1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook