Thinking about volume...
Looking at the equation for volume of a sphere: \(\large\rm v_{\circ}=\dfrac{4}{3}\pi r^3\) This value 4/3 seemed kind of ... random to me. But I remembered that the volume of a cone is given by: \(\large\rm v=\dfrac{1}{3}\pi r^2h\) So if we look at the cone with height equal to radius we have: \(\large\rm v=\dfrac{1}{3}\pi r^3\) Which is precisely a fourth of the volume of a sphere!
So hmm, I'm wondering if there is some clever way to deform a cone, (Like cutting it in a clever way or something) to show that it's shape is the same as half of a hemisphere.
@dan815 @Empty just a weird thought :)
I have heard that this relation was first discovered by Archimedes and was put on his tombstone! I am interested in trying to figure it out, but I need to think a little.
I'm thinking about like putting 4 cones in this awkward shape or playing with them |dw:1445420363944:dw|
Cause that's the same volume as a sphere rearranged I guess, sorta odd.
That idea seems clever and intuitive, I know that's how I would think of a square pyramid in relation to a cube.
Yeah I don't really see this getting us anywhere I just needed to stick the 4 cones up there in as symmetrical of a way as possible but if I had to I'd use 6 cones in a cube. So weird. hmm
The great circle around the sphere of radius r has circumference 2pi*r. Which is the same as the base of the cone. What kind of shape do we get if we cut a cone along it's diagonal? I can't seem to visualize it... if we unfold it, is it half of a circle?
Oh wait, what if we place the tips of the cone at the center, and have them so that they are at the points of a regular tetrahedron?
|dw:1445420730474:dw| I am trying to get as much symmetry as possible since a sphere as a lot of it, it will help to simplify the idea of deforming this in a single kind of a step but... this is admittedly pretty weird to imagine, it's just sorta some idea, but I think we gotta like chop it up somehow so idk
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