MATH 251 Lecture 35

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$C$ is a path through space, $R$ is a 3D region in space, $\mathrm{\Omega }$ represents the boundary surface of the enclosed region, and $D$ represents the domain of $\mathrm{\Omega }$.

Vector Analysis Theorems

1. FTC: $\underset{a}{\overset{b}{\int }}{f}^{\prime }\left(x\right)\mathrm{d}x=f\left(b\right)-f\left(a\right)$
2. FTLI: $\underset{C}{\int }\overrightarrow{\mathrm{\nabla }}f\cdot \mathrm{d}\overrightarrow{r}=f\left(Q\right)-f\left(P\right)$
3. Green's Theorem: $\underset{D}{\iint }(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y})\mathrm{d}A={\oint }_{C}P\mathrm{d}x+Q\mathrm{d}y$
4. Stokes' Theorem: $\underset{\mathrm{\Omega }}{\iint }\overrightarrow{\mathrm{\nabla }}\times \overrightarrow{F}\cdot \hat{n}\mathrm{d}S=\underset{C}{\int }\overrightarrow{F}\cdot \mathrm{d}\overrightarrow{r}$
5. Divergence Theorem: $\underset{R}{\iiint }\overrightarrow{\mathrm{\nabla }}\cdot \overrightarrow{F}\mathrm{d}V=\underset{\mathrm{\Omega }}{\iint }\overrightarrow{F}\cdot \hat{n}\mathrm{d}S$

All have something in common: Integrating over the boundary (rhs) means something about integrating over the interior:

$\underset{R}{\int }\mathrm{d}\omega =\underset{\partial R}{\int }\omega$

Divergence Theorem

(See Vector Analysis Theorems#Gauss' (Divergence) Theorem→)

Example

Given $\overrightarrow{F}=3x^2\hat{ı}-2y\hat{ȷ}+\sin z\hat{k}$, find its flux over the sphere of radius 3 centered at the origin:

$\mathrm{\Phi }=\underset{\mathrm{\Omega }}{\iint }\overrightarrow{F}\cdot \hat{n}\mathrm{d}S$.

We could plug and chug this integral, but the divergence theorem gives us another option:

$\overrightarrow{\mathrm{\nabla }}\cdot \overrightarrow{F}=6x-2+\cos z$, and therefore,

$\mathrm{\Phi }=\underset{R}{\iiint }6x-2+\cos z\mathrm{d}V$ such that $R=\{(x,y,z)\mid x^2+y^2+z^2\le 9\}$

1. $\underset{R}{\iiint }6x\mathrm{d}V=0$ since it's the average value of $x$ over a symmetric region.
2. $-2\underset{R}{\iiint }\mathrm{d}V=-72\pi$ since it's the volume of the region $V=\frac{4}{3}\pi r^3$
3. $\underset{R}{\iiint }\cos z\mathrm{d}V=\dots$ (let's not evaluate this integral)

Another Example

Let $\overrightarrow{F}=(y^3-2z)\hat{ı}+\left(3x{y}^2\right)\hat{ȷ}+\hat{k}$.

Find $\underset{C}{\int }\overrightarrow{F}\cdot \mathrm{d}\overrightarrow{r}$, where $C$ represents the curve $x^2+y^2=1$, $z=0$ counter-clockwise when viewed from above.

We could parameterize this with $x=\cos t,y=\sin t$, but I wouldn't want to integrate $3\cos t{\sin }^{}t$.

According to Stokes' theorem, the integral of a curve is related to the integral of the curl over any surface bounded by that curve.

$\overrightarrow{\mathrm{\nabla }}\times \overrightarrow{F}=-2\hat{ȷ}$.

We know that $\overrightarrow{F}\ne \overrightarrow{\mathrm{\nabla }}f$, but $\overrightarrow{F}=\overrightarrow{\mathrm{\nabla }}(x{y}^3+z)-2z\hat{ı}$. Taking the curl wipes out the conservative parts of a vector field.

Now we have $\underset{\mathrm{\Omega }}{\iint }-2\hat{ȷ}\cdot \hat{n}\mathrm{d}S$, where $\mathrm{\Omega }$ represents the simplest surface bounded by $C$: a unit disk in the $xy$ plane.

$\overrightarrow{X}(u,v)=⟨u,v,0⟩$, but we can easily tell that $\hat{n}=\hat{k}$ (follows RHR since the curve is going counter-clockwise).

$\underset{\mathrm{\Omega }}{\iint }-2\hat{ȷ}\cdot \hat{k}\mathrm{d}A=0$

Yet Another Example

Calculate $\underset{\mathrm{\Omega }}{\iint }\overrightarrow{\mathrm{\nabla }}\times \overrightarrow{F}\cdot \hat{n}\mathrm{d}S$ over the upper hemisphere $x^2+y^2+z^2=1$, $z\ge 0$, where $\hat{n}$ points radially outward.

$\overrightarrow{F}=(y^3-2z)\hat{ı}+(3x{y}^2)\hat{ȷ}+\hat{k}$

According to Stokes' theorem, $\underset{\mathrm{\Omega }}{\iint }\overrightarrow{\mathrm{\nabla }}\times \overrightarrow{F}\cdot \hat{n}\mathrm{d}S=\underset{C}{\int }\overrightarrow{F}\cdot \mathrm{d}\overrightarrow{r}=0$, where $\mathrm{\Omega }$ is any other surface with same boundary curve $C$.