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Silly theoretical area question

  1. Jan 17, 2016 #1
    Just humour me, if you had an infinitely thin cone, would the surface area inside the cone be the same surface area on the outside of the cone? It must be right?
    Is there a formula for the surface areas of the inside and outside of a cone WITH thickness?
     
  2. jcsd
  3. Jan 17, 2016 #2

    HallsofIvy

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    You would have to define "infinitely thin cone". If you are talking about a mathematical cone, rather than a real conical object, then there is NO thickness. There are no "inside" and "outside" surfaces to talk about, just the one surface. I don't think there is a regularly given formula for a "cone with thickness" but it is not too difficult to come up with one. The thickness, I presume, is measured perpendicular to the two, "inside" and "outside", surfaces. That makes the radii and heights of the two cones a little tricky to calculate. Letting [itex]r_1[/itex] and [itex]h_1[/itex] be the radius and height of the "inside" surface and [itex]r_2[/itex] and [itex]h_2[/itex] be the radius and height of the "outside" surface, looking from the side we see two right triangles, one with legs of length [itex]r_1[/itex] and [itex]h_1[/itex], the other with legs of length [itex]r_2[/itex] and [itex]h_2[/itex]. Since the altitude of any right triangle is the "geometric mean" of the lengths of the legs, the altitudes of the two right triangles are given by [itex]a_1= \sqrt{r_1h_1}[/itex] and [itex]a_2= \sqrt{r_2h_2}[/itex] and the "thickness" of the cone is the difference [itex]d= \sqrt{r_1h_1}- \sqrt{r_2h_2}[/itex].
    So if we are given the radius and height, say, of the inner cone and thickness, d, we can calculate the radius and height of the outer cone and the find the surface area of the two cones. But it is NOT an easy calculation!
     
  4. Jan 17, 2016 #3
    I googled "how do you calculate the surface of a cone" and a cool calculator popped up. Outside radius - thickness = inside radius.
     
  5. Jan 17, 2016 #4
    That is an interesting reply, but what does the geometric mean length have to do with the area?

    I see that the area is A=πr(r+(h^2+r^2)^0.5)
    why is the area not equal to 2*π*(r/2)*h ?
    assuming r/2 is the mean radius of the cone.
     
  6. Jan 18, 2016 #5

    HallsofIvy

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    That's not true and why I said "it is NOT an easy calculation". The thickness of the cone is measured perpendicular to the "inside" and "outside" surfaces. That is NOT the difference of the two base radii because the two side are not perpendicular to the base.
     
  7. Jan 18, 2016 #6
    Why isn't it the difference between radii, wouldn't the only difference be a sort of triangular circle at the base? (like revolving a triangle 360degrees)
     
  8. Jan 18, 2016 #7
    I'm sorry I didn't even see your post before I posted mine and yes I know it is complicated to get specific. I was simply "humoring" the question with the intention of isolating the most significant portion of the cone.
    Seems like I could say the same thing about a simple circle drawn with an ultrafine pen and the "outside" edge circumference will always be larger than the "inside" edge. Even if the circle was an atom thick.
     
  9. Jan 19, 2016 #8
    Yeah but isn't that covered by what I said, on that 2D surface the width of the circle is the atom thickness.

    What I was saying is that the volume of the thickness of the cone is a triangle revolved 360deg, at the base, from the bottom of the internal hypotenuse up to the altitude of the exterior surface. Then the rest is the difference between radii?
     
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