Face

N

$\mathbf{F}·\mathbf{N} \mathrm{d\σ}$

value of $\mathbf{F}·\mathbf{N} \mathrm{d\σ}$

$x\=0$

$-\mathbf{i}$

$-x y^3 \mathrm{dA}$

$0$

$y\=0$

$-\mathbf{j}$

$-y z \mathrm{dA}$

$0$

$z\=0$

$-\mathbf{k}$

$-x^{2}z \mathrm{dA}$

$0$

$z\=1-x-y$

$\left(\mathbf{i}\+\mathbf{j}\+\mathbf{k}\right)\/\sqrt{3}$

$\frac{\left(x y^3\+y z\+x^{2}z\right)}{\sqrt{3}}\left(\sqrt{3} \mathrm{dA}\right)$

$x y^3\+\left(y\+x^2\right)\left(1-x-y\right)$

Table 9.8.10(a)   $\mathbf{F}·\mathbf{N} \mathrm{d\σ}$ on the four faces of $R$ 

Initialize

Loading Student:-VectorCalculus

Define the vector field F

$⟨x y^3\,y z\,x^{2}z⟩$ = $\stackrel{\text{to Vector Field}}{\to }$ $\stackrel{\text{assign to a name}}{\to }$$F$

Obtain $\nabla ·\mathbf{F}$, the divergence of F

$\nabla ·\mathbf{F}$ = $y^3\+x^2\+z$$\stackrel{\text{assign to a name}}{\to }$$\mathrm{divF}$

Obtain the volume integral of the divergence of F

Loading Student:-MultivariateCalculus

$\mathrm{divF}$ = $y^3\+x^2\+z$$\stackrel{\text{MultiInt}}{\to }$${\∫}_0^1{\∫}_0^{1-x}{\∫}_0^{1-x-y}\left(y^3\+x^2\+z\right)\ⅆz\ⅆy\ⅆx$$\=$$\frac{1}{15}$

Tools≻Tasks≻Browse:
Calculus - Vector≻Integration≻Flux≻3-D≻Through a Surface Defined over a Triangle

Initialize

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

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

$\mathbf{F}≔\mathrm{VectorField}\left(⟨x y^3\,y z\,x^{2}z⟩\right)\:$

Obtain $\nabla ·\mathbf{F}$, the divergence of F

$\mathrm{divF}≔\mathrm{Divergence}\left(\mathbf{F}\right)$

$y^3\+x^2\+z$

Use the int command to integrate the divergence of F over $R$

$\mathrm{int}\left(\mathrm{divF}\,\left[x\,y\,z\right]\=\mathrm{Region}\left(0..1\,0..1-x\,0..1-x-y\right)\,\mathrm{output}\=\mathrm{integral}\right)$

${\∫}_0^1{\∫}_0^{1-x}{\∫}_0^{1-x-y}\left(y^3\+x^2\+z\right)\ⅆz\ⅆy\ⅆx$

$\mathrm{int}\left(\mathrm{divF}\,\left[x\,y\,z\right]\=\mathrm{Region}\left(0..1\,0..1-x\,0..1-x-y\right)\right)$ = $\frac{1}{15}$

Use the Flux command to obtain the flux of F through the plane $z\=1-x-y$ 

$\mathrm{Flux}\left(\mathbf{F}\,\mathrm{Surface}\left(⟨x\,y\,1-x-y⟩\,\left[x\,y\right]\=\mathrm{Triangle}\left(⟨0\,0⟩\,⟨0\,1⟩\,⟨1\,0⟩\right)\right)\,\mathrm{output}\=\mathrm{integral}\right)$

${\∫}_0^1{\∫}_0^{1-x}\left(x{y}^3\+y\left(1-x-y\right)\+x^2\left(1-x-y\right)\right)\ⅆy\ⅆx$

$\mathrm{Flux}\left(\mathbf{F}\,\mathrm{Surface}\left(⟨x\,y\,1-x-y⟩\,\left[x\,y\right]\=\mathrm{Triangle}\left(⟨0\,0⟩\,⟨0\,1⟩\,⟨1\,0⟩\right)\right)\right)$ = $\frac{1}{15}$