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Divergence Theorem

  1. May 9, 2012 #1
    1. The problem statement, all variables and given/known data

    Folks,

    Verify the divergence theorem for

    F(x,y,z)=zi+yj+xk and G the solid sphere x^2+y^2+z^2<=16
    2. Relevant equations

    ##\int\int\int div(F)dV##

    3. The attempt at a solution

    My attempt

    The radius of the sphere is 4 and div F= 1, therefore the integral becomes

    ##\int\int\int div(F)dV=\int_0^{2\pi} \int_0^{\pi} \int_0^{4} 1dV=\int_0^{2\pi} \int_0^{\pi} \int_0^{4} 2\rho^2 sin (\phi) d\rho d\phi d \theta##

    Is this correct so far?
    Thanks
     
  2. jcsd
  3. May 9, 2012 #2
    where did the 2 come from in the integrand? Other than that it's right.

    EDIT: Also, since divF is a constant and the volume of a sphere can easily be found, you don't actually need to evaluate an integral.
     
  4. May 9, 2012 #3
    Hi, that should be a 1, ie

    ##...=\int_0^{2\pi} \int_0^{\pi} \int_0^{4} \rho^2 sin (\phi) d\rho d\phi d \theta##
    But it says I need to verify the Divergence Theorem...so I guess I can continue and verify?

    Thanks
     
  5. May 9, 2012 #4

    sharks

    User Avatar
    Gold Member

    [itex]\int_0^{2\pi} \int_0^{\pi} \int_0^{4} dV[/itex] = volume of sphere with radius 4.

    Just use the well-known formula for calculating volume of sphere. The answer should obviously be the same as the evaluation of your triple integral expression above.

    Now, when you're asked to verify the Gauss Divergence Theorem, you should independently calculate the total flux using the surface integral formula, to confirm your answer.
    [tex]\iint_S \vec F \hat n \,.d\sigma[/tex]
    Hint: To make it easier, split the sphere exactly in half laterally and evaluate the surface integral for one half. Then multiply by 2 to get the total flux.
     
    Last edited: May 9, 2012
  6. May 9, 2012 #5
    Ok, thanks guys. Will respond hopefully at some stage.

    Cheers
     
  7. May 9, 2012 #6

    sharks

    User Avatar
    Gold Member

    Note: [itex]d\sigma[/itex] is the differential area.
     
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