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Solid of revolution question: verify that the volume of the cone is παβh/3

  1. Sep 14, 2010 #1
    1. The problem statement, all variables and given/known data

    Consider a vertical cone of height h whose horizontal cross-section is an ellipse and whose base is the ellipse with major and minor semi-axes α and β. Verify that the volume of the cone is παβh/3.
    [ Hint: The area of an ellipse with major and minor semi-axes α and β is παβ. ]

    2. Relevant equations

    V = ∫A(y) dy (from c to d)
    V = ∫π(radius)² dy (from c to d)

    3. The attempt at a solution

    It says that the cone is upright, so I'm assuming it wants the cone rotated about the y-axis.
    V = ∫A(y) dy
    V = ∫π(radius)² dy

    Using similar triangles:
    x/y = r/h
    x = ry/h

    V = π∫(ry/h)² dy (the integral is now from 0 to h (c = 0, d = h))
    V = π∫(r²y²/h²) dy
    V = (πr²/h²)∫(y²) dy (since pi, r, and h are all constants)

    At this point I'm not sure where to go. Do I take the integral of y²? How do I incorporate α and β into this integral? (As a side note, I'm very new to these forums and if I've done anything wrong I apologize. I'm not used to writing out integrals on the computer and if the notation is not optimal I'm sorry!)
  2. jcsd
  3. Sep 15, 2010 #2
    Personally, I think you need to go out of your way to draw this thing, nicely. The volume via slices is:

    [tex]V=\int_0^h A(y)dy[/tex]

    with [itex]A(y)=\pi u(y) v(y)[/itex]

    and u and v are the minor and major axes as functions of y as you go up the cone starting from the base up to h.

    Since the axes at the base are [itex]\alpha,\beta[/itex], then as you said, using similar triangles, I get:



    Alright, just integrate now.
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