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: Evaluate integral of second order differential

  1. Aug 9, 2010 #1
    1. The problem statement, all variables and given/known data

    Evaluate
    [tex]\int^{2\pi}_{0}\frac{1}{r}\frac{d^{2}f}{d\varphi^{2}}d\varphi[/tex]

    2. Relevant equations

    n/a

    3. The attempt at a solution

    I get the integral as

    [tex]\frac{1}{r}\frac{df}{d\varphi}[/tex]

    Not sure how to evaluate this.
     
    Last edited: Aug 9, 2010
  2. jcsd
  3. Aug 9, 2010 #2
    Why don't you do a simple one first. Say:

    [tex]f(x)=x^3+2x[/tex]

    so that:

    [tex]\frac{df}{dx}=3x^2+2[/tex]

    and so if I want to integrate:

    [tex]\int_0^{2\pi}\frac{df}{dx}dx=\int_0^{2\pi} (3x^2+2)dx=(x^3+2x)\biggr|_0^{2\pi}=f(2\pi)-f(0)[/tex]

    edit: so I see you got that part. Then it's just the first derivative at the end points right? So just write it as such:

    [tex]\frac{df}{d\phi}\biggr|_{0}^{2\pi}[/tex]
     
  4. Aug 9, 2010 #3
    Thanks for quick response. (I had some terrible trouble with Latex, I don't know if other users were seeing what I was, but when I tried to preview my post it would not refresh the Latex even though I had changed the coding.)

    Is there not any way that I can take this further. This is actually a small part of a larger problem in which I have to show that this integral is negligible.

    Hey, I think I get it now! The function is defined on a disk, so it must be the same at 0 and 2 pi. Therefore this term is negligible.

    Does that sound right?
     
  5. Aug 9, 2010 #4
    I'd say if the derivative is analytic throughout the disk and you're integrating over a closed contour, then the integral is zero.
     
  6. Aug 9, 2010 #5
    Thanks. This is part of a problem on harmonic functions (I still don't really know what they are) -- I guess that term puts some constraints on the function, hopefully the ones you mentioned above.

    Cheers
     
  7. Aug 9, 2010 #6

    Mark44

    Staff: Mentor

    This is a known problem that has yet to be fixed. The problem seems to be that when you preview a post with LaTeX in it, the previewer grabs whatever is in a cache somewhere, so if you have made changes, they won't show up in a preview. The only way around this that I know is to refresh the page after you have clicked Preview Post.
     
  8. Aug 10, 2010 #7
    [tex]\sqrt{Thanks Mark}[/tex]

    Thant trick works for me! :smile:
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook