# 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.

• If you project the area of each triangle on a sphere with radius $r$ and height $2r$, you can show that the heigh of each triangle is $r$. That will give $2\pi r^2$ as the surface area of the hemisphere. Commented Dec 31, 2021 at 6:23
• @1123581321 did you mean a cylinder? Commented Dec 31, 2021 at 6:49
• Yes - projections from the sphere to a cylinder Commented Dec 31, 2021 at 6:49
• Duplicate of this question: math.stackexchange.com/questions/3031320/…. Though the answer given is not very convincing, in my opinion, and it would be nice to see others. Commented Dec 31, 2021 at 8:05

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.

• Yeah, the little tiny pieces must not be triangles, when unrolled. If they were, then this algorithm would give the correct answer. So, if they’re not triangles, what are they? Are they shapes whose area we can easily calculate? Commented Dec 31, 2021 at 8:15
• @bubba. Their sides are curved; they look like the region between $y=-\sin x$ and $y=+\sin x.$ Commented Dec 31, 2021 at 9:36
• I'm struggling to understand the same - why they are not triangles. What exactly are these pieces. Maybe a 3D animation might help. Commented Dec 31, 2021 at 16:27
• @bubba The issue is arguably more fundamental than the pieces not being triangles: The pieces cannot be unrolled (or flattened, or "developed") at all without introducing area distortion proportional to the area of the piece. This is what Parcly Taxel referred to in the question you linked, about the sphere "not being developable," and is why cutting the hemisphere into more pieces does not make the approximation better. :) Commented Dec 31, 2021 at 17:13
• @AndrewD.Hwang — A sphere is not developable, but these little pieces certainly are. I think the real problem is that they’re not triangles, as md2perpe said. I eventually convinced myself of this, too. Commented Jan 1, 2022 at 13:31

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, $$(2\pi r)(2r) = 4\pi r^{2}$$.

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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: $$\pi^2 R^2 \times 4/\pi = 4\pi R^2$$.