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

# 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.
$$25\pi$$