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Volume of a revolved area

  1. Jan 18, 2009 #1
    Hello. I am a calculus student currently studying integrals. I am able to logically break down most problems, but I came to a breaking point today. I am trying to understand how to revolve the area between two trigonometric functions about a vertical line. The conditions I am trying to satisfy is when these functions form an area that is apparently symmetric and end on identical y coordinates. To simplify my question, I'll create an example.

    Revolve the area enclosed by f(x) and g(x) about the line x= -1 on the interval [0, [tex]\pi[/tex]]
    f(x) = sin(x)
    g(x) = -sin(x)

    I am having incredible difficulty merely setting up the integral. Clearly, I'll need something like this:

    V = [tex]\pi[/tex] [tex]\int[/tex] ( something + 1)^2 - ( something + 1)^2 dy

    I understand the concept of selecting f(y) and g(y) and subtracting the axis' x-coordinate, but I do not understand how to choose which one is an outer radius and which one is an inner radius. Additionally, because the points that they intersect at are identical, I am not sure what bounds to use without getting a bad answer (v=0). I appreciate your time and hope that you will forgive my English.
  2. jcsd
  3. Jan 19, 2009 #2
    To find area between two functions you have to find integral of module of differences of them.

    But in current situation it's a bit easier:
    [tex]sin(x) > -sin(x)[/tex] on [tex][0, \pi][/tex]
    [tex]v = \int_o^\pi sin(x)-(-sin(x)) dx = 2 \int_0^\pi sin(x) dx[/tex]

    My English is also not really good :wink:
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