circlesgeometryspherestrigonometry
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?
In this answer, it is shown geometrically that the area of a strip of sphere between two parallels of latitude is the same as the area of the orthogonal projection of that strip onto the cylinder with the same radius as the sphere and whose axis is parallel to the north-south axis of the sphere.
Finding the area of this cylinder is just multiplying its height, $2r$, by its circumference, $2\pi r$. Thus, the area of the sphere is $4\pi r^2$.
Your son's procedure seems to be summing the circumferences of the green strips on the sphere, but there must be some accounting for the widths of these strips. This seems to be the problem he is encountering.
The centroid of semi-circle should be $\dfrac{4r}{3\pi}$ from the origin.
So the volume is
$$V=2\pi \times \frac{4r}{3\pi}\times \frac{\pi r^2}{2}=\frac{4}{3}\pi r^3$$
The centroid of semi-circular arc should be $\dfrac{2r}{\pi}$ from the origin.
So the surface area is
$$S=2\pi \times \frac{2r}{\pi}\times \pi r=4\pi r^2$$
Best Answer
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).