Intuitive demonstration of the formula giving the surface of a sphere.

Question

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 :-)

Answer 1

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)

Answer 2

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.

Answer 3

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.