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Y = sin pi * x Arc Length/Surface Revolution

  1. Nov 27, 2008 #1
    1. The problem statement, all variables and given/known data

    y = sin [tex]\pi[/tex]x Using arc length and surface revoultion on x-axis 0 <= x <= 1

    3. The attempt at a solution

    d/dx sin [tex]\pi[/tex]x = [tex]\pi[/tex] cos [tex]\pi[/tex]x
    ([tex]\pi[/tex] cos[tex]\pi[/tex]x)^2 = [tex]\pi[/tex]^2 cos^2[tex]\pi[/tex]x

    [tex]\int sin pi * x * 2 * pi * \sqrt{1 + pi^2 * cos^2 (pi*x)} [/tex]
    u = pi cos (pi * x)
    du = -pi^2 * sin (pi * x) dx

    -1/2pi[tex] \int \sqrt{1 + u^2}[/tex]
    u = tan [tex]\alpha[/tex]
    du = sec^2 [tex]\alpha[/tex]

    We get the integral of sec^3,

    This doesn't seem to be right, and if it is, the limits of integration don't work out . . .
    Last edited: Nov 27, 2008
  2. jcsd
  3. Nov 28, 2008 #2


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    Welcome to PF!

    Welcome to PF! :smile:

    (have a pi: π :wink:)

    (and in tex, it's \pi)

    … what is the actual question? :confused:
  4. Nov 28, 2008 #3
    Sorry, I'm just getting used to all the quirks of latex.

    The question is to find the length using surface area revoultion forumla of y = sin (pi * x) (rotated about the x-axis) from 0 - 1.

    [tex]\int2\pi*sin(\pi*x) * \sqrt{1 + (sin(\pi*x)')^2}[/tex]

    I think sec^3 is actually correct. You have to reverse substitute using reverse trigonometric identites. [tex]u = tan \theta , sec \theta = \sqrt{1 + u^2}[/tex]
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