# Path integral over a function with image in a circunference

1. May 9, 2012

### carlosbgois

1. The problem statement, all variables and given/known data
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 $r>0$.

3. 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

Last edited: May 9, 2012
2. May 9, 2012

### 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?

Last edited: May 9, 2012
3. May 9, 2012

### 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

4. May 9, 2012

### Dick

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

5. May 9, 2012

### 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

6. May 9, 2012

### Dick

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)).

7. May 9, 2012

### 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

8. May 9, 2012

### Dick

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.