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!

Surface integrals (without real integration)

  1. Mar 9, 2006 #1
    Given F= [tex](ix+jy) Ln(x^2+y^2)[/tex]

    and given S, which is a cylinder of radius r, and height h(in the z axis) evaluate [tex] \int\int_s F.n \,ds[/tex]. It says that you shouldn't need to do any work if you think about it enough. I figured I could find the area of the main part to be [tex]2 \pi r h[/tex] then multiply that by [tex]Ln(r^2)=2Ln(r)[/tex] to get the answer but I am off by a factor of r in my answer. I don't think the caps to the cylinder contribute to this as the normal is orthogonal to F.

    One more question, what exactly does a surface integral return? I feel stupid but I can't seem to find out exactly what the physical meaning of the result of a surface integral is. Thanks for the help.
     
    Last edited: Mar 9, 2006
  2. jcsd
  3. Mar 9, 2006 #2
    A surface integral is the integral of a function over a surface, as opposed to a line, etc. Unit-wise you are gaining an extra meter^2. As far as a physical meaning, it depends on what you are integrating.

    What's happening in this integration would be easier to understand if you switch everything to polar coordinates.

    -Dan
     
  4. Mar 9, 2006 #3
    I think i'm just confusing myself more :uhh: If I integrate that vector function around a cylinder than it will be 0. I can imagine that we would have constant vectors of magnitude [tex]Ln[r^2][/tex] all in the radial direction around the cylinder for any z value. I guess I still don't have a good enough understanding of what a surface integral really is to be able to find an easy way to express this. I think its time to re-read some div,grad,curl.
     
    Last edited: Mar 9, 2006
  5. Mar 9, 2006 #4

    siddharth

    User Avatar
    Homework Helper
    Gold Member

    As topsquark said, switch over to polar.

    In fact, I suggest you work this out the hard way and then look back.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Surface integrals (without real integration)
  1. Surface integrals (Replies: 10)

Loading...