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Line integral

  1. Nov 16, 2009 #1
    1. The problem statement, all variables and given/known data

    What is the result of this? [tex]\oint[/tex]r.dr=?

    2. Relevant equations



    3. The attempt at a solution

    [tex]\oint[/tex]r.dr =[tex]\oint[/tex]rdr=[tex]\int[/tex][tex]^{a}_{a}[/tex]rdr = [tex]\frac{r^2}{2}[/tex] [tex]\left|[/tex] [tex]^{a}_{a}[/tex] = 0

    Is it correct?
     
  2. jcsd
  3. Nov 16, 2009 #2

    HallsofIvy

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    What argument do you have for asserting that [itex]\vec{r}\cdot d\vec{r}= r dr[/itex]? Generally, the dot product of two vectors is equal to the product of their lengths when the vectors are parallel. Are [itex]\vec{r}[/itex] and [itex]d\vec{r}[/itex] parallel?

    In Cartesian coordinates, [itex]\vec{r}= x\vec{i}+ y\vec{j}[/itex] and [itex]d\vec{r}= dx\vec{i}+ dy\vec{j}[/itex] so your integral becomes [itex]\oint xdx+ ydy[/itex] around some closed path.

    A circle of radius R can be written as [itex]\vec{r}= Rcos(\theta)\vec{i}+ R sin(\theta)\vec{j}[/itex] and then [itex]d\vec{r}= -Rsin(\theta)d\theta\vec{i}+ Rcos(\theta)d\theta\vec{j}[/itex] so your integral becomes [itex]R^2\int_{\theta= 0}^{2\pi} -sin(\theta)cos(\theta)+ cos(\theta) sin(\theta) d\theta[/itex] which is even simpler.
     
  4. Nov 16, 2009 #3
    Dear "HallsofIvy" thank you very much for your wonderful help. I thought the angle between r and dr is always zero because the angle between two curves are defined as the angle between their tangent lines.
     
  5. Nov 16, 2009 #4

    HallsofIvy

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    Okay. And what are the tangent lines of "r" and "dr"?
     
  6. Nov 16, 2009 #5
    Well now I think that the tangent lines are different because we have two different vectors, but I didn't notice it!!!
     
  7. Nov 17, 2009 #6

    HallsofIvy

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    I may be pulling your chain a little too hard here! Yes, r and dr are parallel (they both point directly away from the origin) and so their dot product is just rdr. My point was that you have to show that, not just assert it.

    But it is simpler to do it as [itex]R^2\int_{\theta= 0}^{2\pi} -sin(\theta)cos(\theta)+ cos(\theta) sin(\theta) d\theta[/itex] because, of course, [itex]-sin(\theta)cos(\theta)+ cos(\theta)sin(\theta)= 0[/itex].
     
  8. Nov 17, 2009 #7
    The tangent line of [tex]\vec{r}[/tex] is placed on d[tex]\vec{r}[/tex] and the tangent line of d[tex]\vec{r}[/tex] is placed on itself because it is a straight line (because [tex]\vec{r}[/tex]= x[tex]\vec{i}[/tex] + y[tex]\vec{j}[/tex]) so the angle between [tex]\vec{r}[/tex] and d[tex]\vec{r}[/tex] is zero.
    Is it correct? If there is a better interpretation please let me know. And thank you very much in advance for your great help.
     
  9. Nov 17, 2009 #8

    lanedance

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    have you looked at the definition of conservative vector fields at all?
     
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