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Surface Integrals

  1. May 12, 2009 #1
    1. The problem statement, all variables and given/known data

    Find the value of the surface integrals by using the divergence theorem
    [tex]\vec{F} = (y^2z)\vec{i} + (y^3z)\vec{j} + (y^2z^2)\vec{z}[/tex]

    S: [tex] x^2 + y^2 + z^2 [/tex]

    Use spherical coordinates.

    2. Relevant equations

    3. The attempt at a solution

    I've gotten the integral I think. I want to make sure before I go along with evaluating it.

    [tex]\int _0^{2\pi} \int _0^{\pi} \int _0^2 { (7\rho^3 \sin^2{\phi} \sin^2{\theta} \cos{\phi} } \rho^2 d \rho d \phi d \theta[/tex]

    My latex is all messed up... maybe a mod can fix it for me?
     
    Last edited by a moderator: May 13, 2009
  2. jcsd
  3. May 12, 2009 #2

    gabbagabbahey

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    Do you mean [itex]x^2+y^2+z^2=4[/itex]?


    That doesn't look quite right....what do you get for the divergence of F (in Cartesians and Sphericals)?
     
  4. May 12, 2009 #3
    Yes, I meant = 4. Thanks.

    I got [tex] 5y^2 2z [/tex]
     
  5. May 12, 2009 #4

    gabbagabbahey

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    I assume you mean [itex]5y^2 z[/itex]?...If so, you're right. What is that in spherical coordinates? What are you using for [itex]dV[/itex] (infinitesimal volume element) in spherical coordinates?
     
  6. May 12, 2009 #5
    Ok, I have [tex]5y^2z[/tex] in my notes but I thought that was wrong. When I take the partial of [tex]y^2z^2[/tex] with respect to z, why does that come out to just z?
     
  7. May 12, 2009 #6

    gabbagabbahey

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    [tex]\frac{\partial}{\partial z} (y^2 z^2)=y^2 \frac{\partial}{\partial z} (z^2)=2y^2 z[/tex]
     
  8. May 12, 2009 #7
    OooOOooOOooooohhh
     
  9. May 13, 2009 #8

    HallsofIvy

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    I fixed the Latex:
    1) Use "\" in front of Greek letters: \theta, \phi, \rho.

    2) [itex]\rho[/itex] is "rho", not "roe".
     
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