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

Divergence Theorem Question (Gauss' Law?)

  1. May 4, 2015 #1
    If F(x,y,z) is continuous and ight%20%7C%5Cleq%20%5Cfrac%7B1%7D%7B%5Csqrt%7B%28x%5E2+y%5E2+z%5E2%29%5E3%7D+1%7D.gif for all (x,y,z), show that R3 gif.latex?%5Cint%20%5Cint%20%5Cint.gif gif.gif dot F dV = 0

    I have been working on this problem all day, and I'm honestly not sure how to proceed. The hint given on this problem is, "Take Br to be a ball of radius r centered at the origin, apply divergence theorem, and let the radius tend to infinity." I tried letting F = 1/((x2 +y2+z2)(3/2))+1), and taking the divergence of that, but it didn't really seem to get me anywhere. If anyone has any suggestions for at least how to set up this proof, I would really appreciate it.
     
    Last edited: May 4, 2015
  2. jcsd
  3. May 5, 2015 #2

    ShayanJ

    User Avatar
    Gold Member

    Well, from the title it seems that you know you should use divergence theorem. So why don't you?
     
  4. May 5, 2015 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Since your region of integration is a ball and the integrand involves [itex]x^2+ y^2+ z^2[/itex], I would recommend changing to spherical coordinates to do that integration over the surface of the ball.
     
  5. May 5, 2015 #4
    Spherical was my thought too. I guess what's been confusing me is that the vector field isn't explicitly given; there's just the inequality which indicates it exists at (0,0,0). My thought was to use Gauss' Theorem for Divergence in Spherical Coordinates: div F = 1/p2*d/dp*(p2Fp)+1/(psin(phi))*d/dphi*(sin(phi)Fphi + 1/(psin(phi))*dFtheta/dtheta. Since you have 1/p terms, div F would go to zero as the radius, p, goes to infinity. Therefore, the integral of 0 is 0.
     
  6. May 5, 2015 #5

    ShayanJ

    User Avatar
    Gold Member

    I don't see any application of divergence theorem in your explanation. This is how it should be:
    ## \int_V \nabla\cdot \vec F dV=\int_{\partial V} \vec F \cdot \hat n dA ##
    Where ## \partial V ## is the boundary of V. Now because in the surface integral, F is evaluated only at the boundary and the boundary is at infinity and we know that F is smaller than a function which goes to zero at infinity, the integral should be zero.
     
  7. May 5, 2015 #6
    Oh okay, that makes sense. I didn't even think of it that way. Thank you so much for the suggestion!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Divergence Theorem Question (Gauss' Law?)
  1. Gauss theorem (Replies: 3)

  2. Divergence Theorem (Replies: 1)

  3. Gauss theorem (Replies: 2)

Loading...