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Line integral and vector fields

  1. Feb 5, 2006 #1
    Hi, I'm having trouble with the following question.

    Q. Let p be a real constant and [itex]\mathop F\limits^ \to = \left( {yz^p ,x^p z,xy^p } \right)[/itex] be a vector field. For what value of p is the line integral

    [tex]\int\limits_{C_2 }^{} {\mathop F\limits^ \to \bullet d\mathop s\limits^ \to } = 0[/tex]

    Where C_2 is any closed path in R^2.

    Firstly, how can C_2 be a path in R^2 when the vector field is '3D'? That doesn't seem to make sense in the context of the line integral. Assuming that C_2 is any closed path in R^3 then it should be sufficient to find the values of p so that curl F = 0.

    I found [itex]curl\mathop F\limits^ \to = \nabla \times \mathop F\limits^ \to [/itex]

    = \left( {x\left( {p - 1} \right)y^{p - 1} - x^p ,y\left( {p - 1} \right)z^{p - 1} - y^p ,z\left( {p - 1} \right)x^{p - 1} - z^p } \right)

    The answer is p = 1 but substituting p = 1 to what I found doesn't give the result curl F = 0. I've checked over my calculation a few times but I still can't see what's wrong with the curlf that I've computed. Can someone help me out?
    Last edited: Feb 5, 2006
  2. jcsd
  3. Feb 5, 2006 #2


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    You're right. It doesn't make sense to define a vector field in R3 and then ask you to integrate it around a curve in R2!

    There are two things you could do- you assume they mean a path in the xy- plane, z= 0. Of course, that makes the problem trivial- I'm sure that is not what was intended. I would assume that R2 was just a typo- that it should be R3.

    You have [tex] = \left( {x\left( {p - 1} \right)y^{p - 1} - x^p ,y\left( {p - 1} \right)z^{p - 1} - y^p ,z\left( {p - 1} \right)x^{p - 1} - z^p } \right)[/tex]

    Good grief! You've correctly analyzed a problem in Green's theorem and messed up the "power law"?? The derivative of xn is nxn-1, not (n-1)xn-1 as you have!
  4. Feb 5, 2006 #3
    Heh, I don't know how I managed to miss my error with the power rule. Thanks for the help HallsofIvy.
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