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 $\frac{1}{3}bh$.

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

Let the conic solid be parametrized by

$\overrightarrow{\sigma }(\lambda ,t)=(1-\lambda )\overrightarrow{v}+\lambda \overrightarrow{\alpha }(t)$

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

For a fixed $\lambda ={\lambda }_0$, the curve $\overrightarrow{\sigma }({\lambda }_0,t)=(1-{\lambda }_0)\overrightarrow{v}+{\lambda }_0\overrightarrow{\alpha }(t)$ is planar. Why? Because if $\overrightarrow{\alpha }(t)$ is planar, then since ${\lambda }_0\overrightarrow{\alpha }(t)$ is just a magnification of $\overrightarrow{\alpha }(t)$, it is also planar, and $(1-{\lambda }_0)\overrightarrow{v}+{\lambda }_0\overrightarrow{\alpha }(t)$ is just a translation of ${\lambda }_0\overrightarrow{\alpha }(t)$, so it is planar.

Moreover, the shape of $\overrightarrow{\sigma }({\lambda }_0,t)$ is similar to the shape of $\alpha (t)$, and the area enclosed by $\overrightarrow{\sigma }({\lambda }_0,t)$ is ${\lambda }_0^2$ of the area enclosed by $\overrightarrow{\alpha }(t)$, 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 $hd\lambda$. Thus, the differential volume of a slice is

$dV=({\lambda }^{2}b)(hd\lambda )$

Now integrate the volume:

$V=\underset{0}{\overset{1}{\int }}dV=\underset{0}{\overset{1}{\int }}bh{\lambda }^{2}d\lambda =bh{\left[\frac{1}{3}{\lambda }^3\right]}_0^1=\frac{1}{3}bh,$

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 $\overrightarrow{\sigma }({\lambda }_0,t)$, 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 ${\overrightarrow{c}}_B$. Then the slice at $\lambda ={\lambda }_0$ has center of mass

${\overrightarrow{c}}_S({\lambda }_0)=(1-{\lambda }_0)\overrightarrow{v}+{\lambda }_0{\overrightarrow{c}}_B$.

Thus, the center of mass of the cone should be

${\overrightarrow{c}}_{cone}=\frac{1}{M}\underset{0}{\overset{1}{\int }}{\overrightarrow{c}}_S(\lambda )dM$

$=\frac{1}{M}\underset{0}{\overset{1}{\int }}[(1-\lambda )\overrightarrow{v}+\lambda {\overrightarrow{c}}_B]\rho bh{\lambda }^{2}d\lambda$

$=\frac{\rho bh}{M}\underset{0}{\overset{1}{\int }}[\overrightarrow{v}{\lambda }^2+({\overrightarrow{c}}_B-\overrightarrow{v}){\lambda }^3]d\lambda$

$=\frac{\rho bh}{M}[\overrightarrow{v}\underset{0}{\overset{1}{\int }}{\lambda }^{2}d\lambda +({\overrightarrow{c}}_B-\overrightarrow{v})\underset{0}{\overset{1}{\int }}{\lambda }^{3}d\lambda ]$

$=\frac{\rho bh}{\frac{1}{3}\rho bh}[\frac{1}{3}\overrightarrow{v}+\frac{1}{4}({\overrightarrow{c}}_B-\overrightarrow{v})]$

$=3(\frac{\overrightarrow{v}}{12}+\frac{{\overrightarrow{c}}_B}{4})$

∴ ${\overrightarrow{c}}_{cone}=\frac{\overrightarrow{v}}{4}+\frac{3}{4}{\overrightarrow{c}}_B$,

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

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

$\frac{1}{2}bh$

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.

$A=\underset{0}{\overset{2\pi r}{\int }}\frac{1}{2}sdC$

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

$A=\pi rs$

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

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