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Help with Green's Theorem in the Plane

  1. Sep 16, 2013 #1
    Hey Y'all, this problem is bugging me, and I can't figure out what exactly I am doing incorrectly.

    1. The problem statement, all variables and given/known data

    So the problem asks to evaluate the integral along a contour of the function (e^x)*cos(y)*dx-(e^x)*sin(y)*dy, where the contour C is a broken line from A = (ln(2),0) to D = (0,1) to B = (-ln(2),0).


    2. Relevant equations

    I know that the theorem states that the integral of a vector field dotted into a small portion dl of the contour is equal to the double integral of the the normal component (in this case the z component) of the curl of the vector field. So ∫V(dot)dl over a closed contour = ∫∫(partial with respect to x of the y-component of the vector field - partial with respect to y of the x-component of the vector field)dσ over the area σ.

    3. The attempt at a solution

    I get stuck once I try to convert the single integral to the double integral. I know that if I connect B to A, I can make a closed contour, which was what I intended to integrate over. So I'm thinking I should try to evaluate the contour but I don't know how to go between x and y in order to do this. At first I wanted to convert to the double integral form in order to evaluate this, however when I do so I get that -e^x*sin(y)-(e^x*-cos(y)) = 0 which I don't think I should get. Am I computing something incorrectly? Am I just thinking about this incorrectly? Do I not want to take the partial of -e^x*sin(y) with respect to x minus the partial with respect to y of e^x*cos(y)?

    Any help at all would be greatly appreciated. And this might go without saying, but I am gonna say it anyways just in case: Please, please please DON'T tell me how to solve this problem, just what I am doing incorrectly. I really want to figure this out on my own. That isn't to say I don't appreciate any help, it's just that I want to understand this problem, not jut get the right answer.

    Thank you :smile:
     
  2. jcsd
  3. Sep 16, 2013 #2
    I think you might have a typo here, because that doesn't equal 0. Nevertheless, if you work out the partials and do get 0 (I checked it out quickly and seem to also get 0), then the closed contour integral should also go to zero. What makes you suggest that it is wrong?

    Also, note in your problem description that the contour doesn't necessarily say to connect B back to A. If it's supposed to be a closed contour, then proceed as above. But you can't just close the contour and apply Green's Theorem and expect the same answer to an open contour (hint!). Be sure the problem isn't asking you to just integrate over the 2 line segments instead, in which case, less shortcuts.


    Helpful?
     
    Last edited: Sep 16, 2013
  4. Sep 16, 2013 #3

    Dick

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    What is the curl of the given vector field? Can you use that to easily compute the contour integral over the closed triangular path ADB? Then can you compute the contour over the segment BA? Try to use that to find the contour over your 'broken line'.
     
  5. Sep 16, 2013 #4

    Dick

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    There is a shortcut. You can easily do the contour BA and use that to deduce the harder part.
     
  6. Sep 16, 2013 #5
    True--not an all-in-one shortcut, but a shortcut nontheless!
     
  7. Sep 16, 2013 #6
    So are you guys saying that because the total curl is zero for the closed contour then the line integral with respect to the vector field from B to A must be the same as the line integral over the broken line of the vector field?
     
  8. Sep 16, 2013 #7
    Good catch! I did mean -sin(y) instead of -cos(y). Thanks for that.
     
  9. Sep 16, 2013 #8

    Dick

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    I wouldn't say they were the same. I would say they must be negatives of each other. Why would I say that?
     
  10. Sep 16, 2013 #9
    Because it makes sense! If the curl in the x-y plane is zero then the closed loop MUST be zero as well, by Green's Theorem. And since the vector field is conservative, we know that the integral of any path between any two points must have the same value as any other path between the same two points. Using that and the fact that the final direction to connect A and B is opposite going from A to D to B, we must say that these two are equal in magnitude but opposite in sign.

    Thank you both, that helped a lot!
     
  11. Sep 16, 2013 #10
    Nice job working it out!
     
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