Famous Theorems of Mathematics/Geometry/Cones

< Famous Theorems of Mathematics‎ | Geometry



  • 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}