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Deriving the volume and surface area of a cone

  1. Aug 7, 2012 #1
    Hello, this is my first time posting on physics forums, so if I do something wrong, please bear with me :)

    I am trying to derive the formula for the lateral surface area of a cone by cutting the cone into disks with differential height, and then adding up the lateral areas of all of the disks/cylinders to find the lateral area of the cone. (Similar to using the volume of revolution, but just taking the surface area). I assumed that the heights of each of the disk was dH, where H = the height of the cone. However, using that method, I got pi*radius*height instead of pi*radius*(slant height).

    On another thread in physics forum (https://www.physicsforums.com/showthread.php?t=354134) , a person used a similar method as I did, and someone replied saying that the height of each disk is dS, where S = the slant height of the cone, not dH, where H = the height of the cone. And using dS instead of dH allowed me to find the correct formula for the lateral area of the cone!

    HOWEVER, using dS instead of dH contradicts with the same method for finding the volume of the cone. I basically used the method shown in another person's video () to find the volume of the cone, but that person assumed that the height of each disk is dH, not dS.

    So is the height of each disk dS or dH? If it is dS, then I understand how to get the formula for the lateral surface area of the cone, but not the volume. But if it is dH, then I understand how to get the formula for the volume of the cone, not the surface area. Can anyone clear up my confusion? I know that there are other methods to find the surface area/volume, but I just want to know why there is a contradiction between the dH/dS thing. Thanks! Any help is appreciated! :)
     
    Last edited by a moderator: Sep 25, 2014
  2. jcsd
  3. Aug 8, 2012 #2
    ds is the rate of change in the lateral area of an infinitesimal disk, while dh is the change of its height. The volume of an infinitesimal cylinder is computed using its height, not its slant height; while when you are computing the surface area, the height of each cylinder is not what you want to use. You can easily apply the Pythagorean theorem to see what I mean. In a pure cylinder, the slant height and the normal height is equivalent. When you are partitioning a cone however, they are not. This is simply because the cone has a slope that can't simply be discarded. Discarding the slope is exactly what you are doing now. Volume is OK to compute like that, but surface area isn't. This solves your problem.
     
  4. Aug 9, 2012 #3
    Hello Millennial, thank you for your reply! I understand that cylinders have a different slope from cones. However, you said, "Discarding the slope is exactly what you are doing now. Volume is OK to compute like that, but surface area isn't." Why is it okay to discard the slope when computing the volume, but not okay when computing the surface area? Thanks!
     
  5. Aug 9, 2012 #4

    arildno

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    The tiny volume bit error is, relatively speaking, vansihingly small to that which is kept.

    The local cylinder volume approximation is Vc= pi*r^2*dz

    Now, let us look at the volume of the tiny element from an expanding cone.

    This is made up of Vc+the volume of a triangular area with the full circular periphery as its radius.

    This bit's volume can be written (2*pi*(r+dr))*1/2*dz*dr, where dr is the tiny radial additon from position (z,r) to position (z+dz, r+dr).
    Now, the biggest part of this volume equals pi*rdzdr (agreed?)
    Since both dz and dr goes to zero in the limit, the PRODUCT of these tiny quantities go much faster to zero than either one of them.
    Agreed?

    Thus, we can discard that volume bit relative to Vc.
    -----
    However, what you lose when computing surface area (with cylindric approxamition), you'll lose area portions that do not vanish faster than the one you keep.
     
  6. Aug 9, 2012 #5

    arildno

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    As for the surface area element, with the cylinder, you get r*dw*dz, where dw is the tiny angle, and r*dw the tiny arc length.

    With the cone, you get r*dw*dS instead (where dS is the length along the skewed plane), so that the ratio of these terms are dS/dz.
    This does NOT equal 1, so that you cannot approximate the surface area locally with that of the cylinder
     
  7. Aug 9, 2012 #6
    Hi arildno, thank you for your detailed answer :) I completely understand your second answer (the one about the surface area), but I have a question about your first answer (I apologize if I'm asking too many questions haha). I understand that the volume difference (2*pi*(r+dr))*1/2*dz*dr) is infinitesimal and can be discarded, so now I know why it is OK to discard the slope when computing the volume.

    However, why can't we just use dS instead of dH when finding the volume? In other words, why can't we let the differential heights of the cylinders add up to the slant height of the cone instead of only height of the cone when calculating the volume? Wouldn't that fix the problem with the volume difference?
     
  8. Aug 10, 2012 #7

    arildno

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    It is no problem with the volume difference.
    You sure could use dS, rather than dH, but you'd get precisely the same answer, if you use dS correctly (and the appropriate volume element in that case).

    That would, essentially, be to invoke the change-of-variables-theorem, and use another set of coordinates than cylindrical coordinates to represent the cone by.
     
  9. Aug 10, 2012 #8
    Thank you, arildno! Out of all of the places I have asked this question, you and Millenial have been the only ones who could answer my question. I understand it now :) Thank you so much.

    P.S., if you're not too busy, can you show me what steps you would need to use in order to use dS correctly?
     
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