Understanding Curvilinear Integrals: A Scientist's Dilemma

[Telemachus]

Homework Statement

Hi. I have a doubt with this exercise. I'm not sure about what it asks me to do, when it asks me for the curvilinear integral. The exercise says:

Calculate the next curvilinear integral:

$$\displaystyle\int_{C}^{}(x^2-2xy)dx+(y^2-2xy)dy$$, C the arc of parabola $$y=x^2$$ which connect the point $$(-2,4)$$ y $$(1,1)$$
I've made a parametrization for C, that's easy: $$\begin{Bmatrix} x=t \\y=t^2\end{matrix}$$ $$\begin{Bmatrix} x'(t)=1 \\y'(t)=2t\end{matrix}$$

And then I've made this integral:
$$\displaystyle\int_{-2}^{1}t^2-2t^3+(t^4-2t^3)2t dt$$
But now I'm not too sure about this. What I did was:

$$\displaystyle\int_{a}^{b}F(\sigma(t))\sigma'(t)dt$$

But now I don't know if I should use the module, I did this: $$\displaystyle\int_{a}^{b}F(\sigma(t))\sigma'(t)dt$$ and I don't know when I should use this: $$\displaystyle\int_{a}^{b}F(\sigma(t)) \cdot ||\sigma'(t)||dt$$

I mean, both are curvilinear integrals, right?

I think that I understand what both cases means, but I don't know which one I should use when it asks me for the "curvilinear integral". The first case represents the area between the curve and the trajectory, and the second case represents the projection of a vector field over the trajectoriy, i.e. the work in a physical sense, but I know it have other interpretations and uses.

Well, that's all. Bye there, thanks for posting.

 

[jackmell]

Homework Statement

Hi. I have a doubt with this exercise. I'm not sure about what it asks me to do, when it asks me for the curvilinear integral. The exercise says:

Calculate the next curvilinear integral:

$$\displaystyle\int_{C}^{}(x^2-2xy)dx+(y^2-2xy)dy$$, C the arc of parabola $$y=x^2$$ which connect the point $$(-2,4)$$ y $$(1,1)$$
I've made a parametrization for C, that's easy: $$\begin{Bmatrix} x=t \\y=t^2\end{matrix}$$ $$\begin{Bmatrix} x'(t)=1 \\y'(t)=2t\end{matrix}$$

And then I've made this integral:
$$\displaystyle\int_{-2}^{1}t^2-2t^3+(t^4-2t^3)2t dt$$
But now I'm not too sure about this. What I did was:

Try not to use tex tags in-line. It messes up the format. For starters, looks like you have a y-prime in that integral. Also, why not just call it a line integral:

$$\displaystyle\int_{C}^{}(x^2-2xy)dx+(y^2-2xy)dy$$

then making the paramaterizations you suggested, obtain:

$$\displaystyle\int_{-2}^{1}t^2-2t^3+(t^4-2t^3)2t dt$$

That looks like it to me.

 

[Telemachus]

Thanks. Sorry for the bad use of tex, didn't know about it. So the last case must only be used when the problem specifically asks me for the area between the curve and the xy plane?

 

[HallsofIvy]

That makes no sense. The curve you give is in the xy-plane. In your first post you asked for an area "between the curve and the trajectory" but I have no idea what "trajectory" you are talking about.

 

[Telemachus]

You're right, it was a misinterpretation from my part. I thought at first that the exercise was asking for that, but now I know that the area down a trajectory is obtained using the integral of arc length, and what it asked is what I actually did, but it isn't what I thought I was doing.

I think its clear now. But I have this other exercise, which asks me to calculate the circulation of a velocity field on V over the indicated path, this is it, maybe I should make another thread, but will see, its pretty much like the same kind of exercise.

$$\vec{V}=xy^2 \hat{i}+xe^{xy} \hat{j},y=x^2,x=0,y=1$$

I think I know what I have to do, which is the same than I did before with the exercise I've posted at first. But the thing is I don't know if I must use the path $$y=x^2$$ from (0,0) to (1,1), or if it reefers to three different paths with $$y=x^2,x=0,y=1$$ In that case I wouldn't know which extremes to use in the line integral. And the exercise that follows its similar, it gives a path, but it don't tells from where to where, but I think that in that case could be because the path is closed.

Bye and thanks!