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:

$$\int_0^1 (t^2-4t^4-16t^4)dt=\frac{-31}{15}$$

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 $y$,

$$\int_0^1 (2t^5-8t^3+8t^3)dt=\frac{1}{3}$$

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


  • $\begingroup$ Break the last one up into a sum of two integrals. $\endgroup$ – amd Nov 24 '15 at 21:00

You are correct for the first two paths.

For the last path note that the path can be parametrized in two parts as: $$ x=t\qquad y=0 \qquad for \quad 0<t<2\\ $$ and

$$ x=2-t\qquad y=2t \qquad for \quad 0<t<1\\ $$

so that the entire path integral is the sum of two integrals.

Here the sum: $$ \int_0^2t^2dt =\frac{8}{3} $$ for the first part : $(0,0)\to(0,2)$ .

$$ \int_0^1-\{\left[(2-t)^2-(2t)^2 \right]-8t(2-t)\} dt=\int_0^1(11t^2-12t-4)dt=-\frac{19}{3} $$ for the second part $(2,0)\to (1,2)$ so the entire path integral is $\frac{8}{3}-\frac{19}{3}=-\frac{11}{3}$

  • $\begingroup$ I think I solved it. I will edit with what I think it should be $\endgroup$ – Jack Armstrong Nov 24 '15 at 21:12
  • 1
    $\begingroup$ Yes, it's correct. I've corrected my answer, you have used a different parametrization but the result is the same. $\endgroup$ – Emilio Novati Nov 24 '15 at 21:29
  • $\begingroup$ Thank you. I also figured I would accept your answer since you were the only one that answered. Maybe post your way with the parameterization? $\endgroup$ – Jack Armstrong Nov 24 '15 at 21:33
  • $\begingroup$ I've added to my answer. $\endgroup$ – Emilio Novati Nov 24 '15 at 22:33

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