then
∮CP dx+Q dy=∬D(∂Q∂x−∂P∂y) dA.\begin{equation*}\oint_C P\,dx+Q\,dy =\iint_D\left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right)\,dA.\end{equation*}

Motivation

Green's Theorem is named after the self-taught English scientist George Green (1793–1841). He worked full-time in his father's bakery from the age of nine and taught himself mathematics from library books. In 1828 he published privately An Essay on the Application of Mathematical Analysis to the Theories of Electricity and Magnetism, but only 100 copies were printed and most of those went to his friends. This pamphlet contained a theorem that is equivalent to what we know as Green's Theorem, but it didn't become widely known at that time. Finally, at age 40, Green entered Cambridge University as an undergraduate but died four years after graduation. In 1846 William Thomson (Lord Kelvin) located a copy of Green's essay, realized its significance, and had it reprinted. Green was the first person to try to formulate a mathematical theory of electricity and magnetism. His work was the basis for the subsequent electromagnetic theories of Thomson, Stokes, Rayleigh, and Maxwell.

Proof of Green's Theorem

Proof for special regions

It is enough to prove the two identities separately
∮CP dx=−∬D∂P∂y dA and ∮CQ dy=∬D∂Q∂x dA.\begin{equation*}\oint_C P\,dx =- \iint_D \frac{\partial P}{\partial y}\,dA \quad \text{ and }\oint_C Q\,dy = \iint_D \frac{\partial Q}{\partial x}\,dA.\end{equation*}
We first prove the identity involving P dxP\,dxfor simpel region D D :
D={(x,y)∣a≤x≤b,  g1(x)≤y≤g2(x)},\begin{equation*}D=\{(x,y)\mid a\leq x\leq b,\; g_1(x)\leq y\leq g_2(x)\},\end{equation*}
where g1g_1 and g2g_2 are continuous functions.
Then
∬D∂P∂y dy dx=∫ab∫g1(x)g2(x)∂P∂y(x,y) dy dx=∫ab[P(x,g2(x))−P(x,g1(x))] dx⏟The Fundamental Theorem of Calculus.\begin{equation*}\iint_D \frac{\partial P}{\partial y}\,dy\,dx= \int_a^b \int_{g_1(x)}^{g_2(x)} \frac{\partial P}{\partial y}(x,y)\,dy\,dx= \underbrace{\int_a^b \left[P(x,g_2(x))-P(x,g_1(x))\right]\,dx}_{\text{The Fundamental Theorem of Calculus}}.\end{equation*}
Now compute the line integral by breaking CC into the four oriented pieces C1C_1, C2C_2, C3C_3, and C4C_4. On C1C_1, we use xx as the parameter and write
∫C1P(x,y) dx=∫abP(x,g1(x)) dx.\begin{equation*}\int_{C_1} P(x,y)\,dx = \int_a^b P(x,g_1(x))\,dx.\end{equation*}
The curve C3C_3 goes from right to left. Therefore −C3-C_3 goes from left to right, so we can parametrize −C3-C_3 by
∫C3P(x,y) dx=−∫−C3P(x,y) dx=−∫abP(x,g2(x)) dx.\begin{equation*}\int_{C_3} P(x,y)\,dx = - \int_{-C_3} P(x,y)\,dx = -\int_a^b P(x,g_2(x))\,dx.\end{equation*}
On C2C_2 and C4C_4, the variable xx is constant, so correspnding integrals are 0 and we get
∮CP(x,y) dx=∫C1P(x,y) dx+∫C2P(x,y) dx⏟=0+∫C3P(x,y) dx+∫C4P(x,y) dx⏟=0=∫abP(x,g1(x)) dx−∫abP(x,g2(x)) dx.\begin{align*}\oint_C P(x,y)\,dx &= \int_{C_1} P(x,y)\,dx + \underbrace{\int_{C_2} P(x,y)\,dx}_{=0} + \int_{C_3} P(x,y)\,dx + \underbrace{\int_{C_4} P(x,y)\,dx}_{=0} \\&= \int_a^b P(x,g_1(x))\,dx - \int_a^b P(x,g_2(x))\,dx.\end{align*}
Comparing this with the double integral computed above gives
∮CP(x,y) dx=−∬D∂P∂y dA.\begin{equation*}\oint_C P(x,y)\,dx = - \iint_D \frac{\partial P}{\partial y}\,dA.\end{equation*}
The identity involving Q dyQ\,dy is proved in the same way by writing DD as
D={(x,y)∣c≤y≤d,  h1(y)≤x≤h2(y)}.\begin{equation*}D=\{(x,y)\mid c\leq y\leq d,\; h_1(y)\leq x\leq h_2(y)\}.\end{equation*}
Here C1C_1 and C3C_3 are horizontal, so dy=0dy=0 on these sides. Integrating first with respect to xx gives
∮CQ dy=∬D∂Q∂x dA.\begin{equation*}\oint_C Q\,dy = \iint_D \frac{\partial Q}{\partial x}\,dA.\end{equation*}
Together, the two identities prove Green's theorem for simple regions.

