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Homework Help: Find Surface Area obtained by rotating a curve?

  1. Apr 27, 2008 #1
    Find the area of the surface obtained by rotating the curve
    y=2e^(2y)
    from y=0 to y=4 about the y-axis.

    Any help on this would be greatly appreciated. This has my whole hall stumped. We know that you have to use the equation 2pi*int(g(y)sqrt(1+(derivative of function)^2), but cannot figure out how to integrate this correctly.

    What I have gotten so far:
    y=2e^(2y)
    [when u=2y, du/2=dx]
    y=e^u
    New bounds: 1 to e^4

    2pi*int(e^u*sqrt(1+(e^u)^2)

    How do you go from there? Any help would be greatly appreciated.
     
  2. jcsd
  3. Apr 27, 2008 #2
    Sorry, the problem is x=2e^(2y)
     
  4. Apr 28, 2008 #3

    rock.freak667

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    [itex]x=2e^{2y} \Rightarrow \frac{dx}{dy}=4e^{2y}[/itex]

    [tex]S=2\pi \int _{0} ^{4} x\sqrt{1+\left(\frac{dx}{dy}\right)^2} dy[/tex]



    [tex]S=2\pi \int _{0} ^{4} 2e^{2y}\sqrt{1+(4e^{2y})^2} dy[/tex]


    Let [itex]u=4e^{2y} \Rightarrow \frac{du}{dy}=8e^2y \Rightarrow \frac{du}{4}=2e^{2y}dy[/itex]

    [tex]\equiv S=2\pi \int \frac{1}{4} \sqrt{1+u^2} du[/tex]

    I think a hyperbolic trig. substitution will work here e.g.t=sinhx (OR if you want t=secx)
     
  5. Apr 28, 2008 #4
    Thank you very much. I got this far, but tried to use normal trig substitution. IT goes without saying that it didn't really work for me.
     
  6. Apr 28, 2008 #5

    rock.freak667

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    [tex]\equiv S=2\pi \int \frac{1}{4} \sqrt{1+u^2} du[/tex]

    Let [itex]u=sect \Rightarrow \frac{du}{dt}=secttant \Rightarrow du=secttant dt[/itex]

    [tex] \frac{\pi}{2}\int \sqrt{1+sec^2t}secttant dt \equiv \frac{\pi}{2}\int \sqrt{tan^2t}secttant dt[/tex]


    [tex]\frac{\pi}{2}\int secttan^2t dt \equiv \frac{\pi}{2}\int sect(sec^2t-1) dt[/tex]

    Long and tedious but it should work.
     
  7. Apr 28, 2008 #6
    you could just use trig substitution where something in the form sqrt(A^2+X^2) should be approached making the substitution u=tangent(theta)

    so you could plus u into sqrt(1+u^2) and end up with sec(theta)
    then have du=sec^2(theta)

    then for 2pi*integtal(0.25sqrt(1+u^2))du you would get

    pi/2*integral(sec(theta)*sec^2(theta))d(theta)

    then just use u substitution and get your answer
     
  8. Apr 28, 2008 #7

    rock.freak667

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    Ahhh yes my mistake..sect would be the wrong trig fn....tant is much better...My mistake....Though I prefer the hyperbolic ones to the trig ones
     
  9. Sep 22, 2009 #8

    rock.freak667

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    it is not good to thread-jack, so start a new thread showing your work


    use this formula


    [tex]S= \int_{y_1} ^{y_2} 2 \pi x ds \ where \ ds=\sqrt{1+ \left( \frac{dx}{dy} \right) ^2} dy[/tex]
     
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