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Areas of surfaces of revolution

  1. Oct 4, 2011 #1
    Hello. I just started to study integral calculus not long ago and I have some confusion when it comes to calculating the areas of surfaces of revolution using integral.
    As from my testbook, when we want to calculate this kind of surface area, we often use the frustum surface area to approximate it in ds slant. Then integrate it along the defining curve( f(x) ) as the following formular states:
    S=∫2πf(x)ds=∫2πf(x)√(1+f'(x)2) dx
    However my problem is why we don't use the cylinder slice with dx thick to approximate it and then integrate over the specified region, as we do when calculating the volumes of solids of revolution.
    And I found in this case the formular will just be:
    S'=∫2πf(x) dx which is different from above.
    So please someone tell me where I made a mistake. I am sorry my post is lengthy and I also wish I could explain clearer.
    Thanks a lot for your time.
     
  2. jcsd
  3. Oct 4, 2011 #2
    ,

    The cylindrical slice of width dx gives the exact volume (not an approximation!) of a cylindrical shell. For a sphere, finding a volume first with cylindrical shells then taking a derivative of the volume will give the surface area of the sphere. But for other non symetric volumes that might not be so easy to do.

    read more at a randomly selected thread on the subject
    https://www.physicsforums.com/showthread.php?t=528358

    That was pure coincidence
     
    Last edited: Oct 4, 2011
  4. Oct 5, 2011 #3
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