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Volume of a cone using cylindrical coordinates and integration

  1. Nov 14, 2009 #1
    Hi all! I was trying to figure out how to find the volume of a cone with radius R and height h using integration with cylindrical coordinates. I first tried to set the the integral as:

    [tex]\int_{0}^{2\pi}\int_{0}^{h}\int_{0}^{R}\rho d\rho dz d\phi[/tex]

    ...but I think that this is setting up the integral for a cylinder and not a cone. Any suggestions so I end up with the correct volume?

    [tex]V_{cone} = \frac{1}{3}\pi r^2 h[/tex]

    Thank you!
     
  2. jcsd
  3. Nov 15, 2009 #2

    HallsofIvy

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    Draw a picture. Assuming the cone has base of radius R and height h, in the xz-plane, it will be a triangle with vertices at (-R, 0), (R, 0), and (0, h). The slant side will be a line from (R,0) to (0,h) and that has equation z= h- (h/R)x= h(1- x/R). Rotating that around the z-axis to get a cone, "x" becomes "r": z= h(1- r/R). Your z integral should go from 0 to that.
     
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