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Surface Area of a Cone

  1. Aug 12, 2012 #1
    Hello, I have actually asked a similar question before, but I just realized something and I want to edit the question now:

    I am trying to derive the formula for the lateral surface area of a cone by cutting the cone into disks with infinitesimal 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. This seems to work. However, it doesn't make sense to me, because it seems like the height of each disk should be dH, not dS. I tried to show my argument/confusion in this picture:

    http://dl.dropbox.com/u/29312856/Cone.jpg [Broken]

    P.S. I basically used the method shown in another person's video () to find the volume of the cone, and that person assumed that the height of each disk is dH, not dS, and in that case, dH worked. :O

    Can anyone tell me why the height of each cylinder is dS, not dH? Or can anyone tell me what I am doing wrong? Thanks in advance! It's been bothering me for days... :(
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Aug 12, 2012 #2

    I think you forgot you're actually trying to do some 3-dimensional stuff with a 2-dimensional drawing...:) .

    Grab a children-party hat, draw on its courved surface the curved rectangles (which aren't actual rectangles at all, of course)

    and then make an actual drawing of a segment of line representing the lateral height, and you'll get convinced

    this line is ACTUALLY perpendicular to the "rectangles'" sides and it is thus their height!

    Last edited by a moderator: May 6, 2017
  4. Aug 13, 2012 #3
    Hi DonAntonio, thank you for replying! :D

    First, to clear things up, my drawing was just a cross-section of a cone to demonstrate what I meant. (Imagine cutting a party hat in half). The rectangles are actually cross-sections of the cylinders :)
    Here's a better cross-section of the cone:

    http://dl.dropbox.com/u/29312856/Cone.jpg [Broken]

    Now here's the problem: The cylinders' lateral sides are perpendicular to their bases, so the sides are parallel to the height (H) of the cone but not the lateral side (S) of the cone. Also, if the cylinder's height was actually dS, then it would contradict the person's method shown here: , because that person used dH as the cylinders' height when calculating the volume.

    So basically, it's a gigantic confusion that I can't get my head around.
    Last edited by a moderator: May 6, 2017
  5. Aug 13, 2012 #4

    You keep on repeating what I think is your main mistake: drawing a 2-D picture of a 3-D situation, and the mistake:

    you DEDUCE stuff from this 2-D picture, and this is wrong...in your cross section you even drew the "rectangles"

    as ACTUAL rectangles and this is wrong: what you get are rings around the cone, which if you translate the cone's

    height to their side it is NOT even close to be perpendicular to their boundaries!

    In short: you're drawing this "circular rectangles" of yours, you're pasting (projecting, in fact) them as actual

    rectangles to a plane where also the cone's hieght is drawn, and then you deduce "hey, these guys are

    perpendicular to each other!"...wrong!

    On the cone, on each ring (your rectangle) around the cone , the LATERAL height is the one which is

    perpendicular to each ring's boundary (call it side if you like). Do build a cone and do what I proposed to you last time.

    Last edited by a moderator: May 6, 2017
  6. Aug 13, 2012 #5
    The problem I think is that you wouldn't get the surface area if you used cylinders whose sides aren't parallel to the sides of the shape. For example, consider trying to "square" the perimeter of a circle. For illustration: http://qntm.org/trollpi

    The sides of the square are cut into many pieces, and then "steps" are created from it, but you can always combine the steps back into the side of the square. Consider the fourth picture in the link, look at the upper half of the circle. The whole upper side of the square is there, it is just in pieces. So you can jag them all you want, the perimeter is the same and doesn't start approximating a circle.

    In the cross section of the cone, considering it as a 2-d object, you can also try to calculate the perimeter of the slice. If you add up the sides of the rectangles, they will always add up to 2H, no matter how small you make them, however you can check using Pythagoras that it should be more. because its twice (for two sides of the slice) the square root of H^2+R^2
  7. Aug 13, 2012 #6
    Ok, thank you DonAntonio and chingel! I completely understand it now. I really really really appreciate the time you guys spent to answer my question. It's been bothering me for days, and I finally get it! :D Thanks!!
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