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Stokes's Theorem

  1. Feb 16, 2008 #1
    1. The problem statement, all variables and given/known data
    For the vector field [tex]\bold{E} = \bold{ \hat x} (xy) - \bold{ \hat y} (x^2 + 2y^2)[/tex], calculate the following:

    [tex]\oint \bold{E} \cdot d\bold{l}[/tex] around the triangular contour shown.

    I don't have a scanner at the moment so I will explain the drawing. The picture is a right triangle. They show an x and y axis. From (0, 0), there is a line going to (1, 0), then from there it goes up to (1, 1), then a diagonal back to (0, 0).

    2. Relevant equations

    3. The attempt at a solution

    I know how to approach it but I seem to be stuck. This is what I have so far:

    I gave each coordinate a name: a (0, 0), b (1, 0), and c (1, 1).

    [tex]\oint \bold{E} \cdot d\bold{l} = \oint_a^b \bold{E}_{ab} \cdot d\bold{l} + \oint_b^c \bold{E}_{bc} \cdot d\bold{l} + \oint_c^a \bold{E}_{ca} \cdot d\bold{l}[/tex]

    At this point I'm stuck. Could someone please point me in the right direction?
  2. jcsd
  3. Feb 16, 2008 #2
    Your confusion probably lies in that you don't know what [tex]\vec{dl}[/tex] is. For example, for the integral from a to b, [tex]\vec{dl_{1}}=\hat{x}dl[/tex], whereas b to c it becomes [tex]\vect{dl_{2}}=\hat{y}dl[/tex]. Just work out the dot products, which are especially easy for a to b and b to c since either x or y is constant.
    Last edited: Feb 16, 2008
  4. Feb 16, 2008 #3
    Thanks for the quick reply. I have 2 questions:

    1) Shouldn't the integral from a to b, be [tex]\oint_a^b \bold{E}_{ab} \cdot \hat x d\bold{l}[/tex] since ab is horizontal?

    2) I'm not sure what to put for [tex]\bold{E}_{ab}[/tex]. Should it be the component that acts only in the direction of ab, in other words:

    [tex]\oint_a^b \bold{ \hat x} (xy) \cdot \hat x d\bold{l}[/tex]

    Am I in the right direction?
  5. Feb 16, 2008 #4
    Yes whoops you are right about 1). I will change it. And 2) yes you are right. Don't forget that y is a constant.
  6. Feb 16, 2008 #5
    Awsome. So now that I set it up, I have [tex]\oint_a^b xy d\bold{l}[/tex], but I don't know l, or am I missing something obvious?
  7. Feb 16, 2008 #6
    Remember that a is (0,0) and b is (1,0). [tex]dl[/tex] is simply dx in that particular case. The only tricky part is going to be c to a, where you're going to have to express all components of dl in terms of one variable to integrate over.
    Last edited: Feb 16, 2008
  8. Feb 16, 2008 #7


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    If the path is along the positive x axis, you must use [tex] d \vec{l} = dx \vec{i} [/tex]

    and so on.

    In general, [tex] d\vec{l} = dx \vec{i} + dy \vec{j} + dz \vec{k} [/tex]

    If you have a parametrized curve, you replace all the variables in terms of the parameter and the integral is only over one variable,
  9. Feb 16, 2008 #8
    Alright I got it. Thanks for all the help guys.
  10. Feb 16, 2008 #9


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    Since you titled this "Stokes Theorem", are you also going to integrate -2x-2y over the triangular region and show that they are the same?
  11. Feb 18, 2008 #10
    Actually yes, and I'm stuck on that too.

    Also, I thought I figured out how to finish the first part, but it's not working out.

    Here's what I'm getting:

    [tex]\oint \bold{E} \cdot d\bold{l} = \oint_0^1 xy dx + \oint_0^1 -x^2-2y^2 dy + \oint_c^a \bold{E}_{ca} \cdot d\bold{l}[/tex]
    but I don't know what to do with the last term. And when I integrate the first two terms, I get

    which doesn't make sense, because the answer for the whole is -1. What should I do with the last term? Shouldn't it be:

    [tex]\oint_0^1 (xy - x^2 +2y^2) dxdy[/tex]
  12. Feb 19, 2008 #11


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    Homework Helper

    I got -1 for both methods. Did you get -1 for the line integral?
  13. Apr 20, 2009 #12
    this is a perfect question to use Green"s theorem. And the answer is -1. Stokes' theorem is just a generalized version of Green's th. and I don't see how that would be useful in this case. Also there is no need to do 3 integrations. Just convert the whole thing to a double integral using Green's theorem, integrate over the triangle and get -1
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