Deriving surface area of a sphere using triangles I was trying to derive surface area of a sphere myself.
I started with a hemi-sphere, sliced it into infinite triangles and then added the area of all the triangles and finally doubled it to get area of a sphere.
I'm getting π²R² as final result. I could not understand what's wrong with my logic.


 A: The problem is that your pieces are not triangles and don't "approach triangles in the limit" in the necessary way. For example, the proportion of shaded area above height $\frac h2$ in your figure doesn't approach $\frac14$ of the total shaded area as you cut the hemisphere into smaller pieces; it remains somewhat less. Your approach actually computes the surface area of a cone.
Another way to see that your pieces are not triangles: the angle between the base and the sides is always 90°. This means the sides are curved, and each piece is fatter at the bottom than a triangle. Cutting it into thinner pieces doesn't change this; the pieces always have more area than the corresponding triangle, by a fixed ratio that does not approach 1.
A: As noted in the comments and Karl's (+1) answer, the issue is in flattening the "gores," which does not preserve area.
One correct reckoning is instead to use a quasi-triangle with two right angles at the base, angle $d\theta$ at the apex, and width $r\sin\varphi\, r\, d\theta$ at height $\varphi$ (below, left), as noted by md2perpe and bubba in the comments. The area of the flat region is correct,
$$
\int_{0}^{\pi/2} r\sin\varphi\, d\varphi\, (r\, d\theta) = r^{2}\, d\theta,
$$
though this region does not wrap without distortion (i.e., not isometrically) onto a gore, but only in an area-preserving way.
A related correct accounting (below, right, which shows a unit sphere; for radius $r$, multiply all lengths by $r$) slices the quasi-triangle by latitudes. The portion of a sphere between longitude $\theta$ and $\theta + d\theta$, and between colatitude $\varphi$ and $\varphi + d\varphi$, is an infinitesimal trapezoid. Project axially (away from the axis through the poles) onto a circumscribed cylinder. Up to an error infinitesimal compared to $d\theta\, d\varphi$, the trapezoid is a rectangle with width $\sin\varphi\, d\theta$ and height (measured along the sphere) $d\varphi$. The image of this quadrilateral under axial projection is a curved (but developable) rectangle of width $d\theta$ and height (along the cylinder) $\sin\varphi\, d\varphi$. Consequently, axial projection from a sphere to a circumscribed cylinder is area-preserving. (!!) Particularly, the area of a sphere is the area of a circumscribed cylinder, $(2\pi r)(2r) = 4\pi r^{2}$.
$\rule{1in}{0pt}$

A: After suitable scaling, the sides of the pseudo-triangle shown in the answer from Andrew Hwang are the graphs of $\cos(x)$ and $-\cos(x)$ curves, over the range $0$ to $\tfrac12 \pi$. By integrating the cosine function over this range, it's easy to show that the pseudo-triangle has $4/\pi$ times the area of the corresponding triangle.
The OP got the answer $\pi^2 R^2$ by adding up the areas of the triangles. If you multiply this by the area ratio $4/\pi$, you get the correct answer: $\pi^2 R^2 \times 4/\pi = 4\pi R^2$.
