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Surface Integrals

  1. Nov 10, 2007 #1
    1. The problem statement, all variables and given/known data

    Compute the surface integral:

    g = xyz on x^2+y^2+z^2 = 1 above z^2=x^2+y^2.

    2. Relevant equations



    3. The attempt at a solution

    I'm only doubtful about the parameterization. Under normal circumstances, since x^2+y^2+z^2 = 1 is a sphere, we can write:

    r = (SinCos[v])i + (SinSin[v])j + (Cos)k.

    However, how do you account for the "above z^2=x^2+y^2."

    Do I simply sum the square of the x and y components and write:

    r = (SinCos[v])i + (SinSin[v])j + (Sin^2)k.

    Is this correct?
     
  2. jcsd
  3. Nov 10, 2007 #2

    HallsofIvy

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    No, of course not! z^2= x^2+ y^2 has nothing to do with the sphere- it is below the sphere!

    z^2= x^2+ y^2 is a cone with axis the positive z-axis. Since you are only concerned with the surface area, you would only use that to determine the limits of integration. In spherical (angular) coordinates, x= SinCos[v], y= SinSin[v] and z= Cos. Obviously, x^2+ y^2= Sin^2(u) and z^2= Cos^2(u). The equaton of the cone is just Sin^2(u)= Cos^2(u) which results in (since we are talking about z> 0 here) u= [itex]\pi/4[/itex]. v ([itex]\theta[/itex] in spherical coordinates) goes from 0 to 2[itex]\pi[/itex] and u ([itex]\phi[/itex] in spherical coordinates) goes from 0 to [itex]\pi/4[/itex].
     
  4. Nov 10, 2007 #3
    Thanks a lot once again

    Thanks very much again HallsofIvy!!!:smile:
     
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