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Homework Help: Line integral (parametric)

  1. Feb 15, 2009 #1
    1. The problem statement, all variables and given/known data

    For the field [tex]\bold{F} = (y+z) \bold{i} - (x+z) \bold{j} + (x+y) \bold{k}[/tex] find the work done in moving a particle around the following closed curve:

    from the origin to (0,0,2π) on the curve x=1-cos t, y=sin t, z=t; and back to the origin along the z-axis. The answer is 2π. (This question is from Mary L Boas' textbook btw, which is why I have the answer.)

    2. Relevant equations

    We get:
    dx = sin t dt
    dy = cos t dt
    dz = dt

    x = 1 - cos t
    y = sin t
    z = t

    I did this two ways and didn't get either right. Could someone point out what was wrong with both methods and give me a hint?

    First method: I assumed that dx = dy = 0. This gives [tex]\int_0^{2\pi} (1-cos t + sin t) dt = 2\pi[/tex]. But going by this method, going back to origin means that the integral for the reverse action is -2π, so the work is 0. This method is probably wrong because, obviously, x=1 - cos t is not a "static" function.

    Second method: Expand everything in full, which means

    [tex]W = \int F . dr = \int (sin t + t)(sin t) dt - (1 - cos t + t)(cos t) dt + (1 - cos t + sin t) dt = \int_0^{2\pi} 2 + sin t - 2 cos t dt = 4\pi[/tex], which is also wrong. For the reverse action, the limits are flipped and I get 0 again.

    I think this is the right method, but there's something wrong with the integration here although its supposed to be pretty simple.
     
  2. jcsd
  3. Feb 15, 2009 #2

    tiny-tim

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    Hi bigevil! :smile:
    what happened to tsint and tcost?

    also, dr2 = … ? :smile:
     
  4. Feb 15, 2009 #3
    Hi Tim, what do you mean dr^2? There shouldn't be a square in the line integral (F.dr) right?

    I think I saw where I got it wrong... I did integration by parts for t cos t and got 0 (between 2π and 0)... then I removed the t cos t / t sin t from the expression. But the integration by parts for t sin t is 2π.

    So, 4π - 2π = 2π, but I have another problem, that integral was just one direction. In the other direction the integral is -2π, so the ans. is 0!!
     
  5. Feb 18, 2009 #4

    tiny-tim

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    You need to know what dr is in terms of t …

    I only asked about dr squared because it's easier to calculate! :smile:
    What do you mean, "the other direction"? :confused:

    That's the negative z-direction, and F.k = x + y.
     
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