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Define the vector field F

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

Obtain Maple's scalar potential

$\mathbf{F}$ = $\stackrel{\text{scalar potential}}{\to }$$-x{y}^3\+x^{2}y$

Table 9.7.3(a)   Context Panel access to the ScalarPotential command

Define the components of F as the functions $f\left(x\,y\right)$ and $g\left(x\,y\right)$ 

$f\left(x\,y\right)\={\mathbf{F}}_1$$\stackrel{\text{assign as function}}{\to }$$f$$$

$g\left(x\,y\right)\={\mathbf{F}}_2$$\stackrel{\text{assign as function}}{\to }$$g$

Implement Recipe 1, Table 9.7.4

${\int }_a^{x}f\left(t\,b\right) \mathit{\ⅆ}t\+{\int }_b^{y}g\left(x\,t\right) \mathit{\ⅆ}t$

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

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

$a{b}^3-x{y}^3-a^{2}b\+x^{2}y$

$\stackrel{\text{evaluate at point}}{\to }$

$-x{y}^3\+x^{2}y$

Table 9.7.3(b)   Implementation of Recipe 1, Table 9.7.4

Initialize

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

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

Define the vector field F and obtain a scalar potential for it

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

$\mathrm{ScalarPotential}\left(\mathbf{F}\right)$ = $-x{y}^3\+x^{2}y$$$

Table 9.7.3(c)   Application of the ScalarPotential command

Define the components of F as the functions $f$ and $g$ 

$f≔\left(x\,y\right)\to 2 x y-y^3\:$

$g≔\left(x\,y\right)\to x^2-3 x y^2\:$

Use the Int command to write the integrals of Recipe 1, Table 9.7.4

$\mathrm{Int}\left(f\left(t\,b\right)\,t\=a..x\right)\+\mathrm{Int}\left(g\left(x\,t\right)\,t\=b..y\right)$

${\∫}_a^x\left(-b^3\+2bt\right)\ⅆt\+{\∫}_b^y\left(-3t^{2}x\+x^2\right)\ⅆt$

Access the top-level int command by prefixing "colon-dash" to int

$u≔:-\mathrm{int}\left(f\left(t\,b\right)\,t\=a..x\right)\+:-\mathrm{int}\left(g\left(x\,t\right)\,t\=b..y\right)$

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

Apply the simplify command to the result obtained by use of the eval command for setting $a\=b\=0$.

$\mathrm{simplify}\left(\mathrm{eval}\left(u\,\left[a\=0\,b\=0\right]\right)\right)$

$-x{y}^3\+x^{2}y$

Table 9.7.3(d)   Implementation of Recipe 1, Table 9.7.4