$\left(\nabla \times \mathbf{F}\right)·\mathbf{N} \mathrm{d\σ}$

$\=-\frac{1}{\sqrt{77}}\left(4 y\+5 z\+6 x\right) \frac{\sqrt{77}}{6} \mathrm{dA}$

$\=-\left(4 y\+5 z\+6 x\right)\/6 \mathrm{dA}$

$\left(\nabla \times \mathbf{F}\right)·\mathbf{N} \mathrm{d\σ}$

$\=-\left(4 y\+5\left(20-2 x\/3-5 y\/6\right)\+6 x\right) \mathrm{dA}$

$\=\left(y\/36-4 x\/9-50\/3\right) \mathrm{dA}$

$L_{\mathrm{ab}}$

$\mathbf{R}\=\mathbf{a}\+t\left(\mathbf{b}-\mathbf{a}\right)$ = $\left[\begin{array}{c}30\(1-t\)\\ 24 t\\ 0\end{array}\right]$

${\int }_0^{1}30\left(1-t\right)\cdot \left(-720 t\right) \ⅆt\= -3600$ 

$L_{\mathrm{bc}}$

$\mathbf{R}\=\mathbf{b}\+t\left(\mathbf{c}-\mathbf{b}\right)$ = $\left[\begin{array}{c}0\\ 24\(1-t\)\\ 20 t\end{array}\right]$

${\int }_0^{1}24\left(1-t\right)\cdot \left(-480 t\right) \ⅆt\= -1920$ 

$L_{\mathrm{ca}}$

$\mathbf{R}\=\mathbf{c}\+t\left(\mathbf{a}-\mathbf{c}\right)$ = $\left[\begin{array}{c}30 t\\ 0\\ 20\(1-t\)\end{array}\right]$

${\int }_0^{1}20\left(1-t\right)\cdot \left(-600 t\right) \ⅆt\= -2000$ 

Table 9.9.4(a)   Parametrization of the sides of $C$

Initialize

Loading Student:-VectorCalculus

Obtain $z\=z\left(x\,y\right)$ 

$4 x\+5 y\+6 z\=120$$\stackrel{\text{solutions for z}}{\to }$$20-\frac{2}{3}x-\frac{5}{6}y$$\stackrel{\text{assign to a name}}{\to }$$Z$

Define the vector field F

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

Obtain $\nabla \times \mathbf{F}$ 

$\nabla \times \mathbf{F}$ =

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

Table 9.9.4(b)   Task template used to evaluate $\int {\int }_S\left(\nabla \times \mathbf{F}\right)·\mathbf{N} \mathrm{d\σ}$

Form and evaluate the line integral of F around $C$ 

$\mathbf{F}$

$\stackrel{\text{line integral}}{\to }$

${\∫}_0^1\left(-600t\left(20-20t\right)\right)\ⅆt\+{\∫}_0^1\left(-720\left(30-30t\right)t\right)\ⅆt\+{\∫}_0^1\left(-480\left(24-24t\right)t\right)\ⅆt$

$\=$

$-7520$

Table 9.9.3(c)   Evaluation of the line integral of F around $C$ 

Initialize

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

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

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

Use the Curl and Flux commands to obtain the flux of $\nabla \times \mathbf{F}$ through $S_1$ 

$\mathrm{Flux}\left(\mathrm{Curl}\left(\mathbf{F}\right)\,\mathrm{Surface}\left(⟨x\,y\,20-\frac{2}{3} x-\frac{5}{6} y⟩\,\left[x\,y\right]\=\mathrm{Triangle}\left(⟨0\,0⟩\,⟨0\,24⟩\,⟨30\,0⟩\right)\right)\,\mathrm{output}\=\mathrm{integral}\right)$

${\∫}_0^{30}{\∫}_0^{-\frac{4}{5}x\+24}\left(\frac{1}{36}y-\frac{50}{3}-\frac{4}{9}x\right)\ⅆy\ⅆx$

