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    find the area of the surface obtained by revolving the curve y=x^2, 0<x<1 about the y axis
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    use the formula attached:
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    y = x^2 => x = y^1/2

    x = 0, y = 0, x = 1, y = 1

    Surface area, S = 2*pi*integral [0, 1] of g(y)(1 + [g'(y)]^2)^1/2

    g(y) = y^1/2
    g'(y) = (1/2)y^-1/2

    [0,1] are the limits.

    You should be able to do it from there.
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    (Original post by elpaw)
    use the formula attached:
    Damn you elpaw!
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    thanx u guys
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    (Original post by Simbo)
    thanx u guys
    sorry, i didnt see te "about y axis". The formula for your case is thus....
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