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I understand how the area of a pyramid is derived and that a cone is just a pyramid with a circular base.

But I can't help but to imagine the circumference being straightened and being a triangular prism. The circumference is $2\pi r$ and the area of the triangle is $\frac 12 r h$. $\frac 12 r h$ is being replicated $2\pi r$ times!

I think my mathematical intuiton is failing me because things like this keeps me up at night.

Following that, the area of the sphere. Damn Archemedis is also telling me that the volume of a sphere is a ratio to the imaginary cylinder it is inscribed in. But my stupid mathematical intuition tells me it is half the circumference times the area of the circle. Which can be written as $\frac 12 \pi r \times \pi r^2$. Purge this arrogance from me by explaining simply why my intuition is false without using the method of exaustion.

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  • $\begingroup$ I think you mean volume of the cone, not area... $\endgroup$ – Martigan May 23 '18 at 15:25
  • $\begingroup$ A cone is a triangle rotated around the axis by 2pi radians, so you could use a surface area of revolution formula to derive this. $\endgroup$ – Henry Lee May 23 '18 at 15:26
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The area of a triangle is $(b\times h)/2.$

If you consider the area of a cone as the area of a triangle, then the base is $2\pi r$ and the height is the length of the slant, that is $ \sqrt {r^2+h^2} $.

Thus you get $ (1/2)(2\pi r )( \sqrt {r^2+h^2})=\pi r \sqrt {r^2+h^2} $ which is the correct formula.

On the other hand when you open a sphere you do not get a flat surface so the triangular argument does not work.

Note that the triangular slices on the surface of a sphere are indeed triangles in hyperbolic geometry which have two right angles.

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Here is a simple way of seing that your formula can't be right.

Imagine that you do not consider the cone at its base (where the cisrcumference is $2\pi r$), but at its summit (where it is $0$).

Then with the same reasoning I would say that the volume is $0 \times \frac 12 r h=0$...

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  • $\begingroup$ Considering r=0? $\endgroup$ – GLaDOS May 29 '18 at 7:35
  • $\begingroup$ @GLaDOS Yes, after all, in your "false intuitive demonstration", you take the circumference of the base. But you could take the circumference at any point, since what you want to do is "turning the triangle around". By "choosing" a specific point for turning the triangle around, you get a different result, which means it is wrong. And there is absolutely no reason why you should choose the circumference at the base of the triangle, since turning the triangle can be done at any point of the hypothenuse. $\endgroup$ – Martigan May 29 '18 at 7:42
  • $\begingroup$ Oh, that is right. If you rotate the triangle around it's second endpoint, you won't get the same volume if you have done it around the first endpoint. I think that is the argument, you are making. But that means the radius of the circle changes which always has to be equal to the base of the triangle which is a condition that must be maintained. $\endgroup$ – GLaDOS May 29 '18 at 7:53
  • $\begingroup$ @GLaDOS Yes, but you have to ask yourself WHY you, in your mind, you should take the circumference at the base. Why not at the middle after all? Since there is no reason to do either, you should not use this reasonning. $\endgroup$ – Martigan May 29 '18 at 8:07

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