# 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 ${1 \over 3}bh$ .

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 $\lambda ,t\in [0,1]$ .

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:

$V=\int _{0}^{1}dV=\int _{0}^{1}bh\lambda ^{2}\,d\lambda =bh\left[{1 \over 3}\lambda ^{3}\right]_{0}^{1}={1 \over 3}bh,$ ## 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 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

$dM=\rho \,dV=\rho bh\lambda ^{2}\,d\lambda$ .

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

${\vec {c}}_{cone}={1 \over M}\int _{0}^{1}{\vec {c}}_{S}(\lambda )\,dM$ $\qquad ={1 \over M}\int _{0}^{1}[(1-\lambda ){\vec {v}}+\lambda {\vec {c}}_{B}]\rho bh\lambda ^{2}\,d\lambda$ $\qquad ={\rho bh \over M}\int _{0}^{1}[{\vec {v}}\lambda ^{2}+({\vec {c}}_{B}-{\vec {v}})\lambda ^{3}]\,d\lambda$ $\qquad ={\rho bh \over M}\left[{\vec {v}}\int _{0}^{1}\lambda ^{2}\,d\lambda +({\vec {c}}_{B}-{\vec {v}})\int _{0}^{1}\lambda ^{3}\,d\lambda \right]$ $\qquad ={\rho bh \over {1 \over 3}\rho bh}\left[{1 \over 3}{\vec {v}}+{1 \over 4}({\vec {c}}_{B}-{\vec {v}})\right]$ $\qquad =3\left({{\vec {v}} \over 12}+{{\vec {c}}_{B} \over 4}\right)$ ${\vec {c}}_{cone}={{\vec {v}} \over 4}+{3 \over 4}{\vec {c}}_{B}$ ,

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

## Dimensional 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

${1 \over 2}bh$ 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

• Claim: The Surface Area of a right circular cone is equal to $\pi rs+\pi r^{2}$ , 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.

$A=\int _{0}^{2\pi r}{\frac {1}{2}}sdC$ $A=\left[{\frac {1}{2}}sC\right]_{0}^{2\pi r}$ $A=\pi rs\!$ $A=\pi r{\sqrt {r^{2}+h^{2}}}$ Thus, the total surface area of the cone is equal to $\pi r^{2}+\pi r{\sqrt {r^{2}+h^{2}}}$ 