Famous Theorems of Mathematics/Geometry/Cones

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 $$ \lambda = \lambda_0 $$, 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 $$ \alpha(t) $$, 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 $$\lambda = \lambda_0 $$ and $$\lambda = \lambda_1 $$, separated by $$ d\lambda = \lambda_1 - \lambda_0 $$ will be $$ h \, d\lambda $$. Thus, the differential volume of a slice is
 * $$ dV = (\lambda^2 b) (h \, d\lambda) $$

Now integrate the volume:
 * $$ V = \int_0^1 dV = \int_0^1 b h \lambda^2 \, d\lambda = b h \left[ {1\over 3} \lambda^3 \right]_0^1 = {1\over 3} b h, $$

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 $$ M = \rho V $$ be the total mass of the conic solid where &rho; 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 $$\lambda = \lambda_0$$, 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 $$\lambda = \lambda_0 $$ 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
 * $$ \vec c_{cone} = {1\over M} \int_0^1 \vec c_S(\lambda) \, dM $$


 * $$ \qquad = {1\over M} \int_0^1 [(1 - \lambda) \vec v + \lambda \vec c_B] \rho b h \lambda^2 \, d\lambda $$


 * $$ \qquad = {\rho b h \over M} \int_0^1 [\vec v \lambda^2 + (\vec c_B - \vec v) \lambda^3] \, d\lambda $$


 * $$ \qquad = {\rho b h \over M} \left[ \vec v \int_0^1 \lambda^2 \, d\lambda + (\vec c_B - \vec v) \int_0^1 \lambda^3 \, d\lambda \right] $$


 * $$ \qquad = {\rho b h \over {1\over 3} \rho b h} \left[ {1\over 3} \vec v + {1\over 4} (\vec c_B - \vec v) \right] $$


 * $$ \qquad = 3 \left( {\vec v \over 12} + {\vec c_B \over 4}\right) $$


 * &there4; $$ \vec c_{cone} = {\vec v \over 4} + {3\over 4} \vec c_B $$,

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

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.

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

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 $$dC$$ and a height $$s$$, which is the slant height.

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

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

$$ A = \left[ \frac{1}{2} s C \right]_0^{2 \pi r}$$

$$ A = \pi r s\! $$

$$ A = \pi r \sqrt{r^2 + h^2}$$

Thus, the total surface area of the cone is equal to $$ \pi r^2 + \pi r \sqrt{r^2 + h^2} $$