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Math Help - Surface Area Of A Parametric Curve?

  1. #1
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    Surface Area Of A Parametric Curve?

    Does anybody know how to get the surface area under this parametric curve about the x axis? I've been trying for hours..

    x= 8(t^2)
    y= 64t - (t^3)/3
    0 <= t <= 8sqrt(3)

    My work:
    S = 2pi(integral(((64t - (t^3)/3)sqrt((16t^2 + (64-t^2)^2)dt
    from 0 to 8sqrt(3)

    S = 2pi(integral(((64t - (t^3)/3)sqrt((128t^2+4096t^4)dt
    from 0 to 8sqrt(3)
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  2. #2
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    Krizalid's Avatar
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    Quote Originally Posted by AlphaRock View Post
    x= 8(t^2)
    y= 64t - (t^3)/3
    0 <= t <= 8sqrt(3)

    My work:
    S = 2pi(integral(((64t - (t^3)/3)sqrt((16t^2 + (64-t^2)^2)dt
    from 0 to 8sqrt(3)
    certainly, x'(t)=16t thus \big(x'(t)\big)^2=256t^2.
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  3. #3
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    didnot understand

    Quote Originally Posted by Krizalid View Post
    certainly, x'(t)=16t thus \big(x'(t)\big)^2=256t^2.
    i didnt get you. could you please elaborate????
    why cant we simply eleminate the parameter then integrate.
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  4. #4
    Math Engineering Student
    Krizalid's Avatar
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    the OP got a typo, that's all.
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