Green’s Theorem

Green's theorem gives a relationship between the line integral of a two-dimensional vector field over a closed path in the plane and the double integral over the region it encloses. The fact that the integral of a (two-dimensional) conservative field over a closed path is zero is a special case of Green's theorem.

Green's theorem is itself a special case of the much more general Stokes' theorem. The statement in Green's theorem that two different types of integrals are equal can be used to compute either type: sometimes Green's theorem is used to transform a line integral into a double integral, and sometimes it is used to transform a double integral into a line integral.

Green's 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 $P$ and $Q$ are functions of $(x, y)$ defined on an open region containing $D$ and have continuous partial derivatives there, then $$\oint_C (P \, dx + Q \, dy) = \iint_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dx \, dy,$$ where the path integral is traversed counterclockwise.

Evaluate the integral $$\oint_C \big(y^2 \, dx + x^2 \, dy\big),$$ where $C$ is the boundary of the upper half of the unit disk, traversed counterclockwise.

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The boundary is defined piecewise, so this integral would be tedious to compute directly. Green's theorem gives $$\oint_C \big(y^2 \, dx + x^2 \, dy\big) = \iint_D (2x-2y) dx \, dy,$$ where $D$ is the upper half disk. This integral can be computed easily as $$\begin{aligned} \int_{-1}^1 \int_0^{\sqrt{1-x^2}} (2x-2y) dy \, dx &= \int_{-1}^1 \big(2xy-y^2\big) \Big|_0^{\sqrt{1-x^2}} \, dx \\ &= \int_{-1}^1 \left( 2x\sqrt{1-x^2} - \big(1-x^2\big) \right) \, dx \\ &= 0 - \int_{-1}^1 \big(1-x^2\big) \, dx \\ &=- \left.\Big(x-\frac{x^3}{3}\Big)\right|_{-1}^1 = -2+\frac{2}{3} = -\frac{4}{3}.\ _\square \end{aligned}$$ (The integral could also be computed using polar coordinates.)

Let $G$ be a continuous function of two variables with continuous partial derivatives, and let ${\bf F} = \nabla G$ be the gradient of $G,$ defined by ${\bf F} = \left( \dfrac{\partial G}{\partial x}, \dfrac{\partial G}{\partial y} \right).$ Show that $$\oint_C {\bf F} \cdot (dx,dy) = \oint_C \left( \dfrac{\partial G}{\partial x} \, dx + \dfrac{\partial G}{\partial y} \, dy \right) = 0$$ for any closed curve $C.$

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This is a straightforward application of Green's theorem: $$\begin{aligned} \oint_C \left( \dfrac{\partial G}{\partial x} \, dx + \dfrac{\partial G}{\partial y} \, dy \right) &= \iint_R \left( \dfrac{\partial^2 G}{\partial y \partial x} - \dfrac{\partial^2 G}{\partial x \partial y} \right) dx \, dy = \iint_R 0 \, dx \, dy = 0, \end{aligned}$$ because the mixed partial derivatives $\dfrac{\partial^2 G}{\partial x \partial y}$ and $\dfrac{\partial^2 G}{\partial y \partial x}$ are equal. $_\square$

Green's theorem can be used "in reverse" to compute certain double integrals as well. It is necessary that the integrand be expressible in the form given on the right side of Green's theorem. Here is a very useful example.

Let $R$ be a plane region enclosed by a simple closed curve $C.$ Show that the area of $R$ equals any of the following integrals (where the path is traversed counterclockwise): $$\oint_C x \, dy, \quad -\oint_C y \, dx, \quad \frac12 \oint_C (x \, dy - y \, dx).$$

As an application, compute the area of an ellipse with semi-major axes $a$ and $b.$

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The proof of the first statement is immediate: Green's theorem applied to any of the three integrals above shows that they all equal $$\iint_R 1 \, dx \, dy,$$ which is the area of $R.$

To compute the area of an ellipse, use the parametrization $x=a \cos t, y = b \sin t, 0 \le t \le 2\pi,$ to get $$\oint_C x \, dy = \int_0^{2\pi} (a \cos t)(b \cos t)\, dt = ab \int_0^{2\pi} \cos^2 t \, dt = \pi ab.\ _\square$$ (The integral of $\cos^2 t$ is a standard trigonometric integral, left to the reader.)

