1. Limited time only! Sign up for a free 30min personal 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!

Volume of paraboloid using divergence theorem (gives zero)

  1. May 28, 2013 #1
    1. The problem statement, all variables and given/known data

    A surface S in three dimensional space may be specified by the equation
    f(x, y, z) = 0, where f(x, y, z) is a real function. Show that a unit vector nˆ normal to
    the surface at point (x0, y0, z0) is given by

    jk9g0g.png

    2. Relevant equations



    3. The attempt at a solution

    r = (x, y, z)

    ∇f = 2(x, y, -z)

    n = 1/r (x, y, -z)

    dS = dx dy √[1 + (x/z)2 + (y/z)2] (1/r) (x, y, -z)

    When i take r (dot) dS it gives x2 + y2 - z2 which = 0..



    Method 2 (Cylindrical Coordinates)

    dS = n (r dr d∅)

    but this still gives 0 when n (dot) r
     
  2. jcsd
  3. May 28, 2013 #2

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    ... which is correct for that part of the surface. But there's another part, governed by z = h.
     
  4. May 29, 2013 #3
    Ah i see, I missed out that surface entirely! (Which happens to be simply a disc)

    Since divergence theorem only works for closed volumes.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Volume of paraboloid using divergence theorem (gives zero)
Loading...