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Math Help - Can't integrate the surface area of revolving curve the normal way

  1. #1
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    Can't integrate the surface area of revolving curve the normal way

    Here's the problem I was given:


    Find the area of the surface generated by revolving the curve


    x=\frac{e^y + e^{-y} }{2}


    from 0 \leq y \leq ln(2) about the y-axis.


    I tried the normal route first...


    g(y) = x = \frac{1}{2} (e^y + e^{-y})
    g'(y) = dx/dy = \frac{1}{2} (e^y - e^{-y})


    S = \int 2\pi \frac{1}{2} (e^y + e^{-y}) \sqrt{1+(e^y - e^{-y})^2} dy


    S = \pi \int (e^y + e^{-y}) \sqrt{\frac{1}{4}e^{2y}+\frac{1}{2}+\frac{1}{4}e^{-2y} } dy


    S = \pi \int (e^y + e^{-y}) \sqrt{\frac{1}{4} (e^y+e^{-y})^2} dy


    S = \pi \int (e^y + e^{-y}) \frac {1}{2} (e^y+e^{-y}) dy


    But then I got stuck here...


    S = \frac{1}{2} \pi \int (e^y + e^{-y})^2 dy


    How should I proceed? Thanks in advance.
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  2. #2
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    Re: Can't integrate the surface area of revolving curve the normal way

    Multiply out:

    S = \dfrac{\pi}{2}\int_0^{\ln 2}(e^{2y} + 2 + e^{-2y})dy = \dfrac{\pi}{2}\left. \left(\dfrac{e^{2y}}{2} + 2y - \dfrac{e^{-2y}}{2} \right) \right]_0^{\ln 2}
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  3. #3
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    Re: Can't integrate the surface area of revolving curve the normal way

    Another way to do this problem:

    x = \cosh y so \dfrac{dx}{dy} = \sinh y. This simplifies the formula to:

    S = 2\pi \int_0^{\ln 2}\cosh y \sqrt{1+\sinh^2 y}dy

    Since \cosh^2 y = 1 + \sinh^2 y, you can rewrite:

    S = 2\pi \int_0^{\ln 2} \cosh^2 y dy

    Using the double angle formula for hyperbolic cosine: \cosh^2 y = \dfrac{1}{2} + \dfrac{1}{2}\cosh(2y)

    S = \left. \pi y + \dfrac{\pi}{2}\sinh(2y) \right]_0^{\ln 2} = \left. \pi y + \dfrac{\pi}{4}\left(e^{2y} - e^{-2y}\right) \right]_0^{\ln 2}
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  4. #4
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    Re: Can't integrate the surface area of revolving curve the normal way

    Quote Originally Posted by SlipEternal View Post
    Another way to do this problem:

    $$x = \cosh y$$ so $$\dfrac{dx}{dy} = \sinh y$$. This simplifies the formula to:

    $$S = 2\pi \int_0^{\ln 2}\cosh y \sqrt{1+\sinh^2 y}dy$$

    Since $$\cosh^2 y = 1 + \sinh^2 y$$, you can rewrite:

    $$S = 2\pi \int_0^{\ln 2} \cosh^2 y dy$$

    Using the double angle formula for hyperbolic cosine: $$\cosh^2 y = \dfrac{1}{2} + \dfrac{1}{2}\cosh(2y)$$

    $$S = \left. \pi y + \dfrac{\pi}{2}\sinh(2y) \right]_0^{\ln 2} = \left. \pi y + \dfrac{\pi}{4}\left(e^{2y} - e^{-2y}\right) \right]_0^{\ln 2}$$
    redone for new LaTex
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  5. #5
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    Re: Can't integrate the surface area of revolving curve the normal way

    Son of a monkey. I forgot how much I hate working with e. Thanks, @SlipEternal, for pointing out how simple that really was.

    Here's what I was doing:

    S = \frac{1}{2}\pi \int (e^{2y} + 2 + e^{-2y}) dy

    Then, breaking it up into three separate integrals and working with just the first one...

    S = \frac{1}{2}\pi \int (e^{2y}

    S = \frac{1}{2}\pi \frac{1}{2y+1} e^{2y+1} from 0 to ln(2)
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