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Evaluate both sides of divergence theorem

  1. Oct 30, 2012 #1
    1. The problem statement, all variables and given/known data

    NOTE: don't know see the phi symbol so I used theta. this is cylindrical coordinates not spherical.

    Given the field D = 6ρsin(θ/2)ap + 1.5ρcos(θ/2)aθ C/m^2 , evaluate both sides of the divergence theorem for the region bounded by ρ=2, θ=0 to ∏, and z = 0 to 5

    2. Relevant equations

    ∫D * dS = ∫ ∇ * D dV

    3. The attempt at a solution

    The answer (to both sides of course) is 225. I figured out how to do the divergence side, but I'm having problems evaluating the left (surface) side. The way I'm doing it is that the z-component is zero since the equation for D only has a rho and phi.

    From here I'm evaluating ∫∫Dp*pdθdZ from 0 to pi for dθ and 0 to 5 for dz.

    I plug in p = 2 as an "initial condition", and when I calculate the integral through I keep getting 240 even though the answer is 225. Any ideas what's going on? Maybe my integral is just wrong
     
  2. jcsd
  3. Oct 30, 2012 #2

    Dick

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    You are only integrating over part of the surface. You have half a cylinder, it also has a top, a bottom and a flat side. You are only doing the rounded side. Can you picture it? What other part of the surface might contribute to your integral?
     
  4. Oct 30, 2012 #3
    Hmm. I thought about that but my answer didn't really change and here's why:

    The cylinder has 4 sides, right? Rounded, opposite of rounded (flat), top, and bottom. However, from the given equation for D, we only have a rho and phi component. That automatically makes the top and bottom zero.

    WAIT, are you saying that I need to do a coordinate conversion and then calculate the surface dxdz?
     
  5. Oct 30, 2012 #4

    Dick

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    Exactly. That's where you'll find the -15 missing units of surface flux. Very sharp of you to figure out how to do it. You are almost there.
     
  6. Oct 30, 2012 #5
    Took a while but got it right :approve:

    Thanks for the help mate!
     
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