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).

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 \(R\).

- Circulation, or Curl, Form \[\oint_C \vec{F} \cdot \vec{r}\ '(t)\ dt = \oint_C f\ dx \ + g\ dy = \iint_R \dfrac{\partial g}{\partial x} - \dfrac{\partial f}{\partial y} \ dA\]
- Flux, or Divergence, Form \[\oint_C \vec{F} \cdot \hat{n}\ ds = \oint_C f\ dy \ - g\ dx = \iint_R \dfrac{\partial f}{\partial x} + \dfrac{\partial g}{\partial y} \ dA,\] 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.

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.

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

\[\begin{align} \oint_C xy\ dx\ + x^2y^3\ dy &= \int_0^1 \int_0^{2x} 2xy^3-x\ dy dx\\ &= \int_0^1 8x^5-2x^2\ dx\\ &= \dfrac{4}{3}x^6-\dfrac{2}{3}x^3 \bigg|_0^1 = \ans{\dfrac{2}{3}} \end{align}\]- Use Green's Theorem to evaluate \(\displaystyle\oint_C x^3\ dy\ - y^3\ dx\), where \(C\) is the positively oriented circle of radius 2 centered at the origin.
- Use Green's Theorem to evaluate \(\displaystyle\oint_C (4x^3+\sin (y^2))\ dy\ - (4y^3+\cos (x^2))\ dx\), where \(C\) is the boundary of \(R = \{(x,y)\ :\ x^2+y^2 \leq 4\}\).
- Use Green's Theorem to evaluate \(\displaystyle\oint_C \left(2x+e^{y^2}\right)\ dy\ - \left(4y^2+e^{x^2}\right)\ dx\), where \(C\) is the boundary of the square with vertices at \((0,0)\), \((1,0)\), \((1,1)\), and \((0,1)\).

\(\displaystyle\int_0^{2\pi} \displaystyle\int_0^2 3r^2 \cdot r \ dr d\theta = 24\pi\)

\(12\displaystyle\int_0^{2\pi} \displaystyle\int_0^2 r^2 \cdot r\ dr\ d\theta = 96\pi\)

6

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.

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:

\[\iint_R 2\ dA = \int_0^{2\pi} \int_0^1 2r\ dr\ d\theta = \ans{2\pi}\]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 circulation on \(R\).

0

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 \(C = \{(x,y)\ : \ x^2+y^2 = 1\}\).

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:

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

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:

\[\iint_R 2x-2y\ dA = \int_0^{\pi} \int_1^3 (2r \ \cos\theta - 2r\ \sin\theta) r \ dr\ d\theta = \ans{-\dfrac{104}{3}}\]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 \(R = \{(x,y)\ :\ x^2+y^2 \leq 16\}.\)

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

\[\begin{align} \dfrac{1}{2} &\oint_C x\ dy - y\ dx\\ &= \dfrac{1}{2} \oint_C (4 \cos t)(4 \cos t) - (4 \sin t)(4 \sin t)\ dt\\ &= 8 \oint_C \cos^2 t + \sin^2 t\ dt\\ &= 8 \int_0^{2\pi} \ dt = \ans{16\pi} \end{align}\]Use Green's Theorem to find the area of a disk of radius 5.

\(25\pi\)