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How to do a contour integral?

  1. Jul 6, 2013 #1
    Hi there! I'm almost sure that somebody has previously make this same question so, if it is like that, i'm sorry.
    I've just been introduced to contour integrals, I've tried to look around the internet and some text books, but i can't find out what do they actually are so, if someone could explicitly explain me what is exactly a "contour integral", i'd be very grateful.

    PD: By explicitly I mean the formula for doing that integral, there're many examples around the internet, but i haven't found a general expression that explains how to do them...

    Anticipated thanks :)
  2. jcsd
  3. Jul 6, 2013 #2


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    The definition of the contour integral, which should be included in your text book, is pretty much all you need:
    A contour is a one-dimensional curve so x, y, and z can be written in terms of a single parameter, t, say. Then write the function to be integrated in terms of that parameter.

    For example, to integrate [itex]x^2dx+ xydy+ z dz[/itex] from (1, 0, 0) to (-1, 0, 0) along the great circle, above the xy-plane, on the sphere [itex]x^2+ y^2+ z^2= 1[/itex], we can write [itex]x= cos(\theta)[/itex], [itex]y= 0[/itex], [itex]z= sin(\theta)[/itex]. Then [itex]dx= -sin(\theta)d\theta[/itex], [itex]dy= 0[/itex], and [itex]dz= cos(\theta)d\theta[/itex].

    So [itex]x^2dx= (cos^2(\theta)(-sin(\theta)d\theta)= -cos^2(\theta)sin(\theta)d\theta[/itex], [itex]xydy= 0[/itex], and [itex]zdz= (sin(\theta))(cos(\theta d\theta)= sin(\theta)cos(\theta)d\theta[tex].

    We go from (1, 0, 0) to (-1, 0, 0) by taking [itex]\theta[/itex] from 0 to [itex]\pi[/itex]. The integral becomes
    [tex]\int_0^\pi (-cos^2(\theta)sin(\theta)+ sin(\theta)cos(\theta)d\theta= \int_0^\pi (cos(\theta)- cos^2(\theta))sin(\theta)d\theta[/tex]
  4. Jul 6, 2013 #3
    Ok, it's been a little hard to follow since i can't see the symbols correctly, but i finally got it, ty a lot :)
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