Famous Theorems of Mathematics/Geometry/Cones

• Claim: The volume of a conic solid whose base has area b and whose height is h is ${\displaystyle {1 \over 3}bh}$.

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

Let the conic solid be parametrized by

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

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

For a fixed ${\displaystyle \lambda =\lambda _{0}}$, the curve ${\displaystyle {\vec {\sigma }}(\lambda _{0},t)=(1-\lambda _{0}){\vec {v}}+\lambda _{0}\,{\vec {\alpha }}(t)}$ is planar. Why? Because if ${\displaystyle {\vec {\alpha }}(t)}$ is planar, then since ${\displaystyle \lambda _{0}\,{\vec {\alpha }}(t)}$ is just a magnification of ${\displaystyle {\vec {\alpha }}(t)}$, it is also planar, and ${\displaystyle (1-\lambda _{0}){\vec {v}}+\lambda _{0}\,{\vec {\alpha }}(t)}$ is just a translation of ${\displaystyle \lambda _{0}\,{\vec {\alpha }}(t)}$, so it is planar.

Moreover, the shape of ${\displaystyle {\vec {\sigma }}(\lambda _{0},t)}$ is similar to the shape of ${\displaystyle \alpha (t)}$, and the area enclosed by ${\displaystyle {\vec {\sigma }}(\lambda _{0},t)}$ is ${\displaystyle \lambda _{0}^{2}}$ of the area enclosed by ${\displaystyle {\vec {\alpha }}(t)}$, which is b.

If the perpendiculars distance from the vertex to the plane of the base is h, then the distance between two slices ${\displaystyle \lambda =\lambda _{0}}$ and ${\displaystyle \lambda =\lambda _{1}}$, separated by ${\displaystyle d\lambda =\lambda _{1}-\lambda _{0}}$ will be ${\displaystyle h\,d\lambda }$. Thus, the differential volume of a slice is

${\displaystyle dV=(\lambda ^{2}b)(h\,d\lambda )}$

Now integrate the volume:

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

• 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 ${\displaystyle M=\rho V}$ 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 ${\displaystyle {\vec {\sigma }}(\lambda _{0},t)}$, of fixed ${\displaystyle \lambda =\lambda _{0}}$, has differential mass

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

Let us say that the base of the cone has center of mass ${\displaystyle {\vec {c}}_{B}}$. Then the slice at ${\displaystyle \lambda =\lambda _{0}}$ has center of mass

${\displaystyle {\vec {c}}_{S}(\lambda _{0})=(1-\lambda _{0}){\vec {v}}+\lambda _{0}{\vec {c}}_{B}}$.

Thus, the center of mass of the cone should be

${\displaystyle {\vec {c}}_{cone}={1 \over M}\int _{0}^{1}{\vec {c}}_{S}(\lambda )\,dM}$

${\displaystyle \qquad ={1 \over M}\int _{0}^{1}[(1-\lambda ){\vec {v}}+\lambda {\vec {c}}_{B}]\rho bh\lambda ^{2}\,d\lambda }$

${\displaystyle \qquad ={\rho bh \over M}\int _{0}^{1}[{\vec {v}}\lambda ^{2}+({\vec {c}}_{B}-{\vec {v}})\lambda ^{3}]\,d\lambda }$

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

${\displaystyle \qquad ={\rho bh \over {1 \over 3}\rho bh}\left[{1 \over 3}{\vec {v}}+{1 \over 4}({\vec {c}}_{B}-{\vec {v}})\right]}$

${\displaystyle \qquad =3\left({{\vec {v}} \over 12}+{{\vec {c}}_{B} \over 4}\right)}$

∴ ${\displaystyle {\vec {c}}_{cone}={{\vec {v}} \over 4}+{3 \over 4}{\vec {c}}_{B}}$,

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

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

${\displaystyle {1 \over 2}bh}$

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.

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

Proof: The ${\displaystyle \pi r^{2}}$ refers to the area of the base of the cone, which is a circle of radius ${\displaystyle r}$. The rest of the formula can be derived as follows.

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

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

${\displaystyle A=\int _{0}^{2\pi r}{\frac {1}{2}}sdC}$

${\displaystyle A=\left[{\frac {1}{2}}sC\right]_{0}^{2\pi r}}$

${\displaystyle A=\pi rs\!}$

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

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