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Line Integral

  1. Jul 29, 2007 #1
    What does line integral really mean, what is it doing?

    Say you have a function f(x,y,z) and you integrate it w.r.t. arc length along some curve C.
    Is this like finding the area under C over f? Like if you are walking along C, and the vertical area covered below you is the integral?

    It's hard to say what I mean, but is this correct?

  2. jcsd
  3. Jul 29, 2007 #2
    Here is an application. Let [tex]f(x,y,z)[/tex] represent the mass at every given point. Then [tex]\int_C f(x,y,z) ds[/tex] along a rectifiable curve [tex]C[/tex] is the total mass of the string/wire (which is represented by the curve).
  4. Jul 29, 2007 #3
    Say you have z = f(x,y). Then [tex]\int_C f(x,y) ds[/tex] represents the area of the sheet that is traces out. That is, the line represented by C in the xy plane, connect it to the surface f(x,y), and the line integral represents the area of this sheet
  5. Jul 30, 2007 #4
    oh right. that makes more sense. cheers,
  6. Jul 30, 2007 #5
    If you are familiar with basic physics, then

    [tex]\int_C{\vec{F}\cdot\ d\vec{r}[/tex]

    This line integral represents the work done by the force F along the path C.

    [tex]\int_a^b{\vec{E}\cdot\ d\vec{r}[/tex]

    This line integral represents the potential difference (voltage) between points b and a, where E is the electric field.
  7. Jul 30, 2007 #6
    I thought the line integral in its mathematical sense represented the "length" of the line between the points of integration.
  8. Jul 31, 2007 #7

    Gib Z

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    Homework Helper

    Nope you are thinking of arclength. The arclength of an integrable function f(x) over [a,b] is given by [tex]\int^b_a \sqrt{ 1+ (f'(x))^2} dx[/tex]
  9. Aug 2, 2007 #8
    OK, thank you.
  10. Aug 3, 2007 #9
    It can represent arc length. If [tex]\bold{R} = x(t)\bold{i}+y(t)\bold{j}[/tex] (for smooth functions) then [tex]\int_C 1 \ ds = \int_a^b \sqrt{[x'(t)]^2+[y'(t)]^2} dt [/tex] where [tex]C[/tex] is the path obtained from [tex]\bold{R}[/tex].
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