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Find the surface of the generated volume

  1. Mar 11, 2015 #1
    1. The problem statement, all variables and given/known data
    The region bounded by the graphs of the curves y=4√x , y=0, x=5, x=12 is revolved about the x-axis.
    a) Find the religion lateral surface area of the generated volume.
    b) Find the total surface area of the generated volume, including both ends.
    2. Relevant equations


    3. The attempt at a solution
    Part a:
    https://lh6.googleusercontent.com/kZIjTDFGufXqICfEMW2J0CpXFuQ8-OfRJlg91DQJRWZlCpsiCzfND45AYeYCfasTzAUca7ZFa9k=w1256-h843

    Part b:

    https://lh3.googleusercontent.com/jguorhs3qoad_UbPduGEyBG79Y9WVLG0vPzC4YV6Y1sTNV-gbxczDdBbPKr-G5jIH3wO0g=w1256-h843

    Thank you so much for taking time.
     
    Last edited: Mar 11, 2015
  2. jcsd
  3. Mar 11, 2015 #2

    Mark44

    Staff: Mentor

    You have a mistake in the integral on the next to the last line of your first page. The radius is y, not x. Also, you have a typo where you wrote ##y = \frac 2 {\sqrt{x}}##. That should have been y', the derivative.
     
  4. Mar 11, 2015 #3
    The formula for arc length is ##2\pi\int{y\sqrt{1 + (y\prime)^{2}}\ dx}## which using that, I got that the arc length is ##8\pi\int{\sqrt{x+4}\ dx}##. The radius r changes with x! :wink:
     
    Last edited: Mar 11, 2015
  5. Mar 11, 2015 #4
    I'm sorry. The question is correct. 2/√x is the derivative of 4√x.
     
  6. Mar 11, 2015 #5

    Mark44

    Staff: Mentor

    Yes, I understand that. My comment was that you wrote ##y = \frac 2 {\sqrt{x}}## when you should have written ##y ' = \frac 2 {\sqrt{x}}##

    More importantly, your integral is wrong. It should be
    ##2\pi \int_5^{12} y\sqrt{\frac{x + 4}{x}}dx##
     
  7. Mar 11, 2015 #6
    Why is the radius y? I thought the radius is x or (x-5), I'm not not so sure though.
     
  8. Mar 11, 2015 #7
    For a given x the cross-sections of the solid you're interested in are circles whose centre is at x & whose radii are ##4\sqrt x##.
     
  9. Mar 11, 2015 #8

    Mark44

    Staff: Mentor

    The curve is being rotated around the x-axis, so the radius is y.
     
  10. Mar 11, 2015 #9
    Oh yes, you're right. So I have to change the whole equation then.
    I really appreciate your help.
     
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