I remember the example that explains how to derive the area of a circle from several years ago, but there is also an analogue for the surface area of a sphere p

I remember the example that explains how to derive the area of a circle from several years ago, but there is also an analogue for the surface area of a sphere provided you know the volume... The volume is 4/3*pi*radius^3 the surface can be broken up into lots of spherical triangles. The vertices of each of these triangles is joined to the sphere's centre to form a load of tetrahedrons. As the number of these triangles increases and their size decreases without limit, their total volume will asymptotically approach the volume of the sphere. The volume of a tetrahedron is 1/3*base*height. sum(tetrahedron_volume)=volume_of_sphere=4/3*pi*radius^3 The tetrahedrons are flattened to end up as triangular columns with equal height (1/3 of the sphere's radius). The resulting shape should be a column with height 1/3 the radius of the sphere whose cross-sectional area is equal to the sphere's surface area. The height is 1/3*r ... (4/3*pi*radius^3)/(1/3*r)=4*pi*radius^2 EDIT: bloomin' asterisks, I should have remembered.