<Dd> V = 1 3 A B h . (\ displaystyle V = (\ frac (1) (3)) A_ (B) h .) </Dd> <P> In modern mathematics, this formula can easily be computed using calculus--it is, up to scaling, the integral ∫ x 2 d x = 1 3 x 3 . (\ displaystyle \ int x ^ (2) dx = (\ tfrac (1) (3)) x ^ (3).) Without using calculus, the formula can be proven by comparing the cone to a pyramid and applying Cavalieri's principle--specifically, comparing the cone to a (vertically scaled) right square pyramid, which forms one third of a cube . This formula cannot be proven without using such infinitesimal arguments--unlike the 2 - dimensional formulae for polyhedral area, though similar to the area of the circle--and hence admitted less rigorous proofs before the advent of calculus, with the ancient Greeks using the method of exhaustion . This is essentially the content of Hilbert's third problem--more precisely, not all polyhedral pyramids are scissors congruent (can be cut apart into finite pieces and rearranged into the other), and thus volume cannot be computed purely by using a decomposition argument . </P> <P> The center of mass of a conic solid of uniform density lies one - quarter of the way from the center of the base to the vertex, on the straight line joining the two . </P> <P> For a circular cone with radius r and height h, the base is a circle of area π r 2 (\ displaystyle \ pi r ^ (2)) and so the formula for volume becomes </P>

Where does the centre of mass of a cone lie