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:
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 \(R\).
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 \(\dfrac{\partial g}{\partial x} = 2xy^3\),
\[\begin{align} \oint_C xy\ dx\ + x^2y^3\ dy &= \int_0^1 \int_0^{2x} 2xy^3-x\ dy dx\\ &= \int_0^1 8x^5-2x^2\ dx\\ &= \dfrac{4}{3}x^6-\dfrac{2}{3}x^3 \bigg|_0^1 = \ans{\dfrac{2}{3}} \end{align}\]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:
\[\iint_R 2\ dA = \int_0^{2\pi} \int_0^1 2r\ dr\ d\theta = \ans{2\pi}\]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:
\[\iint_R 2\ dA = \int_0^{2\pi} \int_0^1 2r\ dr\ d\theta = \ans{2\pi}\]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
\[\begin{align} \dfrac{1}{2} &\oint_C x\ dy - y\ dx\\ &= \dfrac{1}{2} \oint_C (4 \cos t)(4 \cos t) - (4 \sin t)(4 \sin t)\ dt\\ &= 8 \oint_C \cos^2 t + \sin^2 t\ dt\\ &= 8 \int_0^{2\pi} \ dt = \ans{16\pi} \end{align}\]