Path integral over a function with image in a circunference

[carlosbgois]

Homework Statement

Let \vec{F}: ℝ^{2}->ℝ^{2} be a continuous vector field in which, for every (x, y), \vec{F}(x, y) is parallel to x\vec{i}+y\vec{j}. Evaluate \int_{γ}\vec{F}\cdot d\vec{r} where γ:[a, b]->ℝ^{2} is a curve of class C^{1}, and it's imagem is contained in the circunference centered in the origin and with radius r>0.

The Attempt at a Solution

We know that \vec{F}(x, y)=a(x\vec{i}+y\vec{j}), where a is a constant, and we need to evaluate \int^{b}_{a}\vec{F}(γ(t))\cdot γ'(t)dt. So?...

Thanks

 

[Dick]

So how is the direction γ'(t) points related to the direction \vec{F}(x, y) points? What does that tell you about the dot product?

 

[carlosbgois]

γ '(t) is a tangent to x^{2}+y^{2}=r at any given (x, y), but not every \vec{F}(x, y) is perpendicular to the circunference, so the dot product has many possible values, and my argument must be wrong HEHE

Thanks

 

[Dick]

I think \vec{F}(x, y) is perpendicular to the circumference, isn't it?

 

[carlosbgois]

Indeed it is, my mistake. So I know that F and γ' are always perpendicular, hence the path integral is equal to 0. Now, how may I write this rigorously, I mean, not using just the geometric description?

Many Thanks

 

[Dick]

Indeed it is, my mistake. So I know that F and γ' are always perpendicular, hence the path integral is equal to 0. Now, how may I write this rigorously, I mean, not using just the geometric description?

Many Thanks

You could parametrize the circle. Write things in term of the angle at the value of the parameter t, θ(t). Then x=r*cos(θ(t)) and y=r*sin(θ(t)).

 

[carlosbgois]

Nice. So let R=(r cos(t), r sin(t)), then R is a parametric form for γ, when we let r constant. Then we have \int_{γ}\vec{F} \cdot d\vec{r}=\int^{d}_{c}F(R) \cdot R' dt=ar^{2}\int^{d}_{c}0 \cdot dt=0 for any d and c.

Is it correct?
Thank you

 

[Dick]

Nice. So let R=(r cos(t), r sin(t)), then R is a parametric form for γ, when we let r constant. Then we have \int_{γ}\vec{F} \cdot d\vec{r}=\int^{d}_{c}F(R) \cdot R' dt=ar^{2}\int^{d}_{c}0 \cdot dt=0 for any d and c.

Is it correct?
Thank you

Basically, yes. But you shouldn't assume the parameter t is the same as the angle θ(t). If you work it out with a general function θ(t) you'll get the same result.