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Divergence Theorem

  1. May 29, 2009 #1
    1. The problem statement, all variables and given/known data

    Use the divergence theorem to evaluate

    [itex]\int\int_{\sigma}F . n ds[/itex]
    Where n is the outer unit normal to [itex]\sigma[/itex]

    we have
    [itex]F(x,y,z)=2x i + 2y j +2z k [/itex] and [itex]\sigma[/itex] is the sphere [itex]x^2 + y^2 +z^2=9[/itex]

    2. Relevant equations

    [itex]\int\int_{s}F . dA = \int\int\int_{R}divF dV[/itex]


    3. The attempt at a solution

    I've worked out [itex]divF[/itex] to be 6.

    so I multyiply that by the Volume of a sphere [itex]6\times\frac{4}{3}\pi r^3 = 216\pi[/itex]


    To calulate this using spherical co-ordinates.

    I would need to calculate a triple integral

    I know theres

    [itex]\int\int\int p^2 sin(\theta) dp d\theta d\phi[/itex]

    I know that p = 3 but what would the values of [itex]\theta [/itex] and [itex]\phi [/itex] be

    I guess the limits would be [itex]0<p<3[/itex][itex] 0<\phi<2pi[/itex] and [itex]0<\theta<\phi[/itex]

    Any help greatly appreciated
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
    Last edited by a moderator: May 30, 2009
  2. jcsd
  3. May 30, 2009 #2

    HallsofIvy

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    No. Your limits on [itex]\rho[/itex] and [itex]\phi[/itex] are correct but [itex]\theta[/itex] goes from 0 to [itex]\pi[/itex].

     
  4. May 30, 2009 #3

    Hootenanny

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    Your limits are correct (assuming of course theta runs from 0 to pi, rather than phi).

    All you need to so now is explicitly compute the integral, which is straightforward.
     
  5. May 30, 2009 #4
    When I calculate the integral I'm getting 36[itex]\pi[/itex]
    I'm not sure where I'm going wrong.

    So I compute.


    [itex]\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{3} p^2 sin(\theta) dp d\theta d\phi[/itex]

    [itex]=\int_{0}^{2\pi}\int_{0}^{\pi}\left[\frac{ p^3 sin(\theta)}{3} \right]_{0}^{3} d\theta d\phi[/itex]

    [itex]=\int_{0}^{2\pi}\int_{0}^{\pi}9sin(\theta) d\theta d\phi[/itex]



    [itex]=\int_{0}^{2\pi}\left[-9cos(\theta) \right]_{0}^{\pi} d\phi[/itex]

    [itex]=\int_{0}^{2\pi}18 d\phi[/itex]

    [itex]=\left[18\phi\right]_{0}^{2\pi[/itex]

    [itex]=36\pi[/itex]

    regards
     
  6. May 30, 2009 #5

    HallsofIvy

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    Yes, [itex]36\pi[/itex] is the volume of that sphere. Now multiply by 6: [itex]6(36\pi)= 216\pi[/itex].
     
  7. May 30, 2009 #6
    Thanks for your help
     
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