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How to integrate int ydx + zdy over a helix?

  1. Feb 3, 2008 #1
    1. The problem statement, all variables and given/known data

    [tex]\int[/tex]ydx + zdy
    C

    where C is the part of the helix r(t)=(sin t)(i) + (cos t)(j) + t(k)
    and 0 < t < pi (those should be greater than or equal to signs)


    [tex]\int[/tex](cost)(cost)-t(sint)
    [tex]\int[/tex](1+cos 2t)/2 - t sint
    with limits from zero to pi.

    this then equaled to
    1/2t + (sin 2t)/4 + tcost -sin t


    and this gave me the answer 3(pi)/2
    but the answer should be 3 pi..
    can anyone else solve this and check please or tell me where i went wrong.
     
  2. jcsd
  3. Feb 3, 2008 #2

    Dick

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    Science Advisor
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    I agree with your indefinite integral. But if i put the limits in I get -pi/2. That's still not 3*pi...? Is there a typo in the problem?
     
    Last edited: Feb 3, 2008
  4. Feb 3, 2008 #3
    i'm sorry you're right...
    that should be pi/2 - pi which is -pi/2
    i accidentally added them...
    yea. then you and i are in agreement. i guess the teacher made a mistake. i've spent hours and this was the only solution i came to.
     
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