Union of Simple Regions

The proof for simple regions extends to finite unions of simple regions. Suppose first that D=D1∪D2D=D_1\cup D_2, where D1D_1 and D2D_2 are simple and their interiors do not overlap. If the boundary of D1D_1 is C1∪C3C_1\cup C_3 and the boundary of D2D_2 is C2∪(−C3)C_2\cup(-C_3), then applying Green's theorem to D1D_1 and D2D_2 separately gives
∮C1∪C3P dx+Q dy=∬D1(∂Q∂x−∂P∂y) dA\begin{equation*}\oint_{C_1\cup C_3} P\,dx+Q\,dy=\iint_{D_1}\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)\,dA\end{equation*}
and
∮C2∪(−C3)P dx+Q dy=∬D2(∂Q∂x−∂P∂y) dA.\begin{equation*}\oint_{C_2\cup(-C_3)} P\,dx+Q\,dy=\iint_{D_2}\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)\,dA.\end{equation*}
When we add these two equations, the line integrals along C3C_3 and −C3-C_3 cancel. Therefore
∮C1∪C2P dx+Q dy=∬D(∂Q∂x−∂P∂y) dA,\begin{equation*}\oint_{C_1\cup C_2} P\,dx+Q\,dy =\iint_D \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right)\,dA,\end{equation*}
which is Green's theorem for DD.
For a finite union of nonoverlapping simple regions, the same cancellation occurs along every interior boundary segment. Each such segment is traversed twice, once in each direction, so the interior line integrals cancel pairwise. What remains is only the integral over the outer boundary of DD, while the double integrals over the simple pieces add up to the double integral over all of DD.

Region with Hole

Green's theorem also applies to regions with holes. In this case the boundary CC consists of the outer boundary curve C1C_1 and the inner boundary curve C2C_2. The positive orientation keeps the region on the left, so C1C_1 is oriented counterclockwise and C2C_2 is oriented clockwise.
Cut the region DD into two simple regions D1D_1 and D2D_2. Applying Green's theorem to D1D_1 and D2D_2 separately gives
∬D(∂Q∂x−∂P∂y) dA=∬D1(∂Q∂x−∂P∂y) dA+∬D2(∂Q∂x−∂P∂y) dA=∮∂D1P dx+Q dy+∮∂D2P dx+Q dy.\begin{align*}\iint_D\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)\,dA&=\iint_{D_1}\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)\,dA+\iint_{D_2}\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)\,dA \\&=\oint_{\partial D_1} P\,dx+Q\,dy+\oint_{\partial D_2} P\,dx+Q\,dy.\end{align*}
The line integrals over the artificial cut lines cancel because each cut is traversed once in each direction. What remains is the integral over the true boundary:
∬D(∂Q∂x−∂P∂y) dA=∮C1P dx+Q dy+∮C2P dx+Q dy=∮CP dx+Q dy.\begin{equation*}\iint_D\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)\,dA=\oint_{C_1} P\,dx+Q\,dy+\oint_{C_2} P\,dx+Q\,dy=\oint_C P\,dx+Q\,dy.\end{equation*}
Thus Green's theorem holds for the region with a hole.

References

  • [StewartEtAl2021]Stewart, James, Daniel K. Clegg, and Saleem Watson. Calculus: Early Transcendentals, Metric Edition. Ninth edition. Cengage, 2021.