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Homework Help: Problem Deriving Volume of a Cylindrical shell

  1. Apr 16, 2010 #1
    I have seen the derivation for the formula:

    h2(pi)rdr used in math textbooks. However, earlier today I had a physics problem where I needed to use the volume of a cylindrical shell of inner radius r and outer radius dr+r and length h. without remembering the formula I tried to derive it starting by determining the area of the base minus the area of the gap:

    and then multiplying by h

    I could not get this in the form h2(pi)rdr starting from where I did.

    I understand the derivation given here:
    http://www.stewartcalculus.com/data...texts/upfiles/3c3-Volums-CylinShells_Stu .pdf

    But I'm just frustrated that I can't understand why the method I tried doesn't reduce to the correct formula. Its simply the area of the base*height!

    Why can't that formula be arrived at by my method???
  2. jcsd
  3. Apr 16, 2010 #2


    Staff: Mentor

    Assuming that dr is positive, what you'll get from your formula is a negative number.

    So instead, you should have A = pi(r + dr)^2 - pi*r^2 = pi*( (r + dr)^2 - r^2).

    Expand the (r + dr)^2 part and combine like terms. What do you get?
  4. Apr 16, 2010 #3
    When I expand I get r^2 +2rdr + dr^2

    so the r^2 drops out. but what happens to the the dr^2 ?????
  5. Apr 17, 2010 #4


    Staff: Mentor

    The idea here is that if dr is small in comparison to r, the (dr)^2 will be very much smaller and can be discarded.

    For example, if r = 1 and dr = .01, then (r + dr)^2 = (1.01)^2 = 1.0201.

    If I expand (r + dr)^2, I get r^2 + 2r*dr + (dr)^2 = 1 + 2(.01) + (.0001). If I omit the (dr)^2 term, I get 1.02, which is pretty close to 1.0201.

    The smaller dr is in comparison to r, the closer (r + dr)^2 is to r^2 + 2r*dr.
  6. Apr 17, 2010 #5
    Alright that's what I thought I just couldn't find any verification!

    I never really encountered a differential element squared before.

    Thanks so much!
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