Intuitive demonstration of the formula giving the surface of a sphere. I've been looking for a way to intuitively find the formula to calculate the surface of a sphere, so here is my process of thinking:


*

*In 2D: you imagine a segment of length R that rotates around an axis which corresponds to the center of the circle (Ω). The total angle formed by the revolution is 2π, so your formula is 2πR.

*In 3D: by analogy with 2D you imagine a circle on a plane, and you apply a similar rotation centered on Ω also directed by your radius of length R. And you get a sphere! The total angle formed by the revolution is π so by this logic the formula for the surface of the sphere you would get would be 2π²R²...
But when I searched on the net about the formula I found that it was 4πR², and all the demonstrations explaining the formula use integrals (which I only know how to use on a plane). 
So I would like to know where I'm wrong in my reasoning and how to demonstrate (not rigorously) this formula using simple geometry tools and a bit of imagination.
Thank you for your help :-)
 A: Roughly speaking, your circle calculation works because
$$ \mathrm{d}l = r \mathrm{d} \theta $$
so length accumulates at a constant rate proportional to angle. However, for the sphere,
$$ \mathrm{d}A = r^2 \cos(\varphi) \mathrm{d} \theta \mathrm{d} \varphi $$
(where I use $\theta$ for the angle around the $z$ axis, and $\varphi$ for the angle above the xy plane)
Roughly speaking, the two things going right are:


*

*$\theta$ and $\varphi$ vary in perpendicular directions, so an "area element" is simply the product of the two "length elements"

*$\varphi$ traces out lengths at a constant rate


but something also goes wrong:


*

*$\theta$ does not trace out length at a constant rate


In particular, at high latitudes, increasing $\theta$ traces out a small distance, but near the equator, increasing $\theta$ traces out a large distance.
Put differently, the East-West distance between longitude lines is much greater when you're near the equator than when you're near the poles.
(also, you made another mistake: you only need to rotate a circle through an angle of $\pi$ to obtain an entire sphere)
A: Your analogy fails because different parts of the circumference of the circle you rotate are different distance from the rotation axis.  That means the area they sweep out is different.
A: To correct the example, instead of thinking of little circles, think of little Lunes which have area $2r^2\theta$.

Summing over the total $2\pi$, you get a total area of $4\pi r^2$.
The way to think about the area of a small spherical lune is:


*

*Imaging the top half, the width at the equator is $r \theta$

*The width at angle $\phi$ above the equator will be $r \theta cos(\phi)$.

*The total area of the top half will be $\int_{0}^{r\pi/2}r\theta cos(\frac{\phi}{r}) d\phi$ = $r \theta \big[r sin(\frac{\phi}{r}) \big]_{0}^{\frac{r \pi}{2}} $ = $r^2\theta$. That is, the area is the integral over the height of the half lune of the width of the half lune.


If you can imagine, I have peeled off a lune, and assumed it was small enough to be flattened, then it's area is twice this integral, $2r^2\theta$
This is disappointingly unintuitive, sorry, I'll think about removing the need for an integral.
