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As far as I know any figure shaped like a cone (pyramid, cones, etc.) has a volume equal to one third of the base area times the height, so I planned on proving this statemente for any figure with such qualities using calculus, but i'm stuck an wish some help (please no full answer).

This is what I've thought of up until now. Say $A(x)$ is a function of the area of a transversal cut in terms of the height. Therefore $A(h)=0$ and $A(0)$ is the area of the base or equivalently $A(h)$ is the area of the base and $A(0)=0$ (I'll use the latter). Using Cavalieri's Principle we determine that the volume of the cone-like figure is given by adding the areas of all the transversal cuts, or in notation: $\int_0^h A(x)\, dx$

I also know there must be a rate of change that relates $A(x)$ to $x$, and this is the part where I'm stuck. :(

Thanks in advance.

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I changed "one third of the base area times volume", by replacing volume with height, assuming that's what you mean? – amWhy Mar 11 '13 at 20:49
Oh sure thank you :) – Sebastian Garrido Mar 11 '13 at 21:07

Hint: When cut with a plane parallel to the base, the upper portion (or, more precisely, the portion including the vertex) is similar to the whole solid. What can be said of the ratio of corresponding base areas $\frac{A(x)}{A(h)}$?

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You want $A(x)=cx^2$ which is true if the sections are homothetic, as for the cone.

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Where can I read more about this "omotethic property" you speak about? I googled it and saw some philosophical term instead of this. – Sebastian Garrido Mar 11 '13 at 21:08
@AndréNicolas: corrected :-) – Emanuele Paolini Mar 12 '13 at 6:36

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