Deriving surface area of a sphere from the circumference

Question

given the circumference of a circle, which is 2πr, how many times do I have to add it to itself to cover a whole surface of a sphere and deriving 4πr^2?

Answer 1

It's a bit trickier than adding a whole bunch of circumferences to get the surface area. That requires integration, and requires you to consider the width (not length) of something like the circumference in some fashion to cover the sphere.

This is in some respects an expression for that. I say "in some respects" because it doesn't add up a bunch of circumferences, but a bunch of parallel slices (a bit like you were slicing a tomato for hamburgers):

$$A = R^2\int_{\theta = -\pi/2}^{\pi/2} \int_{\phi = 0}^{2\pi} \cos \theta d \theta d \phi = 4\pi R^2.$$

If you need the surface area in terms of that circumference, then that's a bit easier. Since $r = C / (2 \pi)$,

$$A = 4 \pi r^2 = \frac{4 \pi C^2}{4 \pi^2} = \frac{C^2}{2 \pi}.$$

Answer 2

Denote by $\theta\in[-\pi/2,\pi/2]$ the geographical latitude on this sphere. Then $z(\theta)=r\sin\theta$, and the radius $\rho$ of the latitude circle at latitude $\theta$ is given by $\rho(\theta)=r\cos\theta$.

Consider now an infinitesimal latitude zone $Z:\ [\theta,\theta+\Delta\theta]$ on this sphere. Its area is given by $${\rm area}(Z)\doteq2\pi\rho(\theta)\,(r\,\Delta\theta)=2\pi r^2\cos\theta\,\Delta\theta\ .\tag{1}$$ On the other hand the $z$-coordinates of the two boundary circles differ by $$ \Delta z:=z(\theta+\Delta\theta)-z(\theta)\doteq r\, \cos\theta \>\Delta\theta\ .\tag{2}$$ Combining $(1)$ and $(2)$ we see that $${\rm area}(Z)\doteq2\pi\,r\>\Delta z\ ,$$ so that the total area of the sphere comes to $${\rm area}(S^2_r)=2\pi\,r\int_{z=-r}^{z=r}\Delta z=4\pi\,r^2\ .$$