$\mathbf{F}·\mathbf{N} \mathrm{dsig}\= -4 x^{3}y-4x{y}^3\+xy$

so the surface integral itself is given by the sum

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

Table 9.6.16(a)   Solution by task template

Initialize

Loading Student:-Precalculus

Obtain the equations of the edges of the triangle

$\left[1\,2\right]\,\left[5\,1\right]$$\stackrel{\text{equation of line}}{\to }$$y\=\frac{9}{4}-\frac{1}{4}x$$\stackrel{\text{right hand side}}{\to }$$\frac{9}{4}-\frac{1}{4}x$$\stackrel{\text{assign to a name}}{\to }$$Y_{\mathrm{AB}}$

$\left[5\,1\right]\,\left[3\,3\right]$$\stackrel{\text{equation of line}}{\to }$$y\=6-x$$\stackrel{\text{right hand side}}{\to }$$6-x$$\stackrel{\text{assign to a name}}{\to }$$Y_{\mathrm{BC}}$

$\left[3\,3\right]\,\left[1\,2\right]$$\stackrel{\text{equation of line}}{\to }$$y\=\frac{3}{2}\+\frac{1}{2}x$$\stackrel{\text{right hand side}}{\to }$$\frac{3}{2}\+\frac{1}{2}x$$\stackrel{\text{assign to a name}}{\to }$$Y_{\mathrm{CA}}$

Initialize

Loading Student:-MultivariateCalculus

Define the vector field F

$⟨y z\,x z\,x y⟩$$\stackrel{\text{assign to a name}}{\to }$$F$$$

Define the surface

$x^2\+y^2$$\stackrel{\text{assign to a name}}{\to }$$f$

Obtain N, a unit normal on the surface

$⟨-\frac{\partial }{\partial x} f\,-\frac{\partial }{\partial y} f\,1⟩$ = $\stackrel{\text{normalize}}{\to }$ $\stackrel{\text{assign to a name}}{\to }$$N$

Obtain $\mathrm{dsig}\=\sqrt{1\+f_x^2\+f_y^2}$ 

$\sqrt{1\+{\left(\frac{\partial }{\partial x} f\right)}^2\+{\left(\frac{\partial }{\partial y} f\right)}^2}$ = $\sqrt{4x^2\+4y^2\+1}$$\stackrel{\text{assign to a name}}{\to }$$\mathrm{dsig}$$$

Form $\mathbf{F}·\mathbf{N} \mathrm{dsig}$ and evaluate it on the surface

$\genfrac{}{}{0ex}{}{\mathbf{F}·\mathbf{N} \mathrm{dsig}}{\phantom{x\=a}}|\genfrac{}{}{0ex}{}{\phantom{\mathrm{f(x)}}}{z\=f}$

$\left(-\frac{4y\left(x^2\+y^2\right)x}{\sqrt{4x^2\+4y^2\+1}}\+\frac{xy}{\sqrt{4x^2\+4y^2\+1}}\right)\sqrt{4x^2\+4y^2\+1}$

$\stackrel{\text{simplify}}{\=}$

$-4\left(x^2\+y^2-\frac{1}{4}\right)yx$

$\stackrel{\text{assign to a name}}{\to }$

$Q$

Form and evaluate the flux integral

${\int }_1^3{\int }_{Y_{\mathrm{AB}}}^{Y_{\mathrm{CA}}}Q \mathit{\ⅆ}y \mathit{\ⅆ}x\+{\int }_3^5{\int }_{Y_{\mathrm{AB}}}^{Y_{\mathrm{BC}}}Q \mathit{\ⅆ}y \mathit{\ⅆ}x$ = $-\frac{10069}{10}$$$

