Path integral over a function with image in a circunference

  • #1

Homework Statement


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


The Attempt at a Solution


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

Thanks
 
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Answers and Replies

  • #2
Dick
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So how is the direction [itex]γ'(t)[/itex] points related to the direction [itex]\vec{F}(x, y)[/itex] points? What does that tell you about the dot product?
 
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  • #3
[itex]γ '(t)[/itex] is a tangent to [itex]x^{2}+y^{2}=r[/itex] at any given [itex](x, y)[/itex], but not every [itex]\vec{F}(x, y)[/itex] is perpendicular to the circunference, so the dot product has many possible values, and my argument must be wrong HEHE

Thanks
 
  • #4
Dick
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I think [itex]\vec{F}(x, y)[/itex] is perpendicular to the circumference, isn't it?
 
  • #5
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
Dick
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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)).
 
  • #7
Nice. So let [itex]R=(r cos(t), r sin(t))[/itex], then R is a parametric form for [itex]γ[/itex], when we let r constant. Then we have [itex]\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[/itex] for any d and c.

Is it correct?
Thank you
 
  • #8
Dick
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Nice. So let [itex]R=(r cos(t), r sin(t))[/itex], then R is a parametric form for [itex]γ[/itex], when we let r constant. Then we have [itex]\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[/itex] 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.
 

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