\(\def\ans#1{\bbox[border:1px solid green,6pt]{#1}}\)

Green's Theorem

Introduction

Recall that \(\displaystyle\int_C\) is a path integral. Now, we define \(\displaystyle\oint_C\) as a path integral along a closed path (meaning that the path returns to where it started).

For instance, using the fundamental theorem:

\[\begin{align} \int_C \nabla \phi \cdot \vec{r}\ '(t)\ dt &= \phi(B) - \phi(A)\\ \oint_C \nabla \phi \cdot \vec{r}\ '(t)\ dt &= 0 \end{align}\]

Green's Theorem relates a path integral on a closed curve to a double integral on the region inside the curve. First, though, we need a few simple definitions:

  • A closed curve is a path where the starting point is the same as the ending point.
  • A simple curve is one that doesn't intersect itself.
  • A connected region is one in which any two points can be connected by a continuous curve in the region.
  • A simply connected region is one in which every simple closed curve can be contracted to a point in the region (i.e. there are no holes in the region).
  • We define positive orientation as the direction in which we travel so that the enclosed region is on the left (this tends to be counterclockwise).

Green's Theorem

There are two forms of Green's Theorem. First, though, let \(C\) be a simple, closed, piecewise-smooth curve, positively oriented, that encloses a connected and simply connected region \(R\) in the plane. Assume that \(\vec{F}=\langle f,g \rangle\), where \(f\) and \(g\) have continuous first partial derivatives in \(R\).

  1. Circulation, or Curl, Form
  2. \[\oint_C \vec{F} \cdot \vec{r}\ '(t)\ dt = \oint_C f\ dx \ + g\ dy = \iint_R \dfrac{\partial g}{\partial x} - \dfrac{\partial f}{\partial y} \ dA\]
  3. Flux, or Divergence, Form
  4. \[\oint_C \vec{F} \cdot \hat{n}\ ds = \oint_C f\ dy \ - g\ dx = \iint_R \dfrac{\partial f}{\partial x} + \dfrac{\partial g}{\partial y} \ dA,\] where \(\hat{n}\) is the outward unit normal on the curve.

Note that if you set up the circulation/curl form with \(\vec{F}=\langle -g,f \rangle\), you get the flux/divergence form.

Examples

Use Green's Theorem to evaluate \[\oint_C xy\ dx\ + x^2y^3\ dy,\] where \(C\) is the triangle with vertices \((0,0)\), \((1,0)\), and \((1,2)\) with positive orientation.

Solution

Since \(f=xy\) and \(g=x^2y^3\), \(\dfrac{\partial f}{\partial y} = x\) and \(\dfrac{\partial g}{\partial x} = 2xy^3\),

\[\begin{align} \oint_C xy\ dx\ + x^2y^3\ dy &= \int_0^1 \int_0^{2x} 2xy^3-x\ dy dx\\ &= \int_0^1 8x^5-2x^2\ dx\\ &= \dfrac{4}{3}x^6-\dfrac{2}{3}x^3 \bigg|_0^1 = \ans{\dfrac{2}{3}} \end{align}\]

Consider the rotation vector field \(\vec{F} = \langle -y,x \rangle\) on the unit disk \[R = \{ (x,y) \ :\ x^2+y^2 \leq 1\}.\] Find the circulation \(\displaystyle\iint_R \dfrac{\partial g}{\partial x} - \dfrac{\partial f}{\partial y} \ dA\) of this field.

Solution

Since \(f=-y\) and \(g=x\), \(\dfrac{\partial f}{\partial y} = -1\) and \(\dfrac{\partial g}{\partial x} = 1\), so

\[\iint_R \dfrac{\partial g}{\partial x} - \dfrac{\partial f}{\partial y} \ dA = \iint_R 2\ dA.\]

Because the region is circular, this integral will be easiest using polar coordinates:

\[\iint_R 2\ dA = \int_0^{2\pi} \int_0^1 2r\ dr\ d\theta = \ans{2\pi}\]

Consider the outward flux \[\oint_C \vec{F} \cdot \hat{n}\ ds = \iint_R \dfrac{\partial f}{\partial x} + \dfrac{\partial g}{\partial y}\ dA\] of the radial field \(\vec{F} = \langle x,y \rangle\) across the unit circle \(C = \{(x,y)\ : \ x^2+y^2 = 1\}\).

