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Surface area of a curve

  1. May 18, 2009 #1
    1. The problem statement, all variables and given/known data
    Find the surface area traced out when the curve 8y^2 = x^2(1-x^2) is revolved around the x-axis.

    3. The attempt at a solution
    x-axis means y = 0
    When y = 0, x = 0, -1 or 1.
    Since this curve is "the infinity symbol", the curve has symmetry at x = 0.

    Isolating y,
    [tex]8y^2 = x^2(1-x^2)[/tex]
    [tex]y = \frac{\sqrt{x^2(1-x^2)}}{8}[/tex]

    Surface area = [tex]2\pi \int_{C} y dr [/tex]
    = [tex]4\pi \int^{1}_{0} y dr [/tex]

    [tex] dr = {\sqrt{1+ ({\frac{\sqrt{2x(1-2x^2)}}{8})^2}} {dx}[/tex]
    [tex] dr = {\frac{\sqrt{64+2x(1-2x^2)}}{8}} {dx}[/tex]

    Surface area = [tex]{\frac{\pi}{8}} \int^{1}_{0} {\sqrt{x^2(1-x^2)(64+2x(1-2x^2))}{dx}[/tex]


    And.......I have no idea how to proceed with the integral.
     
    Last edited: May 18, 2009
  2. jcsd
  3. May 18, 2009 #2

    tiny-tim

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    Hi compliant! :smile:

    (have a square-root: √ and an integral: ∫ and try using the X2 tag just above the Reply box :wink:)
    Nooo … using vertical slices, surface area = ∫2πydx times a factor to take account of the slope, which is … ? :smile:
     
  4. May 18, 2009 #3
    The factor is [tex]\sqrt{1+(dy/dx)^2}[/tex], which I'm sure I accounted for above.

    [tex] dr = {\sqrt{1+ ({\frac{\sqrt{2x(1-2x^2)}}{8})^2}} {dx}[/tex]

    It might not be "dr", but I'm 99% sure that's the calculation of the factor.
     
  5. May 18, 2009 #4

    tiny-tim

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    oh i see … yes that's right … i didn't recognise it from your:
    start again … if y = (1/√8)x√(1 - x2), then dy/dx = … ? :smile:
     
  6. May 18, 2009 #5
    Ah, I must've forgotten to chain rule it.



    y = (1/√8)√(x2 - x4)
    dy/dx = (1/2√8)(x2 - x4)-1/2(2x - 4x3)
    dy/dx = (1/√8)(x2 - x4)-1/2(x - 2x3)
     
  7. May 18, 2009 #6

    tiny-tim

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    yes, that looks right … (you had a factor 2 earlier) …

    now square it and add the 1, and I think it'll be a perfect square on the top :wink:
     
  8. May 18, 2009 #7
    Ok, got the answer. Thanks.
     
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