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Finding formula for normal vector of surface

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

    S is surface of upper half of sphere (rad 3) at (0,0,0). Find the formula for n (in Cartesian, spherical, and cylindrical coord syst), then evaluate

    [tex] \int\int\textbf{F}\cdot\textbf{n}dA [/tex]

    Where F(x,y,z) = k

    2. Relevant equations



    3. The attempt at a solution

    in a previous problem, if I let z = f(x,y) be the surface and take

    [tex] \psi(x,y,z) = f(x,y) - z [/tex]

    Since [tex] \nabla\psi [/tex] is orthogonal to the surface on which [tex] \psi [/tex] is constant, then I can use this to get the formula for the normal to the surface.

    Problem is, I'm stuck trying to figure out what the formula for the surface is to start with.

    A sphere with radius 3 is just x2+y2+z2 = 3

    so

    [tex] z = 3 - \sqrt{x^{2}+y^{2}} [/tex]

    where 0< x,y < 3

    now

    [tex] \textbf{n} = \frac{\nabla\psi}{\left\|\nabla\psi\right\|} [/tex]

    Am I on the right track? The unit normal vector field gets pretty ugly after this step :-/
     
  2. jcsd
  3. Nov 29, 2009 #2

    Hurkyl

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    It can't be that bad, can it?

    BTW, using implicit differentiation then solving for the derivative is often much simpler that solving for the variable and differentiating that.


    By the way, surely you know enough about spheres to have an easy way of finding the unit normal?
     
  4. Nov 29, 2009 #3
    I must have done something wrong because I've gotten something pretty heinous.

    [tex] \nabla\psi = -x(x^{2} + y^{2})^{-1/2}\textbf{i} - y(x^{2} + y^{2})^{-1/2}\textbf{j} - \textbf{k} [/tex]

    [tex]\left\|\nabla\psi \right\| = (x^{2} + y^{2}) + 1 [/tex]
     
  5. Nov 29, 2009 #4

    Hurkyl

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    I believe that is wrong, but what's so "heinous" about that expression? :confused: The correct one has the same sort of appearance.
     
  6. Nov 29, 2009 #5
    Maddeningly unhelpful :(

    "By the way, surely you know enough about spheres to have an easy way of finding the unit normal? "

    Obviously not, can you expound a little?
     
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