I have
$
F(x,y) = x^{2}\boldsymbol{i} + xy\boldsymbol{j}
$
and two parametric representation
$ r_{1}(t) = t\boldsymbol{i} + t^{2}\boldsymbol{j}\quad 0\leq t \leq 1 \ $
$ r_{2}(t) = \sin \Theta \boldsymbol{i}+ \sin^{2}\Theta \boldsymbol{j} \quad 0\leq \Theta \leq \frac{\pi}{2} \\\ $
I have no problem finding the answer for this two parametric representations wich is $\frac{11}{15} $ , that is what the book is asking for, but I have a problem is finding another parametric representation that does not throw me that value; based on what the theorem says about PATH AND CONSERVATIVE VECTOR FIELDS: that if F is continue on an open connected region , then the line integral $\int \boldsymbol{F} \cdot d\boldsymbol{r}$ is independent of the path if and only if $\boldsymbol{F}$ is conservative , then in know that $F(x,y)$ is not conservative because
$ \frac{\partial M}{\partial y} = 0 \neq \frac{\partial N}{\partial x } = y $
the excersise is not asking me for another parametric representation, but I want to see the theorem in action. I have tried using mathematica, what I have done soo far is plotting $r_{1}(t)$, $r_{2}(t)$ with this code
ListPointPlot3D[Table[{t, t^2, t^2 + t^3}, {t, 0, 1, 0.01}], Filling -> 0]
ListPointPlot3D[Table[{Sin [x], Sin [x]^2, Sin [x] * Cos [x] + 2*Sin [x] *Cos [x]} , {x, 0, \[Pi]/2, 0.01}], Filling -> 0]
which graphically seems to be a different trajectory, but I need another parametric representation that does not give me $\frac{11}{15}$, according to the theorem
the exercise is taken from larson calculus 9th, 15.3.1
Let us try to move from $(0,0)$ to $(1,1)$ on a cubic curve. Set
$$ r_3(t) = (t,t^3), \,\,\, t \in [0,1]. $$ Then
$$ \int_{r_3} x^2 \, dx + xy \, dy = \int_0^1 (t^2 \cdot 1+t\cdot t^3\cdot 3t^2) \, dt = \int_0^1 (t^2 + 3t^6) \, dt = \frac{1}{3} + \frac{3}{7} = \frac{16}{21}. $$
BTW, there is nothing special about the vector field $F$ you are working with. The reason you have
$$ \int_{r_1} G(x,y) \, dx + H(x,y) \, dy = \int_{r_2} G(x,y) \, dx + H(x,y) \, dy $$
is that $r_2$ is a reparametrization of the path $r_1$ (namely, $r_2(t) = r_1(\sin(t))$) and the line integrals are invariant under reparametrizations of curves whether or not the vector field $(G(x,y),F(x,y))$ you are integrating is conservative (or locally conservative). This is just an expression of the substitution rule of integration.