Visualization of surface area of a sphere

One way to proceed is to make use of the well-known (well, it should be well-known) property of a sphere: If you inscribe a unit sphere within a right cylinder, and slice them "horizontally" (i.e., perpendicular to the axis of the cylinder) the corresponding strips of the sphere and of the cylinder have equal areas.

That this is true can be seen by examining the strips in the limit. Each strip of the sphere has smaller radius than the corresponding strip of the cylinder, by an amount equal to the cosine of the "latitude" of the strip, but by the same token, the sphere's strip is wider than the corresponding strip of the cylinder, by an amount equal to the secant of the latitude. The two factors cancel each other out.

Since the entire cylinder, neglecting the ends (which don't correspond to any portion of the sphere), has height $2$ and circumference $2\pi$, its area—and therefore the area of the sphere—is $2 \times 2\pi = 4\pi$.

I think the idea of trying to do this without calculus is misguided. Instead, try to understand the steps in the calculus. The surface area formula is derived from the volume formula so maybe the question should be: can I get the volume formula from the formula for circle area $A=\pi r^2$? Consider a pile of disks, with the bottom one of diameter $2r$ and the top one of diameter just more than zero. Then we have (using the cylinder volume formula $V=\pi R^2h$): $$V=2\times\int_{\theta=0}^{\theta=\pi/2}\pi (r\cos(\theta))^2\times r\cos(\theta)\tan(\theta_2)$$ The $R$ and $h$ equivalents are the adjacent and opposite sides respectively of the implicit right angle triangles, which move up through the sphere $(\theta_2$ is the angle within those triangles, which is infinitesimal, as is $\tan(\theta_2))^1$. So: $$V=2\pi r^3\int_{\theta=0}^{\theta=\pi/2}\cos^3(\theta) d\theta$$ The remaining integral evaluates to $2/3$, so: $$V=\frac{4}{3}\pi r^3$$

$^1$More intuitively, since $\tan$ is opposite over adjacent, what this means is that the ratio of disk height to disk radius must be as low as possible in order to get the result for a smooth sphere.

This is definitely not an answer, but if you intend to use the "four parts of an orange peel" analogy, this might be relevant.

This might be entertaining for some, especially those who enjoy confusing people with numerical coincidences (like Randall Munroe's XKCD comics 217 and 1047).

I was thinking about the regular simplex in 3D -- the regular tetrahedron.

If you put a regular tetrahedron inside the unit sphere, i.e with the four vertices of the tetrahedron on the unit sphere, the six edges of the tetrahedron are $\sqrt{8/3}$. Each of the four faces is an equilateral triangle (each side being, of course, said $\sqrt{8/3}$). The area of each triangular face is therefore $\sqrt{4/3} = 2/\sqrt{3}$.

The area of the unit sphere is $4\pi$, so the area of each triangular quadrant on the unit sphere -- as if the tetrahedron was puffed up into a sphere -- is $\pi$.

(The edges of each puffed-up face is a great circle.)

The ratio between the area of the unit sphere quadrant and the tetrahedral faces is $\pi:2/\sqrt{3} \approx 2.72070$, which is within one thousandth of $e \approx 2.71828$.
(With three significant digits, both round to $2.72$.)

This is a coincidence; the ratios in lower and higher number of dimensions is quite different.