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calculus 2

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The curve y=sinh(x),0<=x<=1, is revolved about the x-axis. Find the area of the resulting surface.

  • calculus 2 - ,

    Revolved about the x-axis~ must use: integral 2*Pi*y ds (Eq 1)
    For ds use:
    ds = sqrt(1+[F'(x)]^2) dx (Eq 2)
    sinh(x) dx = cosh(x) = F'(x)

    sub cosh(x) into Eq 2 and you get:
    ds = sqrt(1+[cosh(x)]^2) dx

    now sub ds into Eq 1
    integral 2*Pi*y*sqrt(1+[cosh(x)]^2) dx
    We are taking the derivative with respect to x so change y to it's equivalent.

    integral 2*Pi*sinh(x)*sqrt(1+[cosh(x)]^2) dx, from 0 to 1
    you now have your final equation to integrate over the interval, 0 to 1
    and you get 5.52989 (final answer)

  • calculus 2 - ,

    by definition
    sinh(x) = (e^x - e^-x)/2

    so the area

    = ∫sinh(x) dx from 0 to 1
    = ∫(e^x - e^-x)/2 dx from 0 to 1
    = [ e^x + e^-x)/2] from 0 to 1

    = (e^1 + e^-1)/2 - (e^0 + e^0)/2
    = (e + 1/e)/2 - (1+1)/2
    = e/2 + 1/(2e) - 1
    = (e^2 + 1 - 2e)/(2e)
    = (e - 1)^2 / (2e)

    you better check my arithmetic

  • totally ignore my post - calculus 2 - ,

    scrap my reply,
    I found the area, silly me

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