Relationship between areas of circles and spheres How can the surface area of a sphere be exactly 4 times the area of it's cross-sectional area? This seems too clean; why 4?
If I sliced 4 thin circles from the middle of a sphere and stuck them together (and was able to morph them)  would I end up with another hollow sphere? 
 A: Here the $4\pi$ should be viewed not as $4$ times $\pi$ but rather as $2$ times $2\pi$ and then there is a reasonably nice geometric explanation.  Namely, if you project the sphere to, say, the $z$-axis, then the inverse image of each small subsegment of $[-1,1]$ will have area $2\pi h$ where $h$ is the length of the subsegment. You see this particularly clearly in the case of a small subsegment around $0$ in which case you get a thin strip which is roughly the product of the equator (of length $2\pi$) by a segment of length $h$. Therefore the total area is $2\times 2\pi=4\pi$.
Four unit disks do have the same area as the unit sphere, but you can't patch them together without distorting them so as to form the sphere. One way of seeing it is in terms of a metric invariant called Gaussian curvature (this invariant vanishes for the disk but is equal to 1 for the sphere).
A: If you consider a cross-section of the sphere at a height $h$ above the equatorial plane, you get a circle of radius $\sqrt{r^2-h^2}$, and circumference $2\pi\sqrt{r^2-h^2}$.
Then you can be tempted to conclude that the area of the sphere is the average perimeter from $-r$ to $r$, which would yield $\pi^2r^2$.
But you must take into account the fact that the surface is not vertical but oblique and there is a correction factor equal to the secant of the angle of the normal, or $\dfrac r{\sqrt{r^2-h^2}}$.
So after simplification you are just integrating $2\pi r$ from $-r$ to $r$, which explains the factor $4$.
A geometric interpretation is by saying that any slice of small height $\delta h$ in a sphere has the same area (!)
