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

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

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

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

$\={\int }_0^1\left(2t^4-t^2-t\right) \mathrm{dt}$

$\=-13\/30$

Tools≻Tasks≻Browse:

Calculus - Vector≻Integration≻Flux≻2-D≻Through a Plane Curve

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

$\mathbf{R}\=⟨t\,t^2⟩$$\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

$⟨2 t\,-1⟩$ = $\stackrel{\text{Euclidean-normalize}}{\to }$ $\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{4t^2\+1}$$\stackrel{\text{assign to a name}}{\to }$$\mathrm{\ρ}$$$

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

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

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

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

$t\left(2t^3-t-1\right)$

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

${\∫}_0^{1}t\left(2t^3-t-1\right)\ⅆt$

$\=$

$-\frac{13}{30}$

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

$\mathrm{Flux}\left(\mathbf{F}\,\mathrm{Path}\left(⟨x\,x^2⟩\,x\=0..1\right)\,\mathrm{output}\=\mathrm{integral}\right)\=\mathrm{Flux}\left(\mathbf{F}\,\mathrm{Path}\left(⟨x\,x^2⟩\,x\=0..1\right)\right)$

${\∫}_0^1\left(2x^4-x^2-x\right)\ⅆx\=-\frac{13}{30}$

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

$\mathrm{Flux}\left(\mathbf{F}\,\mathrm{Path}\left(⟨x\,x^2⟩\,x\=0..1\right)\,\mathrm{output}\=\mathrm{plot}\,\mathrm{scaling}\=\mathrm{constrained}\right)$

Define the path parametrically, as the position vector R

$\mathbf{R}≔⟨t\,t^2⟩$

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

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

$\sqrt{4t^2\+1}$

Obtain N, a unit normal vector along the path

$\mathbf{N}≔\mathrm{Normalize}\left(⟨2 t\,-1⟩\right)$

Form and evaluate the requisite line integral

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

${\∫}_0^{1}t\left(2t^3-t-1\right)\ⅆt\=-\frac{13}{30}$