${\int }_C\mathbf{F}\mathbf{·}\mathbf{N}\mathbf{ }\mathrm{ds}$

$\={\int }_a^{b}f \mathrm{dy}- g \mathrm{dx}$

$\={\int }_0^1\left[\left(\genfrac{}{}{0ex}{}{x y}{\phantom{x\=a}}|\genfrac{}{}{0ex}{}{\phantom{\mathrm{f(x)}}}{x\=x\left(t\right)\,y\=y\left(t\right)}\right)\cdot \frac{d}{\mathrm{dt}}y\left(t\right)-\left(\genfrac{}{}{0ex}{}{x\+y}{\phantom{x\=a}}|\genfrac{}{}{0ex}{}{\phantom{\mathrm{f(x)}}}{x\=x\left(t\right)\,y\=y\left(t\right)}\right)\cdot \frac{d}{\mathrm{dt}}x\left(t\right)\right] \mathrm{dt}$

$\={\int }_0^1\left[\left(1\+2 t\right)\left(2\+3 t\right)\cdot 3-\left(\left(1\+2 t\right)\+\left(2\+3 t\right)\right)\cdot 2\right] \mathrm{dt}$

$\={\int }_0^1\left(18t^2\+11t\right) \mathrm{dt}$

$\=23\/2$

${\int }_C\mathbf{F}\mathbf{·}\mathbf{N}\mathbf{ }\mathrm{ds}$

$\={\int }_a^{b}f \mathrm{dy}- g \mathrm{dx}$

$\={\int }_0^1\left[\left(\genfrac{}{}{0ex}{}{x y}{\phantom{x\=a}}|\genfrac{}{}{0ex}{}{\phantom{\mathrm{f(x)}}}{x\=x\left(t\right)\,y\=y\left(t\right)}\right)\cdot \frac{d}{\mathrm{dt}}y\left(t\right)-\left(\genfrac{}{}{0ex}{}{x\+y}{\phantom{x\=a}}|\genfrac{}{}{0ex}{}{\phantom{\mathrm{f(x)}}}{x\=x\left(t\right)\,y\=y\left(t\right)}\right)\cdot \frac{d}{\mathrm{dt}}x\left(t\right)\right] \mathrm{dt}$

$\={\int }_0^1\left[\left(3\+2 t\right)\left(5-2 t\right)\cdot \left(-2\right)-\left(\left(3\+2 t\right)\+\left(5- t\right)\right)\cdot 2\right] \mathrm{dt}$

$\={\int }_0^1\left(8t^2-8 t-46\right) \mathrm{dt}$

$\=-142\/3$

${\int }_C\mathbf{F}\mathbf{·}\mathbf{N}\mathbf{ }\mathrm{ds}$

$\={\int }_a^{b}f \mathrm{dy}- g \mathrm{dx}$

$\={\int }_0^1\left[\left(\genfrac{}{}{0ex}{}{x y}{\phantom{x\=a}}|\genfrac{}{}{0ex}{}{\phantom{\mathrm{f(x)}}}{x\=x\left(t\right)\,y\=y\left(t\right)}\right)\cdot \frac{d}{\mathrm{dt}}y\left(t\right)-\left(\genfrac{}{}{0ex}{}{x\+y}{\phantom{x\=a}}|\genfrac{}{}{0ex}{}{\phantom{\mathrm{f(x)}}}{x\=x\left(t\right)\,y\=y\left(t\right)}\right)\cdot \frac{d}{\mathrm{dt}}x\left(t\right)\right] \mathrm{dt}$

$\={\int }_0^1\left[\left(5\+2 t\right)\left(3\+3 t\right)\cdot 3-\left(\left(5\+2 t\right)\+\left(3\+3 t\right)\right)\cdot 2\right] \mathrm{dt}$

$\={\int }_0^1\left(18t^2\+53t\+29\right) \mathrm{dt}$

$\=123\/2$

Tools≻Tasks≻Browse:

Calculus - Vector≻Integration≻Flux≻2-D≻Through a Polygonal Line

Initialize

Loading Student:-VectorCalculus

Define the vector field F

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

Define the path parametrically, as a position vector

$P\=\left[⟨1\,2⟩\,⟨3\,5⟩\right]$$\stackrel{\text{assign}}{\to }$

$\mathbf{R}\={\mathbf{P}}_1\+t \left({\mathbf{P}}_2-{\mathbf{P}}_1\right)$$\stackrel{\text{assign}}{\to }$

Obtain $\stackrel{\mathit{\.}}{\mathbf{R}}$, a tangent vector along the path

$\frac{\ⅆ}{\ⅆ t} \mathbf{R}$ = $$

Obtain N, a unit normal vector along the path

$⟨3\,-2⟩$ = $\stackrel{\text{Euclidean-normalize}}{\to }$ $\stackrel{\text{assign to a name}}{\to }$$N$$$

