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

  1. Feb 18, 2014 #1
    1. The problem statement, all variables and given/known data

    Find the area of the surface generated by revolving the curve

    [itex]x=\frac{e^y + e^{-y} }{2}[/itex]

    from 0 [itex]\leq[/itex] y [itex]\leq[/itex] ln(2) about the y-axis.

    3. The attempt at a solution

    I tried the normal route first...

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

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

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

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

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

    But then I got stuck here...

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

    How should I proceed? Thanks in advance.
     
  2. jcsd
  3. Feb 18, 2014 #2

    pasmith

    User Avatar
    Homework Helper

    I think this should be
    [tex]
    S = \int_0^{\ln 2} 2\pi \frac 12 (e^y + e^{-y}) \sqrt{1 + \frac14 (e^y - e^{-y})^2}\,dy
    [/tex]

    And now missing factor of 1/4 has appeared (but the limits are still missing).

    [itex](e^y + e^{-y})^2 = e^{2y} + 2 + e^{-2y}[/itex]...
     
  4. Feb 18, 2014 #3
    @pasmith, thanks for your response.

    I forgot how much I hate working with e. Thanks for pointing out how simple that really was.

    Here's what I was doing:

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

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

    S = [itex]\frac{1}{2}\pi \int (e^{2y}) [/itex]

    S = [itex]\frac{1}{2}\pi \frac{1}{2y+1} e^{2y+1} [/itex] from 0 to ln(2)
     
  5. Feb 19, 2014 #4

    pasmith

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    Homework Helper

    No.
    [tex]
    \int e^{ay}\,dy = \frac{e^{ay}}a
    [/tex]
     
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