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Line integral over a Vector Field

  1. Jun 11, 2013 #1
    1. The problem statement, all variables and given/known data

    Given a vector field

    [itex] F(x,y,z) = (yz + 3x^{2})\hat{i} + xz\hat{j} + xy\hat{k}[/itex]

    Calculate the line integral

    [itex]∫_{A}^{B}F\bullet dl[/itex]

    where A = (0,1,3) and B = (1,2,2)

    2. Relevant equations

    Right, first of all, what is dl ? I've gone over all my course notes and don't see dl anywhere. I managed to find on this site that dl = (dx, dy, dz). Is that the case?

    If so, my attempt at the solution is

    [itex]∫_{A}^{B}(yz + 3x^{2})dx + xzdy + xydz[/itex]

    Now what? I'm not really sure how to set the intervals from those 2 points in space. Am I supposed to do something like

    [itex]∫_{0}^{1}(yz + 3x^{2})dx + ∫_{1}^{2}xzdx +∫_{3}^{2}xydz[/itex] ?

    Thanks for any assistance.
     
  2. jcsd
  3. Jun 11, 2013 #2

    Dick

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    Science Advisor
    Homework Helper

    No, dl isn't equal to (dx,dy,dz) in general. dl depends on the path you are integrating along. They didn't give you a path. Any idea why not?
     
  4. Jun 11, 2013 #3
    Here is the whole question

    http://imgur.com/KGI28pJ

    http://imgur.com/KGI28pJ
     
  5. Jun 11, 2013 #4
     
    Last edited by a moderator: May 6, 2017
  6. Jun 11, 2013 #5
    Sure. But I don't understand what dl is in this question. How does one determine dl and also what are the limits of integration..? What I find confusing is distinguishing between what dl, ds, and dr all mean in various contexts. Such as

    [itex]∫_{C}F.ds[/itex]

    [itex]∫^{B}_{A}F.dl[/itex]

    and

    [itex]∫_{C}F.dr[/itex]

    what are the differences between all these...?
     
  7. Jun 11, 2013 #6

    CAF123

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    Gold Member

    You get ##\vec{dl}## from the parametrisation of your path. So what path are you going to take? Why do you think the question has not specified a path?
     
  8. Jun 11, 2013 #7
  9. Jun 11, 2013 #8

    CAF123

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    Gold Member

    Yes, that's right, but what justifies using the straight line parametrisation?
     
  10. Jun 11, 2013 #9
    Is that because it's a conservative field and calculations are therefore path independent?
     
  11. Jun 11, 2013 #10

    CAF123

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    Gold Member

    Exactly.
     
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