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Line integrals

  1. Jan 18, 2014 #1
    Find the work done by the Force Field F to make a displacement on the curve C.

    F= <-y^2 , x>

    C: semicircle x^2 + y^2 = 1 , y<=0 , from (-1,0) to (1,0)


    since y<=0 , then it's the semi circle under the x-axis. and according to the solution I have:

    Work=integral[sin t - sin t cos^2 t +(1+cos 2t)/2] dt form t= -pi to t=0

    Now I understand everything going on except the limits. why from -pi to 0? why not from pi to 2pi?

    And it may not make a difference here, but if the integral had a constant inside it would.
     
  2. jcsd
  3. Jan 18, 2014 #2

    tiny-tim

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    hi user3! welcome to pf! :smile:
    either will do
    no, if the field is a function of x and y, the integrand would always be the same whenever you add 2π to the angle :wink:
     
  4. Jan 18, 2014 #3
    Ok...let's say you have the following double integral: integral[integral(x^2 + y)] in region D, where D is the semicircle below the x axis.

    If one decides to use polar coordinates: x=rcos(theta) and y=rsin(theta)

    the integral becomes integral[ integral(r^3 cos^2 (theta) + r^2 sin(theta)) dr) d(theta)] , what are the proper limits of the integral?

    please note that if you rewrite the cos^2(theta) term using the half angle rule, you would have a free (theta) term in the outer integral, in which case the choice of limits would make a difference.

    Thank you :)
     
  5. Jan 18, 2014 #4

    tiny-tim

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    hi user3! :smile:
    you mean, 1/2 + 1/2cos2θ ?

    yes, the 1/2 makes a difference then, but so does the 1/2cos2θ (because it's 2θ not θ) …

    and they should cancel out! :wink:
     
  6. Jan 18, 2014 #5
    Thank you! I was pretty confused.
     
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