Cone (geometry) proofs

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This mathematics article is devoted entirely to providing mathematical proofs and support for claims and statements made in the article Cone (geometry). This article is currently an experimental vehicle to see how we might be able to provide proofs and details for math articles without cluttering up the main article itself. See Wikipedia:WikiProject Mathematics/Proofs for current discussion. This article is "experimental" in that it is a proposal for one way that we might be able to deal with expressing proofs.

[edit] Volume

  • Claim: The volume of a conic solid whose base has area b and whose height is h is {1\over 3} b h.

Proof: Let \vec \alpha (t) be a simple planar loop in \mathbb{R}^3. Let \vec v be the vertex point, outside of the plane of \vec \alpha.

Let the conic solid be parametrized by

\vec \sigma (\lambda, t) = (1 - \lambda) \vec v + \lambda \, \vec \alpha (t)

where \lambda, t \isin [0, 1].

For a fixed λ = λ0, the curve \vec \sigma (\lambda_0, t) = (1 - \lambda_0) \vec v + \lambda_0 \, \vec \alpha (t) is planar. Why? Because if \vec \alpha(t) is planar, then since \lambda_0 \, \vec \alpha(t) is just a magnification of \vec \alpha(t), it is also planar, and (1 - \lambda_0) \vec v + \lambda_0 \, \vec \alpha(t) is just a translation of \lambda_0 \, \vec \alpha(t), so it is planar.

Moreover, the shape of \vec \sigma (\lambda_0, t) is similar to the shape of α(t), and the area enclosed by \vec \sigma(\lambda_0, t) is \lambda_0^2 of the area enclosed by \vec \alpha(t), which is b.

If the perpendicular distance from the vertex to the plane of the base is h, then the distance between two slices λ = λ0 and λ = λ1, separated by dλ = λ1 − λ0 will be h \, d\lambda. Thus, the differential volume of a slice is

dV = (\lambda^2 b) (h \, d\lambda)

Now integrate the volume:

V = \int_0^1 dV = \int_0^1 b h \lambda^2 \, d\lambda = b h \left[ {1\over 3} \lambda^3 \right]_0^1 = {1\over 3} b h,

Q.E.D.

[edit] Center of mass

  • Claim: the center of mass of a conic solid lies at one-fourth of the way from the center of mass of the base to the vertex.

Proof: Let M = ρV be the total mass of the conic solid where ρ is the uniform density and V is the volume (as given above).

A differential slice enclosed by the curve \vec \sigma(\lambda_0, t), of fixed λ = λ0, has differential mass

dM = \rho \, dV = \rho b h \lambda^2 \, d\lambda.

Let us say that the base of the cone has center of mass \vec c_B. Then the slice at λ = λ0 has center of mass

\vec c_S(\lambda_0) = (1 - \lambda_0) \vec v + \lambda_0 \vec c_B.

Thus, the center of mass of the cone should be

\vec c_{cone} = {1\over M} \int_0^1 \vec c_S(\lambda) \, dM
\qquad = {1\over M} \int_0^1 [(1 - \lambda) \vec v + \lambda \vec c_B] \rho b h \lambda^2 \, d\lambda
\qquad = {\rho b h \over M} \int_0^1 [\vec v \lambda^2 + (\vec c_B - \vec v) \lambda^3] \, d\lambda
\qquad = {\rho b h \over M} \left[ \vec v \int_0^1 \lambda^2 \, d\lambda + (\vec c_B - \vec v) \int_0^1 \lambda^3 \, d\lambda \right]
\qquad = {\rho b h \over {1\over 3} \rho b h} \left[ {1\over 3} \vec v + {1\over 4} (\vec c_B - \vec v) \right]
\qquad = 3 \left( {\vec v \over 12} + {\vec c_B \over 4}\right)
∴ \vec c_{cone} = {\vec v \over 4} + {3\over 4} \vec c_B,

which is to say, that \vec c_{cone} lies one fourth of the way from \vec c_B to \vec v, Q.E.D.

[edit] Dimensional comparison

Note that the cone is, in a sense, a higher-dimensional version of a triangle, and that for the case of the triangle, the area is

{1\over 2} b h

and the centroid lies 1/3 of the way from the center of mass of the base to the vertex.

A tetrahedron is a special type of cone, and it is also a stricter generalization of the triangle.

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