- #1

TranscendArcu

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## Homework Statement

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)

^{2}A, where 0 ≤ t ≤ h, and

(b) the volume of the cone is Ah/3.

## The Attempt at a Solution

Before 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}/h

^{2}* (t)

^{3}/3, evaluated from 0 to h. So I have pi*(radius of S)

^{2}/h

^{2}* h

^{3}= 1/3 * pi * (radius of S)

^{2}* h,

which is also what I wanted.

Look okay?