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Surface integral of a cone

  1. Nov 27, 2011 #1
    1. The problem statement, all variables and given/known data

    g(x,y,z) = z2; Ʃ is the part of the cone z = [itex]\sqrt{x2+y2}[/itex] between the planes z = 1 and z = 3.

    2. Relevant equations

    Conversion to polar coordinates
    ∫∫Ʃg(x,y,z)dS = ∫∫Rg(x,y,f(x,y)) [itex]\sqrt{fx2 + fy2+1}[/itex]

    3. The attempt at a solution
    If we're talking in terms of r and θ, r goes from 1 to 3 and θ goes from 0 to 2π. I converted to polar coordinates from cartesian and I got 8*π*√2 *z2. If you were converting to polar/cylindrical coordinates, z2 wouldn't change, correct?

    Sorry, for some reason the square root LaTeX command didn't want to work...
     
  2. jcsd
  3. Nov 27, 2011 #2

    LCKurtz

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    Don't use the subscript and superscript buttons in a tex expression use f_x for subscripts and f^2 for superscripts in tex as I have done for you.

    You shouldn't have any z in your answer; it should be a number. Of course z depends on x and y or r and θ because on the surface [itex]z=\sqrt{x^2+y^2}[/itex].
     
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