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Surface Area by rotating a curve

  1. Sep 22, 2009 #1
    1. Find the surface area generated by rotating the curve about the y-axis

    2. X=(1/3)Y^(3/2)-Y^(1/2) between 1 and 3

    3. x' = (1/2)Y^(1/2) - (1/2)Y^(-1/2)
    x'^2 = (y^2-2y+1)/4y

    I'm not sure if that derivative is correct or if implicit differentiation is required. I also can't get any further using the surface area formula. Any help would be appreciated.
  2. jcsd
  3. Sep 22, 2009 #2
    Your math is correct; you shouldn't need to use implicit differentiation. Which formula are you using for surface area?
  4. Sep 22, 2009 #3
    https://www.physicsforums.com/latex_images/23/2361260-0.png [Broken]

    After substituting my data into the equation I can't simplify it far enough to take the integral
    Last edited by a moderator: May 4, 2017
  5. Sep 22, 2009 #4
    The first x in that integral should be a y. :wink:

    Combine the 1 with (y^2-2y+1)/4y and try to get the quadratic part into (y + something)2 so you can take its square root.
  6. Sep 22, 2009 #5
    I didn't think to do that, I think I've got it now. Thanks!
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