1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    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].
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Line Integral
  1. Line integral (Replies: 1)

  2. Line integral (Replies: 3)

  3. Line Integral (Replies: 11)

  4. Line integrals (Replies: 4)