Why do 4 circles cover the surface of a sphere?

Is there a geometric explanation for why a sphere has surface area $4 \pi r^2$ ?
Ie equal to 4 times its cross-section (a circle of radius r).

Solution 1:

Let $Z$ be a cylinder of height $2r$ touching the sphere $S_r$ along the equator $\theta=0$. Consider now a thin plate orthogonal to the $z$-axis having a thickness $\Delta z\ll r$. It intersects $S_r$ at a certain geographical latitude $\theta$ in a nonplanar annulus of radius $\rho= r\cos\theta$ and width $\Delta s=\Delta z/\cos\theta$, and it intersects $Z$ in a cylinder of height $\Delta z$. Both these "annuli" have the same area $2\pi r \Delta z$. As this is true for any such plate it follows that the total area of the sphere $S_r$ is the same as the total area of $Z$, namely $4\pi r^2$.

Solution 2:

One geometric explanation is that $4\pi r^2$ is the derivative of $\frac{4}{3}\pi r^3$, the volume of the ball with radius $r$, with respect to $r$. This is because if you enlarge $r$ a little bit, the volume of the ball will change by its surface times the small enlargement of $r$.

So why is the volume of the full ball $\frac{4}{3}\pi r^3$? By slicing the ball into disks, using Pythagoras, you get that its volume is $$ \int_{-r}^r \pi (r^2-x^2)\mathrm{d}x $$ which is indeed $\frac{4}{3}\pi r^3$.

Solution 3:

In the first step consider a hemi-sphere radius $R$ and its enveloping cylinder at the equator of sphere of length $R$.

$ \cos \phi = r/R = dz/dl $

$ d Area = 2 \pi r dl $

This results in same area

$$ 2 \pi R^2 $$

so a finite slice width has the same area either for a cone or cylinder of same width $\Delta z $.

Now consider the second step. it is comfortable to calculate flat areas, use the well known method of area calculation:

Cut the free boundary into an infinity of divisions of triangles, cut and discard every alternate infinitesimal narrow triangle,this will again cut into half its area, now to:

$$ 2 \pi R^2. $$

EDIT1

The area of a sphere segment is seen from integration as $ 2 \pi R $ times axial length of the spherical segment, which is the diameter for the full sphere.