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I Surface Area of Volume of Revolution

  1. Oct 10, 2017 #1
    The problem is, find the surface area of the volume of revolution generated by rotating the curve y=e2x between x=0 and x=2 about the x-axis.

    Here's what I have so far...


    and from here I'm not really sure what to do. Any help would be appreciated.
  2. jcsd
  3. Oct 10, 2017 #2


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    You want to solve the integral ##I = \int e^{2x} \sqrt{1+4e^{4x}}dx = \int e^{2x} \sqrt{1+(2e^{2x})²}dx##

    To do this, I suggest you perform the substitution ##u = 2e^{2x}##. The integral you then obtain can be easily solved using an appropriate trigonometric substitution.
    Last edited: Oct 10, 2017
  4. Oct 10, 2017 #3
    OK, so setting u=2e2x, I get du=4e2x dx, and dx=1/4e2x du. Substituting that in, I get...

    I=2π∫(e2x)(1/4)(1/e2x)√(1+u2) and then when I simplify...

    I=1/2π∫√(1+u2) using a trig sub,

    I=1/2π∫sec3θ dθ

    Is this correct?
  5. Oct 10, 2017 #4


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    The result you obtain is correct.
  6. Oct 10, 2017 #5
    Thank you very much for your help!
  7. Oct 10, 2017 #6


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    Little disclaimer, I just helped you solve the integral. I don't know the formula of revolution by heart, so I don't know whether you have that part of the question correct. Also, you should post questions like these in the homework sections ("calculus and beyond").
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