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Homework Help: 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?
  2. jcsd
  3. Dec 4, 2005 #2


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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?

    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


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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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