MATH 251 Lecture 34

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Last Episode: $\underset{\mathrm{\Omega }}{\iint }f\mathrm{d}S$

This Week: $\underset{\mathrm{\Omega }}{\iint }(\overrightarrow{F}\cdot \overrightarrow{n})\mathrm{d}S$ called "flux"

Surface Area

$\text{S. A.}=\underset{\mathrm{\Omega }}{\iint }1\mathrm{d}S$

Example

Surface area of torus of mean radius $R$ and cross-sectional radius of $a$: position given by $\theta$ (polar angle) and $\varphi$ (angle formed between vector from center of cross-sectional circle and point on surface and the radial vector)

Parameterization

$X(\varphi ,\theta )=\{\begin{array}{l}x=(R+a\cos \varphi )\cos \theta \\ y=(R+a\cos \varphi )\sin \theta \\ z=a\sin \varphi \end{array}$

Where $0\le \varphi \le 2\pi$ and $0\le \varphi \le 2\pi$.

Area Element

$\begin{array}{rl}{\overrightarrow{X}}_{\varphi }& =⟨-a\sin \varphi \cos \theta ,-a\sin \varphi \sin \theta ,a\cos \varphi ⟩\\ {\overrightarrow{X}}_{\theta }& =⟨-(R+a\cos \varphi )\sin \theta ,(R+a\cos \varphi )\cos \theta ,0⟩\\ {\overrightarrow{X}}_{\varphi }\times {\overrightarrow{X}}_{\theta }& =a(R+a\cos \varphi )⟨-\cos \theta \cos \varphi ,-\sin \theta \cos \varphi ,-\sin \varphi {\cos }^{}\theta -\sin \varphi si{n}^2\theta ⟩\\ \Vert {\overrightarrow{X}}_{\varphi }\times {\overrightarrow{X}}_{\theta }\Vert & =a(R+a\cos \varphi )\end{array}$

Evaluate

$\underset{0}{\overset{2\pi }{\int }}\underset{0}{\overset{2\pi }{\int }}a(R+a\cos \varphi )\mathrm{d}\varphi \mathrm{d}\theta =(2\pi a)(2\pi R)$

Another Example

$z=f(x,y)$, so $x(u,v)=u$, $y(u,v)=v$, and $z(u,v)=f(u,v)$.

$\begin{array}{rl}\overrightarrow{X}(u,v)& =⟨u,v,f(u,v)⟩\\ {\overrightarrow{X}}_u& =⟨1,0,f_u(u,v)⟩\\ {\overrightarrow{X}}_v& =⟨0,1,f_v(u,v)⟩\\ {\overrightarrow{X}}_u\times {\overrightarrow{X}}_v& =⟨-f_u,-f_v,1⟩\\ \Vert {\overrightarrow{X}}_u\times {\overrightarrow{X}}_v|& =\sqrt{{f_u}^2+{f_v}^2+1}\end{array}$

Flux

Vector field $\overrightarrow{F}$ flowing across a surface $\mathrm{\Omega }$.

Assume $\mathrm{\Omega }$ is orientable (has a well-defined normal vector field $\hat{n}$ (unit length that points outward))

Therefore $\overrightarrow{F}\cdot \hat{n}$ is the component of $\overrightarrow{F}$ that is perpendicular to $\mathrm{\Omega }$.

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

Example

Calculate the flux of $\overrightarrow{F}=⟨x-y,x^2,1⟩$ through a paraboloid $z=x^2+y^2$ over $[-1,1]\times [-1,1]$ in $xy$ plane.

Parameterize

$\overrightarrow{X}(u,v)=⟨u,v,u^2+v^2⟩$

Find Normal Unit

$\begin{array}{rl}{\overrightarrow{X}}_u& =⟨1,0,2u⟩\\ {\overrightarrow{X}}_v& =⟨0,1,2v⟩\end{array}$

We want to have $\hat{n}$ pointing outward, so we take $\overrightarrow{n}={\overrightarrow{X}}_v\times {\overrightarrow{X}}_u$.

$\hat{n}=\frac{{\overrightarrow{X}}_v\times {\overrightarrow{X}}_u}{\Vert {\overrightarrow{X}}_v\times {\overrightarrow{X}}_u\Vert }$

Find Area Element

$\mathrm{d}S=\Vert {\overrightarrow{X}}_u\times {\overrightarrow{X}}_v\Vert$ This is true always.

Set up integral

Notice how $\mathrm{d}S$ and $\hat{n}$ can simplify to just $\overrightarrow{n}$

$\underset{-1}{\overset{1}{\int }}\underset{-1}{\overset{1}{\int }}\overrightarrow{F}\cdot {\overrightarrow{X}}_v\times {\overrightarrow{X}}_u\mathrm{d}u\mathrm{d}v=\underset{-1}{\overset{1}{\int }}\underset{-1}{\overset{1}{\int }}(2u^2-2uv+2v{u}^2-1)\mathrm{d}u\mathrm{d}v$