Tools≻Tasks≻Browse:
Calculus - Vector≻Integration≻Flux≻3-D≻Through a Parametric Surface

Table 9.6.13(a)   Solution by task template

Initialize

Loading Student:-MultivariateCalculus

Define the parametric surface via the position vector R

$⟨u\+v\,u-v\,u v⟩$$\stackrel{\text{assign to a name}}{\to }$$R$

Obtain the vector ${\mathbf{R}}_u\times {\mathbf{R}}_v$, its length, and the unit normal N

$\left(\frac{\partial }{\partial u} \mathbf{R}\right)\times \left(\frac{\partial }{\partial v} \mathbf{R}\right)$ = $\stackrel{\text{normalize}}{\to }$ $\stackrel{\text{simplify}}{\=}$ $\stackrel{\text{assign to a name}}{\to }$$N$

Evaluate $\mathbf{F}$ on the parametric surface

$⟨x\,y\,z⟩\,\mathbf{R}$ = $\stackrel{\text{equate}}{\to }$$\left[x\=u\+v\,y\=u-v\,z\=uv\right]$$\stackrel{\text{assign to a name}}{\to }$$S$$$

$\genfrac{}{}{0ex}{}{⟨x\,y\,z⟩}{\phantom{x\=a}}|\genfrac{}{}{0ex}{}{\phantom{\mathrm{f(x)}}}{S}$ = $\stackrel{\text{assign to a name}}{\to }$$F$$$

Form and evaluate the flux integral

${\int }_0^1{\int }_0^u\mathbf{F}·\mathbf{N} \sqrt{4\+2 u^2\+2 v^2} \mathit{\ⅆ}v \mathit{\ⅆ}u$ = $\frac{1}{4}$$$$$

Table 9.6.13(b)   Solution from first principles

Initialize

$\mathrm{with}\left(\mathrm{Student}:-\mathrm{VectorCalculus}\right)\:$

$\mathbf{F}≔\mathrm{VectorField}\left(⟨x\,y\,z⟩\right)\:$

$\mathrm{Flux}\left(\mathbf{F}\,\mathrm{Surface}\left(⟨u\+v\,u-v\,u v⟩\,u\=0..1\,v\=0..u\right)\,\mathrm{output}\=\mathrm{integral}\right)$

${\∫}_0^1{\∫}_0^u\left({\left(u\+v\right)}^2\+\left(u-v\right)\left(-u\+v\right)-2uv\right)\ⅆv\ⅆu$

$\mathrm{Flux}\left(\mathbf{F}\,\mathrm{Surface}\left(⟨u\+v\,u-v\,u v⟩\,u\=0..1\,v\=0..u\right)\right)$

$\frac{1}{4}$

Table 9.6.13(c)   Flux computed with the Flux command

Initialize

$\mathrm{BasisFormat}\left(\mathrm{false}\right)\:$

$\mathbf{R}≔⟨u\+v\,u-v\,u v⟩\:$

$\mathrm{dsig}≔\mathrm{Norm}\left(\mathrm{CrossProduct}\left(\mathrm{diff}\left(\mathbf{R}\,u\right)\,\mathrm{diff}\left(\mathbf{R}\,v\right)\right)\right)$

$\sqrt{2u^2\+2v^2\+4}$

$\mathbf{N}≔\mathrm{Normalize}\left(\mathrm{CrossProduct}\left(\mathrm{diff}\left(\mathbf{R}\,u\right)\,\mathrm{diff}\left(\mathbf{R}\,v\right)\right)\right)$

$$\begin{gathered} \mathrm{Int}\left(\mathrm{simplify}\left(\mathrm{eval}\left(\mathrm{DotProduct}\left(\mathbf{F}\,\mathbf{N}\right) \mathrm{dsig}\,\mathrm{Equate}\left(\mathbf{F}\,\mathbf{R}\right)\right)\right)\,\left[v\=0..u\,u\=0..1\right]\right)\= \\ :-\mathrm{int}\left(\mathrm{simplify}\left(\mathrm{eval}\left(\mathrm{DotProduct}\left(\mathbf{F}\,\mathbf{N}\right) \mathrm{dsig}\,\mathrm{Equate}\left(\mathbf{F}\,\mathbf{R}\right)\right)\right)\,\left[v\=0..u\,u\=0..1\right]\right) \end{gathered}$$

${\∫}_0^1{\∫}_0^{u}2uv\ⅆv\ⅆu\=\frac{1}{4}$

Table 9.6.13(d)   Solution from first principles