This method is especially useful for regions bounded by parametric curves.

A cardioid is a curve traced by a fixed point on the perimeter of a circle of radius $r$ which is rolling around another circle of radius $r.$ It is parameterized by the equations $$\begin{aligned} x &= r(2\cos t-\cos 2t) \\ y &= r(2\sin t - \sin 2t), \end{aligned}$$ where $0 \le t \le 2\pi.$ What is the area inside the cardioid?

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Here $dy = r(2\cos t-2\cos 2t)\, dt,$ so $$\begin{aligned} \oint_C x \, dy &= \int_0^{2\pi} r^2(2\cos t-\cos 2t)(2\cos t-2\cos 2t) \, dt \\ &= r^2 \left( \int_0^{2\pi} 4 \cos^2 t \, dt + \int_0^{2\pi} 2 \cos^2 2t \, dt - \int_0^{2\pi} 4\cos t\cos 2t \, dt \right). \end{aligned}$$ The first two integrals are straightforward applications of the identity $\cos^2(z) = \frac12(1+\cos 2t).$ They are equal to $4\pi$ and $2\pi,$ respectively. The third integral is simplified via the identity $\cos 2t \cos t = \frac12(\cos 3t+\cos t),$ and equals $0.$ So the final answer is $6\pi r^2.$ $_\square$

One of the fundamental results in the theory of contour integration from complex analysis is Cauchy's theorem:

Let $f$ be a holomorphic function and let $C$ be a simple closed curve in the complex plane. Then $\oint_C f(z) dz = 0.$

The proof reduces the problem to Green's theorem.

Write $f = u+iv$ and $dz = dx + i dy.$ Then the integral is $$\oint_C (u+iv)(dx+i dy) = \oint_C (u \, dx - v \, dy) + i \oint_C (v \, dx + u \, dy).$$ These integrals can be evaluated by Green's theorem: $$\begin{aligned} \oint_C (u \, dx - v \, dy) &= \iint_R \left(- \frac{\partial v}{\partial x} - \frac{\partial u}{\partial y} \right) dx \, dy \\ \oint_C (v \, dx + u \, dy) &= \iint_R \left(\frac{\partial u}{\partial x} - \frac{\partial v}{\partial y} \right) dx \, dy. \end{aligned}$$

But holomorphic functions satisfy the Cauchy-Riemann equations $\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}$ and $\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}.$ Therefore, both integrals are 0 and the result follows. $_\square$

Green's theorem is a special case of the three-dimensional version of Stokes' theorem, which states that for a vector field $\bf F,$ $$\oint_C {\bf F} \cdot d{\bf s} = \iint_R (\nabla \times {\bf F}) \cdot {\bf n} \, dA,$$ where $\bf n$ is the normal vector to the region $R$ and $\nabla \times {\bf F}$ is the curl of $\bf F.$ When ${\bf F} = (P,Q,0)$ and $R$ is a region in the $xy$-plane, the setting of Green's theorem, ${\bf n}$ is the unit vector $(0,0,1)$ and the third component of $\nabla \times {\bf F}$ is $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y},$ so the theorem becomes $$\oint_C (P,Q,0) \cdot (dx,dy,dz) = \iint_R \left( \dfrac{\partial Q}{\partial x} - \dfrac{\partial P}{\partial y} \right) dA$$ and the left side is just $\oint_C (P \, dx + Q \, dy)$ as desired.

This proof is the reversed version of another proof; watch it here.

