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Surface area of a solid of revolution

  1. Jul 22, 2012 #1
    1. The problem statement, all variables and given/known data
    Having recently learned the disk/shell/washer method for finding the volume of a solid of revolution, I'm trying to apply similar methods to derive the formula for the surface area of a cone (and hopefully after that, that of a sphere).
    The region that is revolved around to form the cone is that under f(x) = (r/h)x from 0 to h, where r is the radius of the base and h is the height of the cone (both are constants).

    2. Relevant equations
    Since V = πr^2h for a cylinder, the volume of the cone is ∫ π[f(x)]^2 dx from 0 to h. When I evaluate that, I get V = πr^2h/3,which is correct.

    3. The attempt at a solution
    I reasoned that since A = 2πrh for the curved surface of a cylinder, evaluating ∫ 2πf(x) dx from 0 to h should result in an expression for the area of the curved surface of the cone (everything but the base). But instead of getting πrs (s being √(r^2 + h^2), I think it's called the lateral height), I get πrh as the area of the curved surface.

    Is my method wrong, or am I just integrating wrong?
     
    Last edited: Jul 22, 2012
  2. jcsd
  3. Jul 22, 2012 #2

    tiny-tim

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    hi newageanubis! :smile:
    yes, but the surface of a cone is sloping, so it's a lot more than that, isn't it? :wink:
     
  4. Jul 22, 2012 #3
    Oh. Now I feel dumb :(.

    Can this "problem" be solved with single-variable integration, though?
     
    Last edited: Jul 22, 2012
  5. Jul 22, 2012 #4

    tiny-tim

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    you'd be surprised how often people ask the same thing on this forum! :biggrin:
    not following you, this is a single ∫, isn't it? :confused:
     
  6. Jul 22, 2012 #5

    HallsofIvy

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    If a curve is given by y= f(x), then a "differential of arc length" is given by [itex]ds= \sqrt{(dx)^2+ (dy)^2}= \sqrt{1+ (dy/dx)^2}dx[/itex].

    And it is the arclength, rotated around an axis, that will give you a surface area.

    In the case of the cone, say y= ax, rotated around the y- axis, the circumference of a small section would be [itex]2\pi x[/itex] so that it area would be [itex]2\pi x\sqrt{1+ (dy/dx)^2}dr[/itex]. Of course, in this example, dy/dx= a so that would be [itex]2\pi x\sqrt{1+ a^2}dx[/itex]. The area from the coordinate plane up to y= aR, so that x= R, would be [itex]\pi \int_0^R x\sqrt{1+ a^2}dx= \pi R^2\sqrt{1+ a^2}/2[/itex].
     
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