Cone (geometry)/Proofs

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Main article: Cone (geometry)

1 Volume
2 Center of mass
3 Dimensional comparison
4 Surface Area

[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 $λ = λ 0$ , 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 perpendiculars distance from the vertex to the plane of the base is h, then the distance between two slices $λ = λ 0$ and $λ = λ 1$ , separated by $d λ = λ 1 - λ 0$ 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 $M = ρ V$ 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 $λ = λ 0$ , 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 $λ = λ 0$ 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.

[edit] Surface Area

Claim: The Surface Area of a right circular cone is equal to $π r s + π r 2$ , where $r$ is the radius of the cone and $s$ is the slant height equal to $\sqrt{r^2+h^2}$

Proof: The $π r 2$ refers to the area of the base of the cone, which is a circle of radius $r$ . The rest of the formula can be derived as follows.

Cut $n$ slices from the vertex of the cone to points evenly spread along its base. Using a large enough value for $n$ causes these slices to yield a number of triangles, each with a width $d C$ and a height $s$ , which is the slant height.

The number of triangles multiplied by $d C$ yields $C = 2π r$ , the circumference of the circle. Integrate the area of each triangle, with respect to its base, $d C$ , to obtain the lateral surface area of the cone, A.

$A = \int_0^{2 \pi r} \frac{1}{2} s dC$