I am trying to find the line integral of the vector field $F(x,y)=(x^2-y^2)$x$-2xy$ y from $(0,0)$ to $(1,2)$ along a few different paths.
First path is along curve $y=2x^2$
Second path is curve described by $x=t^2$ and $y=2t$
Last path is the path that goes from $(0,0)$ to $(2,0)$ along the x-axis and then along line $(2,0)$ to $(1,2)$.
This is what I have so far.
For the first one:
$$\int F(x,y)\cdot dr=\int ((x^2-y^2)+(-2xy)4t) dt$$
I said $r(t)=x(t)$x + $y(t)$ y. Therefore $dr=(1$x+$4t$y)dt. This part is algebra and derivatives.
Thus by setting $x=t$ and $y=2x^2=2t^2$, we can solve the integral as follows:
For the second one: Same process, but this time, $r(t)=t^2$x+$2t$y. Therefore, $dr=(2t$ x+$2$ y)$dt$. Now the integral becomes $$\int F(x,y)\cdot dr=\int ((x^2-y^2)t^2+(-2xy)2) dt$$
Now with substituting for $x$ and $y$,
For the last one: I broke this up first along the x-axis then along the line $y=-2x+4$. Now we have
$$\int F_x dx+\int F \cdot dr=\int_0^1(x^2-y^2)dx+\int_2^1((x^2-y^2)+2xy*-2)dx$$ I found out that $dy=-2$ since the line from $(2,0)$ to $(1,2)$. So for the first integral, y=0, but the second one, $y=-2x+4$. Thus left with