Exposition:
In two dimensions, there is a (are many) straightforward explanation(s) of the fact that the perimeter (i.e. circumference) and area of a circle relate to the radius by $2\pi r$ and $\pi r^2$ respectively. One argument proceeds by approximating these quantities using regular cyclic polygons (equilateral, equiangular, on the circle of radius $r$), noting that such a polygon with $n$ sides can be decomposed into $n$ isosceles triangles with peak angle $\frac{2\pi}{n}$, base length $~2r\sin\frac{\pi}{n}$, and altitude $~r \cos \frac{\pi}{n}$ . Then, associating the circle with the limiting such polygon, we have, $$ P = \lim_{n\to\infty} n \cdot \text{base length } = \lim_{n\to\infty}2r \cdot \pi \frac{n}{\pi} \sin \frac{\pi}{n} = 2\pi r ~~, $$ and similarly, (via trig identity) $$ A = \lim_{n\to\infty} n\left(\frac{1}{2} \text{ base } \times \text{ altitude }\right) = \lim_{n\to\infty}\frac{r^2\cdot 2\pi}{2} \frac{n}{2\pi} \sin \frac{2\pi}{n} = \pi r^2 ~~. $$ Question:
Could someone offer intuition, formulas, and/or solutions for performing a similarly flavored construction for the surface area and volume of a sphere?
Images and the spatial reasoning involved are crucial here, as there are only so many platonic solids, so I am not seeing immediately the pattern in which the tetrahedra (analogous to the 2D triangles) will be arranged for arbitrarily large numbers of faces. Thus far my best result has been a mostly-rigorous construction relying on this formula (I can write up this proof on request). What I'd like to get out of this is a better understanding of how the solid angle of a vertex in a polyhedron relates to the edge-edge and dihedral angles involved, and perhaps a "dimension-free" notion for the ideas used in this problem to eliminate the need to translate between solid (2 degrees of freedom) and planar (1 degree) angles.