Let $C$ be a piecewise smooth, simple closed curve in the plane. Let this smooth curve be enclosed in the region $R$, and assume that $P$ and $Q$ and their first partial derivatives are continuous at each point in the region $R$ containing $C$. We need to prove that

$$\oint_C \mathbf F\cdot d\mathbf s =\oint_C P \, d\hat{\mathbf x}+Q \, d\hat{\mathbf y} =\iint_R \left(\dfrac{\partial Q}{\partial x}-\dfrac{\partial P}{\partial y}\right) \, dx \, dy.$$

Let us say that the curve $C$ is made up of two curves $C_1$ and $C_2$ such that

$$\begin{aligned} C_1: y &= f_1(x) \ \forall a\leq x\leq b\\ C_2: y &= f_2(x) \ \forall b\leq x\leq a. \end{aligned}$$

Now, under the following conditions, integrating $\frac{\partial P}{\partial y}$ with respect to $y$ between $y=f_1(x)$ and $y=f_2(x)$ yields

$$\int_{f_1(x)}^{f_2(x)}\dfrac{\partial P}{\partial y} \, dy = P(x,f_2(x))-P(x,f_1(x)).$$

Integrating the resulting integrand over the interval $(a,b)$, we obtain

$$\begin{aligned} \int_a^b\int_{f_1(x)}^{f_2(x)}\dfrac{\partial P}{\partial y} \, dy \, dx &=\int_a^b \big(P(x,f_2(x))-P(x,f_1(x))\big) \, dx\\ &=\int_a^b (P(x,f_2(x)) \, dx -\int_a^b (P(x,f_1(x)) \, dx\\ &=-\int_b^a (P(x,f_2(x)) \, dx -\int_a^b (P(x,f_1(x)) \, dx\\ &=-\int_{C_2} P \, dx-\int_{C_1} P \, dx \\ &=-\oint_{C} P \, dx.\\ \end{aligned}$$

Thus, we arrive at the first half of the required expression

$$\oint_{C} P \, dx = -\int_a^b\int_{f_1(x)}^{f_2(x)}\dfrac{\partial P}{\partial y} \, dy \, dx=\iint_R \left(-\dfrac{\partial P}{\partial y}\right) \, dx \, dy.$$

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Similarly, we can arrive at the other half of the proof. Let us say the curve $C$ is made up of two curves $C'_1$ and $C'_2$ such that

$$\begin{aligned} C'_1: x &= g_1(y) \ \forall d\leq x\leq c\\ C'_2: x &= g_2(y) \ \forall c\leq x\leq d. \end{aligned}$$

Now, integrating $\frac{\partial Q}{\partial x}$ with respect to $x$ between $x=g_1(y)$ and $x=g_2(y)$ yields

$$\int_{g_1(y)}^{g_2(y)}\dfrac{\partial Q}{\partial x} \, dx = Q(g_2(y),y)-Q(g_1(y),y).$$

Integrating the resulting integrand over the interval $(c,d)$ we obtain

$$\begin{aligned} \int_c^d\int_{g_1(y)}^{g_2(y)}\dfrac{\partial Q}{\partial x} \, dx \, dy &=\int_c^d \big(Q(g_2(y),y)-Q(g_1(y),y)\big) \, dy\\ &=\int_c^d (Q(g_2(y),y) \, dy -\int_c^d (Q(g_1(y),y) \, dy\\ &=\int_c^d (Q(g_2(y),y) \, dy +\int_d^c (Q(g_1(y),y) \, dy\\ &=\int_{C'_2} Q \, dy +\int_{C'_1} Q \, dy \\ &=\oint_{C} Q \, dy.\\ \end{aligned}$$

Thus, we arrive at the second half of the required expression. $$\oint_{C} Q \, dy = \int_c^d\int_{g_1(x)}^{g_2(x)}\dfrac{\partial Q}{\partial x} \, dy \, dx= \iint_R \left(\dfrac{\partial Q}{\partial x}\right) \, dx \, dy.$$

Summing both the results finishes the proof of Green's theorem:

$$\begin{aligned} \oint_{C} P \, dx +\oint_{C} Q \, dy=\oint_C \mathbf F\cdot d\mathbf s &=\iint_R \left(-\dfrac{\partial P}{\partial y}\right) \, dx \, dy + \iint_R \left(\dfrac{\partial Q}{\partial x}\right) \, dx \, dy\\ &=\iint_R \left(\dfrac{\partial Q}{\partial x}-\dfrac{\partial P}{\partial y}\right) \, dx \, dy. \ _\square \end{aligned}.$$

• Stokes' Theorem

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