Cone (geometry)/Proofs

Volume

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

"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 $dlambda\; =\; lambda\_1\; -\; lambda\_0$ will be $h\; ,\; dlambda$. Thus, the differential volume of a slice is:$dV\; =\; (lambda^2\; b)\; (h\; ,\; dlambda)$

Now integrate the volume::$V\; =\; int\_0^1\; dV\; =\; int\_0^1\; b\; h\; lambda^2\; ,\; dlambda\; =\; b\; h\; left\; [\; \{1over\; 3\}\; lambda^3\; ight]\; \_0^1\; =\; \{1over\; 3\}\; b\; h,$"Q.E.D."

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 $M\; =\; ho\; V$ 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:$dM\; =\; ho\; ,\; dV\; =\; ho\; b\; h\; lambda^2\; ,\; dlambda$.

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:$vec\; c\_\{cone\}\; =\; \{1over\; M\}\; int\_0^1\; vec\; c\_S(lambda)\; ,\; dM$

:$qquad\; =\; \{1over\; M\}\; int\_0^1\; [(1\; -\; lambda)\; vec\; v\; +\; lambda\; vec\; c\_B]\; ho\; b\; h\; lambda^2\; ,\; dlambda$

:$qquad\; =\; \{\; ho\; b\; h\; over\; M\}\; int\_0^1\; [vec\; v\; lambda^2\; +\; (vec\; c\_B\; -\; vec\; v)\; lambda^3]\; ,\; dlambda$

:$qquad\; =\; \{\; ho\; b\; h\; over\; M\}\; left\; [\; vec\; v\; int\_0^1\; lambda^2\; ,\; dlambda\; +\; (vec\; c\_B\; -\; vec\; v)\; int\_0^1\; lambda^3\; ,\; dlambda\; ight]$

:$qquad\; =\; \{\; ho\; b\; h\; over\; \{1over\; 3\}\; ho\; b\; h\}\; left\; [\; \{1over\; 3\}\; vec\; v\; +\; \{1over\; 4\}\; (vec\; c\_B\; -\; vec\; v)\; ight]$

:$qquad\; =\; 3\; left(\; \{vec\; v\; over\; 12\}\; +\; \{vec\; c\_B\; over\; 4\}\; ight)$

: ∴ $vec\; c\_\{cone\}\; =\; \{vec\; v\; over\; 4\}\; +\; \{3over\; 4\}\; vec\; c\_B$,

which is to say, that $vec\; c\_\{cone\}$ lies one fourth of the way from $vec\; c\_B$ to $vec\; v$, "Q.E.D."

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:$\{1over\; 2\}\; b\; h$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

* 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.

$A\; =\; int\_0^\{2\; pi\; r\}\; frac\{1\}\{2\}\; s\; dC$

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

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

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

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