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).
 A: 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 $\frac{4}{3}\pi r^3$.
A: In the first step consider a hemi-sphere radius $R$ and its enveloping cylinder at the equator of sphere of length/ height $R$, $\phi=p=$ slope and $dz$ is differential length along axis as shown.

$$ d(Area)= 2\pi r dl,\; \text{since}(\cos \phi = r/R  = dz/dl) , d (Area) = 2 \pi R dz\;; $$
Integrating for an axial length for sphere
$$\Delta Area= 2 \pi R \Delta z $$
Next step consider the cylinder from which this segment projects between same parallels. Directly its prismatic area is
$$\Delta Area =   2 \pi R \cdot \Delta z $$
The difference areas $abfe$ and $dcfe$ are equal.
For the full sphere/tangent cylinder accordingly the same area.
It was became known from Archimedes's time , he last-wished it to be inscribed on his tombstone when attacked by soldiers.
A: 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 $4\pi r^2$.
