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Sphere Volume to Surface Area, Why not for Cone?

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  1. Nov 30, 2015 #1

    MDH

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    1. The problem statement, all variables and given/known data
    Wikipedia tells me that I can obtain the surface area of a sphere by realizing that the volume of a sphere is equivalent to the infinite sum of the surface areas of hollow, nested spheres, sort of like little Russian dolls. That makes sense, and then differentiating both sides immediately yields the usual formula for the surface area of a sphere.

    However when I attempt to use this approach for the lateral surface area of a right circular cone it produces an obviously wrong answer. I'm assuming we can similarly identify the volume of a cone as the infinite sum of the surface areas of hollow, nested cones that 'sit' inside the overall cone.

    Why is this approach not fungible in this case?

    2. Relevant equations

    Case of the sphere:
    V = 4/3 πr3 = ∫ SA dr where the integral is over 0 to r, and SA is the function of r that represents the surface area of a sphere of a particular radius,

    and therefore
    dV/dr = 4πr2 = SA(r)




    3. The attempt at a solution

    Case of the right circular cone:

    V = 1/3 πr2h= ∫ SA dr, holding h constant, where the integral is over the base of the cone from 0 to r, and SA is the function of r that represents the lateral surface area of a sphere of a particular radius,

    and therefore
    dV/dr = 2/3πrh = SA(r), where obviously it should be πrl, where l2 = (r2 + h2)
     
  2. jcsd
  3. Nov 30, 2015 #2

    andrewkirk

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    I presume you mean a cone with a particular radius.

    THat said, I think a reason it doesn't work is that by changing r but not h, you are changing the shape of the cone. The incremental volume you are calculating is of maximum thickness dr at the base and zero thickness at the top, so it is not proportional to the surface area.

    I expect you could make it work by expressing h as a function of r in order to maintain the same shape, and then differentiating wrt r.

    Give it a try and see what happens.
     
  4. Nov 30, 2015 #3

    MDH

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    Thank you AndrewKirk, that was very intuitive and helpful.

    My first new attempt used the similarity between a representative 'proportional nested cone' and the overall cone, getting that H/R = h/r. I then integrated from 0 to r, the lateral surface area of the representative cone which is then πR(R/r)(h2+r2)½ with respect to dR. This unfortunately gave me the volume as ⅓πlr2.

    Then I realized it's not correct that the 'representative shell' has thickness dR. It actually has thickness dS, where the relationship is that their ratio is the same as the overall ratio of h and l. That is, dR is measured in the plane of the base, but the shell thickness is at (90-α) to that base, where α is the angle between long axis and slant axis. So I shoulda integrated wrt dS, where dS = (h/l)dR.

    Making that correction turns the integral into the right shape: ∫ πR(R/r)(h2+r2)½ (h/l)dR. Now (h2+r2)½ cancels with (1/l) and we end up with the correct volume of 1/3 πhr2.

     
  5. Nov 30, 2015 #4

    LCKurtz

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    Since this apparently isn't a homework question, I'm going to add more detail. Consider the following figure which is a cross section of a portion of a cone.
    cone cross section.jpg
    Here I am supposing the height ##h## is proportional to the radius ##r## so ##h=kr##. The surface area of this cone is $$S(r) = \pi r \sqrt{k^2r^2 + r^2} = \pi r^2 \sqrt{k^2+1}$$ To get an incremental volume you have to take the surface area an multiply it by the incremental thickness ##dw##, which is not equal to ##dr##. From the figure you can see that ##w = r\sin a## so $$dw = dr \sin a = dr \frac{kr}{r\sqrt {k^2+1}}
    =\frac{k}{\sqrt {k^2+1}}dr$$Consequently your element of volume becomes$$
    dV(r) = S(r)dw = \pi r^2 \sqrt{k^2+1}\cdot \frac{k}{\sqrt {k^2+1}}dr= \pi r^2 k dr$$If you integrate this with respect to ##r## and remember what ##k## is, you will get the correct volume for a cone.
     
  6. Nov 30, 2015 #5

    MDH

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    Totally agree, and sitting in a Starbucks I was able to sketch that out, as per the reply to AndrewKirk. Thanks much!
     
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