Solution

Since \(f=x\) and \(g=y\), \(\dfrac{\partial f}{\partial x} = 1\) and \(\dfrac{\partial g}{\partial y} = 1\), so

\[\oint_C \vec{F} \cdot \hat{n}\ ds = \iint_R 2\ dA.\]

Because the region is circular, this integral will be easiest using polar coordinates:

\[\iint_R 2\ dA = \int_0^{2\pi} \int_0^1 2r\ dr\ d\theta = \ans{2\pi}\]

Try it yourself:

(click on the problem to show/hide the answer)

Suppose \(\vec{F} = \nabla\left(\sqrt{x^2+y^2}\right)\) and \(R\) is the half annulus \(R = \{(r,\theta)\ :\ 1 \leq r \leq 3, 0 \leq \theta \leq \pi\}\). Calculate the outward flux across the boundary of \(R\).
\(\displaystyle\int_0^{\pi} \displaystyle\int_1^3 \left(-\dfrac{r^2}{r^3}+\dfrac{2}{r}\right) r\ dr\ d\theta = 2\pi\)

Consider the vector field \(\vec{F} = \langle y^2,x^2 \rangle\) on the half annulus \[R = \{(x,y)\ : \ 1 \leq x^2+y^2 \leq 9, y \geq 0\}.\] Find the circlulation on \(C\), the boundary of \(R\).

Solution

Recall that the circulation is \(\displaystyle\oint_C \vec{F} \cdot \vec{r}\ '(t)\ dt = \displaystyle\iint_R \dfrac{\partial g}{\partial x} - \dfrac{\partial f}{\partial y} \ dA\).

Since \(f=y^2\) and \(g=x^2\), \(\dfrac{\partial f}{\partial y} = 2y\) and \(\dfrac{\partial g}{\partial x} = 2x\), so

\[\oint_C \vec{F} \cdot \vec{r}\ '(t)\ dt = \iint_R 2x-2y\ dA.\]

Again, this integral will be easiest using polar coordinates:

\[\iint_R 2x-2y\ dA = \int_0^{\pi} \int_1^3 (2r \ \cos\theta - 2r\ \sin\theta) r \ dr\ d\theta = \ans{-\dfrac{104}{3}}\]

Sidenote: using Green's Theorem to find area

If \(\vec{F}=\langle 0,x \rangle\), \[\oint_C \vec{F} \cdot \vec{r}\ '(t) \ dt = \oint_C x\ dy = \iint_R \ dA = \textrm{ area of } R\]

On the other hand, if \(\vec{F}=\langle y,0 \rangle\), \[\oint_C \vec{F} \cdot \vec{r}\ '(t) \ dt = \oint_C y\ dx = -\iint_R \ dA = -\textrm{area of } R\]

Therefore, the area of \(R\) is \[\oint_C x\ dy = -\oint_C y\ dx,\] so \[\textrm{Area of } R = \dfrac{1}{2} \oint_C x\ dy - y\ dx.\]

Find the area of \(R = \{(x,y)\ :\ x^2+y^2 \leq 16\}.\)

Solution

The boundary of this region can be expressed parametrically as

\[\begin{align} x &= 4 \cos t \longrightarrow dx = -4 \sin t\ dt\\ y &= 4 \sin t \longrightarrow dy = 4 \cos t\ dt \end{align}\]

The area of this region, then, is

\[\begin{align} \dfrac{1}{2} &\oint_C x\ dy - y\ dx\\ &= \dfrac{1}{2} \oint_C (4 \cos t)(4 \cos t) - (4 \sin t)(4 \sin t)\ dt\\ &= 8 \oint_C \cos^2 t + \sin^2 t\ dt\\ &= 8 \int_0^{2\pi} \ dt = \ans{16\pi} \end{align}\]

Try it yourself:

(click on the problem to show/hide the answer)

Use Green's Theorem to find the area of a disk of radius 5.
\(25\pi\)