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Line Integral question

  1. Dec 5, 2012 #1
    Suppose I have a vector V and I want to compute for the line integral from point (1,1,0) to point (2,2,0) and I take the path of the least distance (one that traces the identity function).

    The line integral is of the form:
    [tex] \int _a ^b \vec{V} \cdot d\vec{l} [/tex]

    Where:

    [tex] x=y, \ d\vec{l} =dx \hat{x} + dx \hat{y} [/tex]

    Thus the integral can be computed purely in terms of x (can also be y), which looks something like this:
    [tex] \int _a ^b V(x)dx [/tex]

    What I don't exactly understand is why is it okay to use the limits like this:
    [tex] \int _1 ^2 V(x)dx [/tex]

    Why can we use the limits from 1 to 2 if we express the line integral in terms purely of x. I have a very vague idea of why it is, but I'd rather take it from people who actually know this to explain this to me. Thanks.
     
  2. jcsd
  3. Dec 5, 2012 #2
    If I understand you correctly, the function you end up integrating is only in terms of ##x##, and therefore you don't need to parametrize it (or you can look at it by saying you are using ##x## as your parameter). Either way, because your function is dependent only on ##x##, all you have to do is integrate along the x-axis, which is from 1 to 2.
     
  4. Dec 7, 2012 #3
    Actually, I've found out that to 'parametrize' the variables into x=t, y=t is a more comforting method to do it. At least intuitively, I see it as tracing the path of integration when we set the x and y variables into that parametric equation.

    Edit: Yes, I didn't see it, but I was using x as the parameter. Thanks.
     
    Last edited: Dec 7, 2012
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