MATH 251 Lecture 34

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

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

Surface Area

${\mbox{S. A.}}=\iint _{\Omega }1\,\mathrm {d} S$

Example

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

Parameterization

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

Where $0\leq \phi \leq 2\pi$ and $0\leq \phi \leq 2\pi$.

Area Element

${\begin{aligned}{\vec {X}}_{\phi }&=\left\langle -a\,\sin {\phi }\,\cos {\theta },\,-a\,\sin {\phi }\,\sin {\theta },\,a\,\cos {\phi }\right\rangle \\{\vec {X}}_{\theta }&=\left\langle -(R+a\cos {\phi })\sin {\theta },(R+a\cos {\phi })\,\cos {\theta },0\right\rangle \\{\vec {X}}_{\phi }\times {\vec {X}}_{\theta }&=a(R+a\cos {\phi })\left\langle -\cos {\theta }\cos {\phi },-\sin {\theta }\cos {\phi },-\sin {\phi }\cos ^{2}{\theta }-\sin {\phi }sin^{2}{\theta }\right\rangle \\\left\|{\vec {X}}_{\phi }\times {\vec {X}}_{\theta }\right\|&=a(R+a\cos {\phi })\end{aligned}}$

Evaluate

$\int _{0}^{2\pi }\int _{0}^{2\pi }a(R+a\cos {\phi })\,\mathrm {d} \phi \,\mathrm {d} \theta =(2\pi a)(2\pi R)$

Another Example

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

${\begin{aligned}{\vec {X}}(u,v)&=\left\langle u,v,f(u,v)\right\rangle \\{\vec {X}}_{u}&=\left\langle 1,0,f_{u}(u,v)\right\rangle \\{\vec {X}}_{v}&=\left\langle 0,1,f_{v}(u,v)\right\rangle \\{\vec {X}}_{u}\times {\vec {X}}_{v}&=\left\langle -f_{u},-f_{v},1\right\rangle \\\left\|{\vec {X}}_{u}\times {\vec {X}}_{v}\right|&={\sqrt {{f_{u}}^{2}+{f_{v}}^{2}+1}}\end{aligned}}$

Flux

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

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

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

$\iint _{\Omega }{\vec {F}}\cdot {\hat {n}}\,\mathrm {d} S$

Example

Calculate the flux of ${\vec {F}}=\left\langle x-y,x^{2},1\right\rangle$ through a paraboloid $z=x^{2}+y^{2}$ over $[-1,1]\times [-1,1]$ in $xy$ plane.

Parameterize

${\vec {X}}(u,v)=\left\langle u,v,u^{2}+v^{2}\right\rangle$

Find Normal Unit

${\begin{aligned}{\vec {X}}_{u}&=\left\langle 1,0,2u\right\rangle \\{\vec {X}}_{v}&=\left\langle 0,1,2v\right\rangle \end{aligned}}$

We want to have ${\hat {n}}$ pointing outward, so we take ${\vec {n}}={\vec {X}}_{v}\times {\vec {X}}_{u}$.

${\hat {n}}={\frac {{\vec {X}}_{v}\times {\vec {X}}_{u}}{\left\|{\vec {X}}_{v}\times {\vec {X}}_{u}\right\|}}$

Find Area Element

$\mathrm {d} S=\left\|{\vec {X}}_{u}\times {\vec {X}}_{v}\right\|$ This is true always.

Set up integral

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

$\int _{-1}^{1}\int _{-1}^{1}{\vec {F}}\cdot {\vec {X}}_{v}\times {\vec {X}}_{u}\,\mathrm {d} u\,\mathrm {d} v=\int _{-1}^{1}\int _{-1}^{1}(2u^{2}-2uv+2vu^{2}-1)\,\mathrm {d} u\,\mathrm {d} v$