$\mathrm{Flux}\left(\mathrm{Curl}\left(\mathbf{F}\right)\,\mathrm{Surface}\left(⟨x\,y\,20-\frac{2}{3} x-\frac{5}{6} y⟩\,\left[x\,y\right]\=\mathrm{Triangle}\left(⟨0\,0⟩\,⟨0\,24⟩\,⟨30\,0⟩\right)\right)\right)$

$-7520$

$\mathrm{LineInt}\left(\mathbf{F}\,\mathrm{LineSegments}\left(⟨0\,0\,20⟩\,⟨30\,0\,0⟩\,⟨0\,24\,0⟩\,⟨0\,0\,20⟩\right)\,\mathrm{output}\=\mathrm{integral}\right)$

${\∫}_0^1\left(-600t\left(20-20t\right)\right)\ⅆt\+{\∫}_0^1\left(-720\left(30-30t\right)t\right)\ⅆt\+{\∫}_0^1\left(-480\left(24-24t\right)t\right)\ⅆt$

$\mathrm{LineInt}\left(\mathbf{F}\,\mathrm{LineSegments}\left(⟨0\,0\,20⟩\,⟨30\,0\,0⟩\,⟨0\,24\,0⟩\,⟨0\,0\,20⟩\right)\right)$ = $-7520$$$

Table 9.9.4(d)   Line integral of the tangential component of F around $C$ 

$\mathbf{a}\,\mathbf{b}\,\mathbf{c}≔⟨30\,0\,0⟩\,⟨0\,24\,0⟩\,⟨0\,0\,20⟩\:$

$\mathbf{R}≔⟨x\,y\,z⟩\:$

Obtain parametric representations of each side of $C$

Obtain $\mathbf{F}·\mathbf{dr}$ on each side of $C$ 

${\mathrm{\λ}}_{\mathrm{ab}}≔\mathrm{eval}\left(\mathrm{DotProduct}\left(\mathbf{F}\,\mathrm{diff}\left(L_{\mathrm{ab}}\,t\right)\right)\,\mathrm{Equate}\left(\mathbf{R}\,L_{\mathrm{ab}}\right)\right)\:$ 

${\mathrm{\λ}}_{\mathrm{bc}}≔\mathrm{eval}\left(\mathrm{DotProduct}\left(\mathbf{F}\,\mathrm{diff}\left(L_{\mathrm{bc}}\,t\right)\right)\,\mathrm{Equate}\left(\mathbf{R}\,L_{\mathrm{bc}}\right)\right)\:$ 

${\mathrm{\λ}}_{\mathrm{ca}}≔\mathrm{eval}\left(\mathrm{DotProduct}\left(\mathbf{F}\,\mathrm{diff}\left(L_{\mathrm{ca}}\,t\right)\right)\,\mathrm{Equate}\left(\mathbf{R}\,L_{\mathrm{ca}}\right)\right)\:$ 

Use the top-level Int and int commands to integrate along each side of $C$, and sum

$$\begin{gathered} \mathrm{Int}\left({\mathrm{\λ}}_{\mathrm{ab}}\,t\=0..1\right)\+\mathrm{Int}\left({\mathrm{\λ}}_{\mathrm{bc}}\,t\=0..1\right)\+\mathrm{Int}\left({\mathrm{\λ}}_{\mathrm{ca}}\,t\=0..1\right)\= \\ :-\mathrm{int}\left({\mathrm{\λ}}_{\mathrm{ab}}\,t\=0..1\right)\+:-\mathrm{int}\left({\mathrm{\λ}}_{\mathrm{bc}}\,t\=0..1\right)\+:-\mathrm{int}\left({\mathrm{\λ}}_{\mathrm{ca}}\,t\=0..1\right) \end{gathered}$$

${\∫}_0^1\left(-600t\left(20-20t\right)\right)\ⅆt\+{\∫}_0^1\left(-720\left(30-30t\right)t\right)\ⅆt\+{\∫}_0^1\left(-480\left(24-24t\right)t\right)\ⅆt\=-7520$

Table 9.9.4(e)   Evaluation of ${∳}_C\mathbf{F}·\mathbf{dr}$ from first principles