- Claim: The volume of a conic solid whose base has area b and whose height is h is .
Proof: Let be a simple planar loop in . Let be the vertex point, outside of the plane of .
Let the conic solid be parametrized by
For a fixed , the curve is planar. Why? Because if is planar, then since is just a magnification of , it is also planar, and is just a translation of , so it is planar.
Moreover, the shape of is similar to the shape of , and the area enclosed by is of the area enclosed by , which is b.
If the perpendiculars distance from the vertex to the plane of the base is h, then the distance between two slices and , separated by will be . Thus, the differential volume of a slice is
Now integrate the volume:
Center of Mass edit
- 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 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 , of fixed , has differential mass
Let us say that the base of the cone has center of mass . Then the slice at has center of mass
Thus, the center of mass of the cone should be
- ∴ ,
which is to say, that lies one fourth of the way from to .
Dimensional Comparison edit
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
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 edit
- Claim: The Surface Area of a right circular cone is equal to , where is the radius of the cone and is the slant height equal to
Proof: The refers to the area of the base of the cone, which is a circle of radius . The rest of the formula can be derived as follows.
Cut slices from the vertex of the cone to points evenly spread along its base. Using a large enough value for causes these slices to yield a number of triangles, each with a width and a height , which is the slant height.
The number of triangles multiplied by yields , the circumference of the circle. Integrate the area of each triangle, with respect to its base, , to obtain the lateral surface area of the cone, A.
Thus, the total surface area of the cone is equal to