Obtain $\mathrm{\ρ}\=\frac{\mathrm{ds}}{\mathrm{dt}}\=∥\stackrel{\.}{\mathbf{R}}∥$ 

$∥\frac{\ⅆ}{\ⅆ t} \mathbf{R}∥$ = $\sqrt{13}$$\stackrel{\text{assign to a name}}{\to }$$\mathrm{\ρ}$$$

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

$\genfrac{}{}{0ex}{}{\mathbf{F}·\mathbf{N} \mathrm{\ρ}}{\phantom{x\=a}}|\genfrac{}{}{0ex}{}{\phantom{\mathrm{f(x)}}}{x\=R_1\,y\=R_2}$

$\left(\frac{3}{13}\left(1\+2t\right)\left(2\+3t\right)\sqrt{13}-\frac{2}{13}\left(3\+5t\right)\sqrt{13}\right)\sqrt{13}$

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

$t\left(18t\+11\right)$

$\stackrel{\text{integrate w.r.t. t}}{\to }$

${\∫}_0^{1}t\left(18t\+11\right)\ⅆt$

$\=$

$\frac{23}{2}$

Initialize

$$\begin{gathered} \mathrm{with}\left(\mathrm{Student}:-\mathrm{VectorCalculus}\right)\: \\ \mathrm{BasisFormat}\left(\mathrm{false}\right)\: \end{gathered}$$

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

Apply the Flux command

$$\begin{gathered} \mathrm{Flux}\left(\mathbf{F}\,\mathrm{LineSegments}\left(⟨1\,2⟩\,⟨3\,5⟩\,⟨5\,3⟩\,⟨7\,6⟩\right)\,\mathrm{output}\=\mathrm{integral}\right)\= \\ \mathrm{Flux}\left(\mathbf{F}\,\mathrm{LineSegments}\left(⟨1\,2⟩\,⟨3\,5⟩\,⟨5\,3⟩\,⟨7\,6⟩\right)\right) \end{gathered}$$

${\∫}_0^1\left(3\left(1\+2t\right)\left(2\+3t\right)-6-10t\right)\ⅆt\+{\∫}_0^1\left(-2\left(3\+2t\right)\left(5-2t\right)-16\right)\ⅆt\+{\∫}_0^1\left(3\left(5\+2t\right)\left(3\+3t\right)-16-10t\right)\ⅆt\=\frac{77}{3}$

Use  the Flux command to generate a graph of F and the path of integration

$\mathrm{Flux}\left(\mathbf{F}\,\mathrm{LineSegments}\left(⟨1\,2⟩\,⟨3\,5⟩\,⟨5\,3⟩\,⟨7\,6⟩\right)\,\mathrm{output}\=\mathrm{plot}\,\mathrm{scaling}\=\mathrm{constrained}\right)$

Define the path parametrically, as the position vector R

$$\begin{gathered} L≔\left[⟨1\,2⟩\,⟨3\,5⟩\,\right]\: \\ \mathbf{R}≔{\mathbf{L}}_1\+t \left({\mathbf{L}}_2-{\mathbf{L}}_1\right) \end{gathered}$$

Obtain $\mathrm{\ρ}\=\frac{\mathrm{ds}}{\mathrm{dt}}\=∥\stackrel{\.}{\mathbf{R}}∥$ 

$\mathrm{\ρ}≔\mathrm{Norm}\left(\mathrm{diff}\left(\mathbf{R}\,t\right)\right)$

$\sqrt{13}$

Obtain N, a unit normal vector along the path

$\mathbf{N}≔\mathrm{Normalize}\left(⟨3 t\,-2 t⟩\right) assuming t\>0$

Form and evaluate the requisite line integral

$$\begin{gathered} \mathrm{Int}\left(\mathrm{simplify}\left(\mathrm{eval}\left(\mathrm{DotProduct}\left(\mathbf{F}\mathbf{\,}\mathbf{N}\right)\cdot \mathrm{\ρ}\,\left[x\=R_1\,y\=R_2\right]\right)\right)\,t\=0..1\right)\= \\ \mathrm{int}\left(\mathrm{simplify}\left(\mathrm{eval}\left(\mathrm{DotProduct}\left(\mathbf{F}\mathbf{\,}\mathbf{N}\right)\cdot \mathrm{\rho }\,\left[x\=R_1\,y\=R_2\right]\right)\right)\,t\=0..1\right) \end{gathered}$$

${\∫}_0^{1}t\left(18t\+11\right)\ⅆt\=\frac{23}{2}$