ON THE SPHERE AND CYLINDER OF ARCHIMEDES

surface of the cylinder with the plane, and by the lines joining their endpoints. Let there be a right cylinder whose bases are the circles $αβ$ and opposite them the circles $γ, δ$. And let $αγ$ and $βδ$ be joined. I say that the cylindrical surface cut off by the straight lines $αγ$ and $βδ$ is greater than the parallelogram $αγ, βδ$. For let each of the arcs $αβ$ and $γδ$ be bisected at points $ε, ζ$. And let $αε, εβ, γζ, ζδ$ be joined. Since $αε, εβ$ are greater than $αβ$, and the height of the parallelograms on them is vertical, the parallelograms whose bases are $αε, εβ$ and whose height is the same as that of the cylinder will be greater than the parallelogram $αβ, γδ$. Therefore, it is greater by some amount; let this be the area $κ$. The area $κ$ is either smaller than the plane segments $αε, εβ, γζ, ζδ$, or not smaller. Let it first be not smaller. And since the cylindrical surface cut off by the straight lines $αγ, βδ$, and the segments $αεβ, γζδ$, has as its boundary the plane of the parallelogram $αγ, βδ$, but also the composite surface—from the parallelograms whose bases are $αε, εβ$ and whose height is the same as the cylinder, and the triangles $αεβ, γζδ$—has as its boundary the plane of the parallelogram $αβγδ$. And one contains the other, and both are concave on the same sides. The cylindrical surface cut off by the straight lines $αγ, βδ$, and the plane segments $αεβ, γζδ$, must be greater than the composite surface formed by the parallelograms whose bases are $αε, εβ$ and whose height is the same as the cylinder, and the triangles $αεβ, γζδ$. Let the triangles $αεβ, γζδ$ be removed as common parts. The remaining cylindrical surface cut off by the straight lines $αγ, βδ$, and the plane segments $αεβ, γζδ$, must be greater than the composite surface formed by the parallelograms whose bases are $αε, εβ$ and whose height is the same as the cylinder. But the parallelograms whose base is $αε, εβ$ and whose height is the same as the cylinder are equal to the parallelogram $αγβδ$ and the area $κ$. Therefore, the remaining cylindrical surface cut off by the straight lines $αγ, βδ$ is greater than the parallelogram $αγβδ$. But let the area $κ$ be smaller than the plane segments $αε, εβ, γζ, ζδ$. And let each of the arcs $αε, εβ, γζ, ζδ$ be bisected at points $θ, η, κ, π, λ, μ$. And let $αθ, θε, εη, ηβ, γκ, κζ, ζλ, λδ$ be joined. From the plane segments $αε, εβ, γζ, ζδ$, there are removed no less than half, namely the triangles $αθε, εηβ, γκζ, ζλδ$. As this proceeds, certain segments will be captured that are smaller than the area $κ$. Let these be left, and let them be $αθ, θε, εη, ηβ, γλ, λζ, ζμ, μδ$. We will say similarly that the parallelograms whose base is $αθ, θε, εη, ηβ$ and whose height is the same as the cylinder will be greater than the parallelograms whose base is $αε, εβ$ and whose height is the same as the cylinder. And since the cylindrical surface cut off by the straight lines $αγ, βδ$, and the plane segments $αεβ, γζδ$, has as its boundary the plane of the parallelogram $αγβδ$, but also the composite surface from the parallelograms whose base is $αθ, θε, εη, ηβ$ and whose height is the same as the cylinder, and the rectilinear segments $αθεηβ, γλζμδ$. Let the rectilinear segments $αθεηβ, γλζμδ$ be removed as common parts. The remaining cylindrical surface cut off by the straight lines $αγ, βδ$, and the plane segments $αθ, θε, εη, ηβ, γλ, λζ, ζμ, μδ$, is greater than the composite surface formed by the parallelograms whose base is $αθ, θε, εη, ηβ$ and whose height is the same as the cylinder. And the parallelograms whose base is $αθ, θε, εη, ηβ$ and whose height is the same as the cylinder are greater than the parallelograms whose base is $αε, εβ$ and whose height is the same as the cylinder. Therefore, the cylindrical surface cut off by the straight lines $αγ, βδ$, and the plane segments $αεβ, γζδ$, is greater than the parallelograms whose base is $αε, εβ$ and whose height is the same as the cylinder. But the parallelograms whose base is $αε, εβ$ and whose height is the same as the cylinder are equal to the parallelogram $αγβδ$ and the area $κ$. Therefore, the cylindrical surface cut off by the straight lines $αγ, βδ$, and the plane segments $αθ, θε, εη, ηβ, γλ, λζ, ζμ, μδ$, is greater than the parallelogram $αγβδ$ and the area $κ$. But the segments $αθ, θε, εη, ηβ, γλ, λζ, ζμ, μδ$ were removed as being smaller than the area $κ$.

A geometric diagram showing a cylindrical segment. Points $α$ and $β$ are the endpoints of a horizontal line, with points $λ$, $ε$, and $μ$ on an arc above it. A parallel base line is defined by points $γ$ and $δ$, with an arc below containing point $ζ$. Lines connect these points to form triangular segments and a central parallelogram, representing areas on the cylindrical surface.

A second, more detailed geometric diagram showing the method of exhaustion. The arcs are further subdivided with points labeled $θ, ε, η$ on the top arc and $κ, ζ, λ, π, μ$ on the lower sections. These subdivisions create smaller triangular segments, illustrating the proof that the cylindrical surface area is greater than the inscribed parallelogram.