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Homework Help: Surface Area of y = e^5x revolved around the x-axis from 0 to ln(4)

  1. Feb 23, 2010 #1
    1. The problem statement, all variables and given/known data


    Note... I used wolfram alpha to get the answer, I did not get it myself... So I still need help. The answer shown is correct, so you'll know if you got it.

    2. Relevant equations

    Integral [0, ln(4)] sqrt(1+(dy/dx)^2)

    3. The attempt at a solution

    2pi Integral [0, ln(4)] y*sqrt(1+(dy/dx)^2)

    2pi Integral [0, ln(4)] (e^5x)*sqrt(1+5e^5x^2)dx

    u = 5e^5x
    du = 25e^5x dx
    dx = du/25e^5x

    2pi Integral [0, ln(4)] (e^5x)*sqrt(1+u^2)du/25e^5x

    2pi Integral [0, ln(4)] sqrt(1+u^2)du

    u = tan(t)

    2pi/25 Integral [0, ln(4)] sqrt(1+tan^2(t))du

    2pi/25 Integral [0, ln(4)] sqrt(sec^2(t))du

    2pi/25 Integral [0, ln(4)] sec(t)du

    du = sec^2(t)dt
    dt = du*cos^2(t)

    2pi/25 Integral [0, ln(4)] cos^2(t)/cos(t)dt

    2pi/25 Integral [0, ln(4)] cos(t)dt

    Edit bounds...

    [arctan(5), arctan(5e^(5*ln(4)))]

    Then get ****ed over with an answer of .0048...

    What did I do wrong.
  2. jcsd
  3. Feb 23, 2010 #2


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    If du=sec^2(t)*dt, then sec(t)*du is sec^3(t)*dt.
  4. Feb 23, 2010 #3
    Right... My mistake, but I'm also having trouble with integration, and that isn't my strong suit, how would I integrate that?
  5. Feb 23, 2010 #4


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    It's kind of a long haul. You start by integrating by parts u=sec(t), dv=sec(t)^2*dt. It probably goes a little easier if you go back to the integral of sqrt(1+x^2)*dx and substitute x=sinh(u), if you are ok with hyperbolic functions.
  6. Feb 23, 2010 #5
    I am, but I'm in a class that doesn't use them yet lol.
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