Surface Integral of quarter-cylinder

  1. 1. The problem statement, all variables and given/known data
    I am taking the surface integral over a quarter cylinder. Everything is fine and I can get the correct answer, it's just a conceptual problem that I need help with.


    2. Relevant equations

    The da for the "curved" outer surface is [tex]da=sd\phi dz\hat{s}[/tex]
    The da for the bottom surface is [tex]da=sdsd\phi (-\hat{s})[/tex]

    I understand why the curved da is multiplied by s, since we are integrating over a surface that is projected into 3-space by a distance s.

    I do not understand why this s occurs in the da for the bottom (and top) surface. We integrate over a dynamic ds to find the surface integral for this piece....so why multiply the differential area by s?
     
  2. jcsd
Know someone interested in this topic? Share a link to this question via email, Google+, Twitter, or Facebook