Table 9.6.16(c)   Solution from first principles

Initialize

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

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

Use the Flux command with the Triangle option

$\mathrm{Flux}\left(\mathbf{F}\,\mathrm{Surface}\left(⟨x\,y\,x^2\+y^2⟩\,\left[x\,y\right]\=\mathrm{Triangle}\left(⟨1\,2⟩\,⟨5\,1⟩\,⟨3\,3⟩\right)\right)\,\mathrm{output}\=\mathrm{integral}\right)$

$-\left({\∫}_1^3{\∫}_{\frac{3}{2}\+\frac{1}{2}x}^{\frac{9}{4}-\frac{1}{4}x}\left(-4y\left(x^2\+y^2\right)x\+xy\right)\ⅆy\ⅆx\right)-\left({\∫}_3^5{\∫}_{6-x}^{\frac{9}{4}-\frac{1}{4}x}\left(-4y\left(x^2\+y^2\right)x\+xy\right)\ⅆy\ⅆx\right)$

$\mathrm{Flux}\left(\mathbf{F}\,\mathrm{Surface}\left(⟨x\,y\,x^2\+y^2⟩\,\left[x\,y\right]\=\mathrm{Triangle}\left(⟨1\,2⟩\,⟨5\,1⟩\,⟨3\,3⟩\right)\right)\right)$

$-\frac{10069}{10}$

Table 9.6.16(b)   Solution via the Flux command with the Triangle option

Initialize

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

Loading Student:-Precalculus

The triangle

$P≔\left[\left[1\,2\right]\,\left[5\,1\right]\,\left[3\,3\right]\right]\:$

$\mathrm{plot}\left(\left[\left[P_1\,P_2\right]\,\left[P_2\,P_3\right]\,\left[P_3\,P_1\right]\right]\, \mathrm{color}\=\left[\mathrm{black}\,\mathrm{red}\,\mathrm{green}\right]\,\mathrm{view}\=\left[0..6\,0..4\right]\,\mathrm{labels}\=\left[x\,y\right]\right)$

Obtain the integrand $\mathbf{F}·\mathbf{N} \mathrm{d\σ}$ 

$f≔x^2\+y^2\:$

Use the diff command to obtain the surface normal field $-f_x i-f_y j\+k$.

Apply the Normalize command to obtain a unit normal field.

$\mathbf{N}≔\mathrm{Normalize}\left(⟨-\mathrm{diff}\left(f\,x\right)\,-\mathrm{diff}\left(f\,y\right)\,1⟩\right)$

$q≔\mathrm{simplify}\left(\mathrm{eval}\left(\mathrm{DotProduct}\left(\mathbf{F}\,\mathbf{N} \sqrt{1\+\mathrm{diff}{\left(f\,x\right)}^2\+\mathrm{diff}{\left(f\,y\right)}^2}\right)\,z\=f\right)\right)$

$-4x^{3}y-4x{y}^3\+xy$

Use the Int and int commands to write and evaluate the flux integral

$$\begin{gathered} \mathrm{Int}\left(q\,\left[y\=Y_{12}..Y_{31}\,x\=1..3\right]\right)\+\mathrm{Int}\left(q\,\left[y\=Y_{12}..Y_{23}\,x\=3..5\right]\right)\= \\ :-\mathrm{int}\left(q\,\left[y\=Y_{12}..Y_{31}\,x\=1..3\right]\right)\+:-\mathrm{int}\left(q\,\left[y\=Y_{12}..Y_{23}\,x\=3..5\right]\right) \end{gathered}$$

${\∫}_1^3{\∫}_{\frac{9}{4}-\frac{1}{4}x}^{\frac{3}{2}\+\frac{1}{2}x}\left(-4x^{3}y-4x{y}^3\+xy\right)\ⅆy\ⅆx\+{\∫}_3^5{\∫}_{\frac{9}{4}-\frac{1}{4}x}^{6-x}\left(-4x^{3}y-4x{y}^3\+xy\right)\ⅆy\ⅆx\=-\frac{10069}{10}$

Table 9.6.16(e)   Solution from first principles