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Calculating line integrals

  1. Dec 4, 2005 #1
    Calculating line integrals....

    Ok, the problem is:
    h(x,y) = 3x (x^2 + y^4)^1/2 i + 6y^3 (x^2 + y^4)^1/2 j;
    over the arc: y = -(1 - x^2)^1/2 from (-1,0) to (1,0).

    In my notes, I had written: if h is a gradient, then the INTEGRAL of g*dr over curve C depends only on the endpoints. Also, if the curve C is closed AND h is a gradient, then the integral of g*dr over curve c is 0.

    So, my question is, when testing to see if a given function like the one above, should you just test to see if:

    partial derivative of the i component with respect to y EQUALS the partial derivative of the j component with respect to x?

    If not equal, then it is not a gradient, right?
    Thanks.
     
  2. jcsd
  3. Dec 4, 2005 #2

    arildno

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    True enough, but you have placed your focus on a SIMPLIFYING SPECIAL CASE, rather than on the general case (which is what you SHOULD focus on).
    So, how are line integrals IN GENERAL computed?

    EDIT:
    As far as your question is concerned, yes that is how you could test for whether h is a gradient field.
     
    Last edited: Dec 4, 2005
  4. Dec 4, 2005 #3
    I couldn't parametrize the arc...and I thought that perhaps taking the integrals by considering only the endpoints would be easier. In this example, I parametrized the straight line connecting the endpoints, r(u) = (2u-1)i + (0)j. This however produces the wrong answer.

    To answer the question on how to calculate the line integral:

    [INTEGRAL on curve C] h*dr, where r is the parametrization of the curve. Right?
     
  5. Dec 4, 2005 #4

    arildno

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    Try the parametrization:
    [tex]x=\cos\theta, y=\sin\theta, \pi\leq\theta\leq2\pi[/tex]

    I'm fairly sure it'll work out if you fiddle about with it for a while.
     
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