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Surface Integral of cylinder.

  1. Jan 8, 2010 #1
    1. The problem statement, all variables and given/known data

    I’m trying to integrate the surface of a cylinder.
    I know when integrating the surface of a cylinder the surface element is:

    Where ρ² + z² = r²

    And for a sphere it is:

    In a sphere r=ρ

    But in a cylinder when I’m integrating its surface, could it be written as:

    (r² - z² )½ . dθ.dz

    For example ∫∫zdS over a cylinder from 0<z<5 and 0<Ø<2π

    Would it be:

    ∫∫z. (r² - z² )½ . dθ.dz ?
  2. jcsd
  3. Jan 9, 2010 #2


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    Science Advisor

    You seem to be confusing "[itex]\rho[/itex]" and "r". In spherical coordinates, [itex]\rho[/itex] is the straight line distance from the origin to a point. In cylindrical coordinates, "r" is the same as in polar coordinates- the straight line distance from the origin to a point in the xy-plane. In three dimensional cylindrical coordinates r is the distance from the origin to t he point (x,y,0) directly "below" the point (x,y,z). The "differential of area", on the surface of a sphere of (fixed) radius r, is [itex]r d\theta dz[/itex] where r is a constant. I can see no reason to introduce "[itex](\rho^2- z^2)^{1/2}[/itex]".
    Last edited by a moderator: Jan 9, 2010
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