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Line integral of a spherical vector field over cartesian path

  1. Sep 28, 2013 #1
    1. The problem statement, all variables and given/known data

    Compute the line integral of

    [itex]\vec{v} = (rcos^{2}\theta)\widehat{r} - (rcos\theta sin\theta)\widehat{\theta} + 3r\widehat{\phi}[/itex]

    over the line from (0,1,0) to (0,1,2) (in Cartesian coordinates)

    3. The attempt at a solution

    Well, I expressed the path as a parametrized vector

    [itex]\vec{r}(t) = \frac{1}{sint} \widehat{r} + t\widehat{\theta} + \frac{\pi}{2} [/itex], t:(arctan(1/2), pi/2)

    the derivative of which is

    [itex]\vec{r}'(t) = -\frac{cost}{sin^{2}t} \widehat{r} + \widehat{\theta}[/itex]

    I'm looking for the integral to be equal to 2, but whenever I work it out I get a mess of logarithms and square roots. Have I parameterized this the wrong way?
     
  2. jcsd
  3. Sep 28, 2013 #2

    tiny-tim

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    Hi PeteyCoco! :smile:
    Wouldn't it be massively easier to use the parameter z ? :confused:

    (and convert v to Cartesian)
     
  4. Sep 28, 2013 #3
    Yeah, it was much simpler. Should I be able to get exactly the same answer if I work it out in terms of spherical coordinates though? When I did the arithmetic on the answer from that process I got something around 1.975 instead of 2. Is that due to poor math on my part or is the result off because of the unnatural fit of a line in spherical coordinates?
     
  5. Sep 29, 2013 #4

    tiny-tim

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    Hi PeteyCoco! :smile:

    (just got up :zzz:)
    Yes, but you have to be very careful about the line element in spherical coordinates (I expect that's where you went wrong) …

    in Cartesian corordinates, it's just dz, with no factors !! :wink:

    If you want us to find your mistake (the answer should have been exactly the same), you'll have to type it out for us. :smile:
     
  6. Sep 29, 2013 #5

    vanhees71

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    The [itex]z[/itex]-axis is a coordinate singularity in standard spherical coordinates!
     
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