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?
3 Answers
$\begingroup$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 $${\rm area}(S^2_r)=2\pi\,r\int_{z=-r}^{z=r}\Delta z=4\pi\,r^2\ .$$
$\endgroup$ $\begingroup$The spherical volume element is $dV=(dr)(rd\theta)(rsin\theta d\phi)$, where the parentheses indicate the differentials for the various coordinates. For area, r is fixed, therefore $dA=(Rd\theta)(Rsin\theta d\phi)$. The integral is thus:$A=\int_{0}^{\pi}\int_{0}^{2\pi} R^2sin\theta d\phi d\theta = R^2\int_{0}^{\pi}sin\theta d\theta \int_{0}^{2\pi}d\phi = R^2 (-cos(\pi)+cos(0))(2\pi-0) = 4\pi R^2$
Note that $C=\int_{0}^{2\pi}R d\theta =2\pi R$ can't be used as the basis for this integral. Issues arise due to the differently defined ranges on $\theta$ in polar vs spherical coordinates, and also expanding to the third dimension requires integrating over $rsin\theta d\phi$ which itself has a dependence on $\theta$ (with different limits than used previously), thus going directly from circumference of a circle to surface are of a sphere is not possible without additional arguments to reintroduce the angular dependence and compensate for limit differences. The idea to use $rd\phi$ to integrate up by rotating the circle $\pi$ in the perpendicular direction creates a circle with another half circle perpendicular to it, not a sphere, as it only rotates the point at polar coordinates $\theta =0$ through the angle $\pi$ in the spherical coordinates $\hat\phi$ direction, not the entire circle.
$\endgroup$ $\begingroup$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 $r = C / (2 \pi)$,
$$A = 4 \pi r^2 = \frac{4 \pi C^2}{4 \pi^2} = \frac{C^2}{2 \pi}.$$
$\endgroup$ 4
