1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Homework Helper

    [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

    User Avatar
    Homework Helper

    [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

    User Avatar
    Homework Helper


    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

    User Avatar
    Homework Helper

    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]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?