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Help with using the Divergence Theorem!

  1. May 26, 2004 #1
    Hi!

    We are nearing the end of our course --- culminating in Stokes and Divergence Theorems for surface integrals, and I am having some difficulty with the following

    1. F(x,y,z) = [tex]<x^3y, -x^2y^2, -x^2yz>[/tex]

    where S is the solid bounded by the hyperboloid x^2 + y^2 - z^2 =1 and the planes z = -2 and z=2.

    I computed div F properly...

    Well I know what the z limits are in my Triple Integral, however what must I use as my radius? Theta should go from 0 to 2pi correct?

    2. F(x,y,z) = [tex]<x^2y, xy^2, 2xyz>[/tex]

    where S is the surface of the tetrahedron bounded by the planes x=0, y=0, z=0, and x+2y+z = 2

    Here must my triple integral be from 0 to 2 for the x limit, then 0 to (2-x)/2 for my y limits, and for z just 0 to 2-x-2y.

    Those seem correct, but a confirmation would be nice! Thanks a lot!
     
  2. jcsd
  3. May 26, 2004 #2
    Whoops -- it's probably implied but the questions ask to compute


    [tex]\int\int_SFdS[/tex] = [tex]\int\int\int_EdivFdV[/tex]
     
    Last edited: May 26, 2004
  4. May 26, 2004 #3

    arildno

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    1. Did you get the divergence to be zero?
    I did, so is this an exercise in verifying that computing the surface integral directly also yields zero?
     
  5. May 26, 2004 #4

    arildno

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    The limits in 2 seems right.
     
  6. May 26, 2004 #5
    Ahh so maybe i didnt compute div F properly-- yup it's 0, making the answer 0. Thanks.
     
  7. May 26, 2004 #6
    Ahh heres another one thats a bit more challenging:


    Its BEGGING spherical coordinates:

    [tex]F(x,y,z) = <x^3+ysinz, y^3+zsinx, 3z>[/tex]

    S: Surface of the solid bounded by the hemispheres [tex]z=sqrt(4-x^2-y^2)[/tex] and [tex]z=sqrt(1-x^2-y^2)[/tex] and the plane [tex]z=0[/tex].

    I set: r(r,phi,theta) = <rsin(phi)cos(theta), rsin(phi)sin(theta), rcos(phi) >
    Are the following the correct limits of integration?

    1 < r < 2
    0 < phi < pi/2
    0< theta < 2pi
     
  8. May 27, 2004 #7

    arildno

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    Are you talking about the limits of integration for the surface here??
    (Remember, if you're talking about the solid, r=0 is certainly included.)

    In order to solve this, I suggest you split your region in two solids, the two hemishperes, both bounded by the plane z=0.

    Then it is easy to apply spherical coordinates on these solids separetely.
     
  9. May 27, 2004 #8

    arildno

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    I'm sorry, total mistak on my parte (I thought there was a minus sign somewhere :eek:)
    Your approach is perfefctly correct
     
  10. May 27, 2004 #9
    So it is from 1 to 2... okay thanks!!!
     
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