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Surface integrals

  1. Feb 9, 2007 #1
    Please Help! Surface integrals

    I am wondering if someone can help me with the following? I am asked to evaluate ∫∫F∙dS where F(x,y,z) = z^2xi + (1/3y^3 +tanz)j + (x^2z+y^2)k and S is the top half of the sphere x^2+y^2+z^2 = 1.

    ∫∫F∙dS = ∫∫∫divFdV. Here, div F = x^2+y^2+z^2. I know that S is not a closed surface and so you would need to evaluate S as the difference between 2 surfaces, S1 as the closed surface that is the top half of the sphere and S2 as the disk that is x^2+y^2≤1 where the orientation is downwards. So, evaluating div F over the whole top half of the sphere I got 2pi/5.

    But I am wondering how I would evaluate ∫∫∫x^2+y^2+z^2 over the disk x^2+y^2≤1? Could I convert to spherical coordinates and I could simplify the expression of the integrant to ρ^4sinφ, but I am wondering what φ would be in this instance? Thanks so much!
  2. jcsd
  3. Feb 9, 2007 #2


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    You wouldn't integrate a volume integral over a surface!

    [itex]\int\int\int (x^2+ y^2+ z^2)dzdydx[/itex] is integrated over the volume- the half-ball- not the surface. If you are using cartesian coordinates, integrate with x from -1 to 1, y from [itex]-\sqrt{1- x^2}[/itex] to [itex]\sqrt{1- x^2}[/itex], z from [itex]-\sqrt{1- x^2- y^2}[/itex] to [itex]\sqrt{1- x^2- y^2}[/itex]. It's simpler in cylindrical coordinates: integrate with r from 0 to 1, [itex]\theta[/itex] from 0 to [itex]\2 pi[/itex], z from 0 to [itex]\sqrt{1- r^2}[/itex]. And, of course, it's much simpler in spherical coordinates: integrate with r from 0 to 1, [itex]\theta[/itex] from 0 to [itex]2\pi[/itex], [itex]\phi[/itex] from 0 to [itex]\pi/2[/itex].
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