# Famous Theorems of Mathematics/Geometry/Cones

## VolumeEdit

• 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 MassEdit

• 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 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 $\lambda = \lambda_0$, 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

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

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

## Dimensional ComparisonEdit

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 AreaEdit

• 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 $\sqrt{r^2+h^2}$

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.

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