A solid cone is obtained by connecting every point of a plane region S with a vertex not in the plane of S. Let A denote the area of S, and let h denote the altitude of the cone. Prove that:
(a) The cross-sectional area cut by a plane parallel to the base and at a distance t from the vertex is (t/h)2A, where 0 ≤ t ≤ h, and
(b) the volume of the cone is Ah/3.
The Attempt at a SolutionBefore I attempt this, I'd just like to know whether or not it is reasonable to assume that S is circular. This would make the A equal to pi*(radius of S)2. I think
the limit as t goes to h of (t/h) is 1. Then, the change in radius of the circle is given by (radius of S)(t/h). This means that the area of the circle given the change in t is
pi*(radius of S)2(t/h)2, which is what I wanted to show. (I find this to be the most unconvincing part of my post)
In order to find the volume, I integrate. I have
∫(0≤t≤h) (pi*(radius of S)2(t/h)2)dt
(pi*(radius of S)2/h2 * (t)3/3, evaluated from 0 to h. So I have pi*(radius of S)2/h2 * h3 = 1/3 * pi * (radius of S)2 * h,
which is also what I wanted.