MATH 251 Lecture 32

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Green's Theorem

(See Vector Analysis Theorems#Green's Theorem→)

When you add up all these little swirly thingies, you get the big swirly thingy.

For $\overrightarrow{F}(x,y)=P(x,y)\hat{ı}+Q(x,y)\hat{ȷ}$ in a region $\mathrm{\Omega }$ bounded by the curve $C$,

${\oint }_C\overrightarrow{F}\cdot \mathrm{d}\overrightarrow{r}=\underset{\mathrm{\Omega }}{\iint }\overrightarrow{\mathrm{\nabla }}\times \overrightarrow{F}\cdot \hat{k}=\underset{\mathrm{\Omega }}{\iint }\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\mathrm{d}A$

This could be an elaborate way to calculate area when the curl is equal to 1.

Example

Find $\underset{C}{\int }x\sin y\mathrm{d}x-y\mathrm{d}y$, where $C$ is the upper half of the unit circle.

$P(x,y)=x\sin y$ and $Q(x,y)=-y$.

$\begin{array}{rl}{Q}_x-P_y& =-x\cos y\\ \underset{-1}{\overset{1}{\int }}\underset{0}{\overset{\sqrt{1-x^2}}{\int }}-x\cos y\mathrm{d}y\mathrm{d}x& =\underset{-1}{\overset{1}{\int }}x\sin \left(\sqrt{1-x^2}\right)\mathrm{d}x\\ & =0\end{array}$

Because $x\sin \left(\sqrt{1-x^2}\right)$ is odd, and we integrate it from -1 to 1.

Conservative Vector Fields

We have $\overrightarrow{F}$ that is known to be conservative. We know that ${\oint }_C\overrightarrow{F}\cdot \mathrm{d}\overrightarrow{r}=0$ over every loop.

By Green's Theorem, this also means that ${\oint }_C\overrightarrow{F}\cdot \mathrm{d}\overrightarrow{r}=\underset{R}{\iint }\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\mathrm{d}A=0$

In order for $\underset{R}{\iint }\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\mathrm{d}A=0$ to be true, the integrand $Q_x-P_y=0$, which also means that $Q_x=P_y$

Let's find a potential function $f(P)$ that starts at a base point (0,0) and connects to $P$ such that $f(P)={\oint }_C\frac{F}{\cdot }\mathrm{d}\overrightarrow{r}$ where $C$ connects the two points and could be any path.

Fact: $\frac{\partial f}{\partial x}=P$ and $\frac{\partial f}{\partial y}=Q$.

Stoke's Theorem

(See Vector Analysis Theorems#Stokes' (Curl) Theorem→)

On a 3D surface, there are many swirlies, but the swirlies may not lie in the $xy$ plane. Let $\hat{n}$ represent the unit vector normal to the surface at a point on the surface.

$\iint \overrightarrow{\mathrm{\nabla }}\times \overrightarrow{F}\cdot \hat{n}\mathrm{d}S=\underset{C}{\int }\overrightarrow{F}\cdot \mathrm{d}\overrightarrow{r}$