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Volume of a cube

  1. Aug 24, 2009 #1
    Note this is not a homework assignment. I was generally learning about integrating volumes and thought that after being able to integrate the volume of a cylinder

    [tex]\int^{L}_{0}\pi r^{2} dL[/tex]

    That I maybe could integrate the volume of a cube in a simple way:

    [tex]\int^{s}_{0}s^{2} ds[/tex]

    Yet that results in [tex]\frac{s^{3}}{3}[/tex] and not the correct [tex]s^{3}[/tex]

    I don't understand why. I am taking the area of a slice of infinitesmall thickness (ds) and integrating it over the length of the cube. Seems to me to be the technique of integration for volumes. Where does my thinking go wrong?

    Please note that I do realize the main difference between the integration of the volume of a cylinder and a cube is that the length of the cylinder is a seperate variable from its radius. Despite being aware of this key difference, I still do not understand why the cube integration would not work even though the same variable is used.

    Thanks!
     
    Last edited: Aug 24, 2009
  2. jcsd
  3. Aug 24, 2009 #2
    The last inegral should be

    [tex]\int_{0}^{s} s^2 \; dx[/tex]

    s is a constant, but the integral is being carried out along the interval [0, s].

    --Elucidus
     
  4. Aug 24, 2009 #3
    Thanks!
     
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