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Line intergrals

  1. Jul 22, 2004 #1
    with out a picture i find this hard to ask, but... i don't really understand what a line intergral represents physically? like is a 2D plane in a 3D space? i don't understand why you take line intergrals in the shape of a square of some function what does that acutally represent, id appreciate any help, thanks all!
     
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  3. Jul 22, 2004 #2

    Galileo

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    As you know, in one dimension the expression [itex]\int_a^bf(x)dx[/itex] can be interpreted as the area bounded by x=a, x=b, y=0 and f(x) when you plot it.
    Suppose you are in two dimensions and you have [itex]\int_{(x_1,y_1)}^{(x_2,y_2)}f(x,y)ds[/itex]. You can plot the graph of f(x,y) in 3 dimensions. The height of the graph wil be z=f(x,y). Now imagine your path from [itex](x_1,y_1)[/itex] to [itex](x_2,y_2)[/itex] in the xy-plane. Then the line integral is the area of the 'curtain' under f(x,y) for those point x,y on the path of integration.

    I find it difficult to find a real simple physical example. Usually line integral in 3-d are used to calculate the work done by a forcefield on a particle when the particle traverses some path in space. In electromagnetism, these line integrals are of fundamental importance.
    You could also line integrals to find the lenght of some curve in space.
     
    Last edited: Jul 22, 2004
  4. Jul 23, 2004 #3
    Phymath,

    If your problem is understanding integrals in general, then that's one thing. Let me know if that's where your difficulty lies.

    But, assuming you are happy with the 1-D case as Galileo described, a line path in 3D is just a generalisation of that.

    So, the 3D path integral [itex]\int {{\bf \vec F} \bullet } {\bf d\vec x}[/itex] becomes [itex]\int {F_x (x)} dx[/itex] along a straight line!
     
  5. Jul 23, 2004 #4

    mathwonk

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    A line integral is an old name for a path integral. It measures the cumulative behavior of some quantity along the path, (which is usually a curve rather than a line).

    E.g. suppose you are swimming in the ocean and the tide starts to carry you out. The best thing to do is not swim back to shore but to swim parallel to the shore. Why? Because you are not fighting the curent that way, and when you get far enough downshore maybe the current there will be weaker and you can swim in to shore.

    The work you do fighting the current is measured by a path integral. I.e. you take the velocity vector of the current and compare it to the velocity vector of your path of swimming. the more parallel they are the more work you do. So in this case you dot those two vectors together and get a number for each point of your path and integrate these numbers to get the total work done.



    Obviously by swimming perpendicular to the current, i.e. parallel to the shore, you minimize the work done by you in bucking the current. So if the x component of the current's velocity is A, and the y component is B, you get an integral like "integral of Adx + Bdy". Then when you plug in some function (x(t),y(t)) for your path of motion, the integral becomes the integral of A(x,y)dx/dt +B(x,y)dy/dt. This is the same as the dot product of your velocity vector (dx/dt,dy/dt) and the velocity vector (A,B) of the water.

    I may misunderstand the physics here and maybe I should have a force vector for the water (in terms of acceleration of the water instead of velocity) to call this "work", but I hope the idea of the path integral is clear enough, even if the terminology is ill conceived.

    Anybody else? please correct me.
     
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