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Proving Properties of a Cone

  1. Nov 20, 2011 #1
    1. The problem statement, all variables and given/known data
    A solid cone is obtained by connecting every point of a plane region S with a vertex not in the plane of S. Let A denote the area of S, and let h denote the altitude of the cone. Prove that:

    (a) The cross-sectional area cut by a plane parallel to the base and at a distance t from the vertex is (t/h)2A, where 0 ≤ t ≤ h, and

    (b) the volume of the cone is Ah/3.

    3. The attempt at a solutionBefore I attempt this, I'd just like to know whether or not it is reasonable to assume that S is circular. This would make the A equal to pi*(radius of S)2. I think

    the limit as t goes to h of (t/h) is 1. Then, the change in radius of the circle is given by (radius of S)(t/h). This means that the area of the circle given the change in t is

    pi*(radius of S)2(t/h)2, which is what I wanted to show. (I find this to be the most unconvincing part of my post)

    In order to find the volume, I integrate. I have

    ∫(0≤t≤h) (pi*(radius of S)2(t/h)2)dt

    (pi*(radius of S)2/h2 * (t)3/3, evaluated from 0 to h. So I have pi*(radius of S)2/h2 * h3 = 1/3 * pi * (radius of S)2 * h,

    which is also what I wanted.

    Look okay?
     
  2. jcsd
  3. Nov 20, 2011 #2

    LCKurtz

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    I would say it is not reasonable to make that assumption. That is a special case and working with might give you insight, but it is not a solution to the question asked.
     
  4. Nov 20, 2011 #3

    I like Serena

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    Hi TranscendArcu! :smile:

    If you assume that S is circular, it looks okay.

    However, the statement also holds if S is not circular.
    For that you would for instance divide S into a partition of (infinitesimal) rectangles.
    Each rectangle scales down in length and width...

    Btw, for part (b) you do not have to assume that S is circular.
    You can use the result of (a) to find the volume of the cone.
     
  5. Nov 20, 2011 #4

    LCKurtz

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    And also note that the "cone" may be slanted.
     
  6. Nov 20, 2011 #5
    So I get part b. That's pretty easy. But I don't really understand how I can work (t/h)2 into part a. I thought my reasoning was pretty weak in the circular example above.

    To me, it seems to me like the the area at any cross section of the cone should just be A(t/h). But clearly this isn't so. How should I be thinking about this differently to see where the squared term comes in?
     
  7. Nov 20, 2011 #6

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    Suppose S is a rectangle with width W and height H.
    Going up (slanted or not), the rectangle's dimensions would scale down.
    By how much?

    Now suppose S consists of many small rectangles...

    Now make the rectangles smaller and keep increasing the number of rectangles to infinity...
     
  8. Nov 20, 2011 #7
    This is so frustrating. I assume that the dimensions scale down by a factor of (t/h)^2. Is there a formula or something that I might need in order to show this? I've tried fiddling around with rectangles and attempted to use the Pythagorean Theorem in vain. The problem is that I can't visually see how, say, W and the new dimension dimension for W (call it W') vary as we move up. I can see W' ≤ W, but that's about it at this point.
     
  9. Nov 20, 2011 #8

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    A rectangle edge and your vertex together form a triangle.

    If you draw lines segments within the triangle, parallel to your rectangle edge, their lengths will decrease linearly until you reach the vertex.
    If you want, this can be proven with vector algebra or trig identities.

    In other words, the width W of a rectangle scales down with t/h.
    That is: W(t) = t/h Wo.

    The same thing for the length of the rectangle.
     
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