Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Proving a force is conservative (via path integrals)

  1. Nov 14, 2004 #1
    It's always the easy questions that get me stuck....

    For some reason, I'm having a mental block on how to answer this one:

    Consider the force function:
    F = ix + jy
    Verify that it is conservative by showing that the integral,
    [tex] \int F \cdot dr [/tex]
    is independent of the path of integration by taking two paths in which the starting point is the origin (0,0), and the endpoint is (1,1). For one path take the line x = y. For the other path take the x-axis out to the point (1,0) and then the line x = 1 up to the point (1,1).

    Now I've already verified that it is conserved by taking the curl of F, but I can't seem to come to a similiar conclusion using the path integrals. Can someone help me out with this one? At the very least, if I could see the integrals themselves for each path, perhaps I could figure out where I've made my mistake. Thanks.
     
  2. jcsd
  3. Nov 14, 2004 #2

    ehild

    User Avatar
    Homework Helper

    Do not forget that the work is the integral of the scalar product of force and displacement .

    [tex] W = \int_{(0,0)}^{(1,1)} \vec F \cdot \vec dr = \int (F_xdx+F_ydy)=\int_{(0,0)}^{(1,1)}(xdx+ydy)[/tex]

    For the first path, y = x and dy = dx.
    [tex] W= \int _0^1{2xdx}=1[/tex]

    To calculate the work along the second path, we integrate along the x axis first, than along the vertical line at x=1.

    [tex]\int_{(0,0)}^{(1,1)}(xdx+ydy)=\int_{(0,0)}^{(1,1)}xdx +\int_{(0,0)}^{(1,1)}ydy=0.5*[x^2]_0^1+0.5*[y^2]_0^1=1[/tex]

    ehild
     
  4. Nov 14, 2004 #3
    Thank's so much.... I see what I was doing wrong now.

    Specifically, instead of
    [tex] \int (F_xdx+F_ydy) [/tex]
    I had been using
    [tex] \int ((F_x + F_y) dx dy) [/tex]

    Thanks again. :smile:
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook