The Complete Works of Archimedes

...[surface] of the cylinder [contained by] the straight lines, and let their extremities be the same. Let there be a right cylinder, whose base is the circle $AB$, and the opposite [base] is $CD$. And let $AC, BD$ be joined. I say that the cylindrical surface contained by the straight lines $AC, BD$ is greater than the parallelogram $ACDB$. For let each of the arcs $AB, CD$ be bisected at points $E, Z$. And let $AE, EB, CZ, ZD$ be joined. Since $AE, EB$ are equal to $CZ, ZD$, and the heights are equal, the parallelograms upon them are also [equal]. And the parallelograms whose bases are $AE, EB$, and whose height is that of the cylinder itself, are greater than the parallelogram $ACDB$. By how much, then, are they greater? Let it be by the area $H$. Now, the area $H$ is either less than the plane segments $AE, EB, CZ, ZD$, or it is not less.

First, let it be not less. And since the cylindrical surface contained by the straight lines $AC, BD$ and the segments $AEB, CZD$ has as its boundary the plane of the parallelogram $ACDB$; and also the composite surface consisting of the parallelograms whose bases are $AE, EB$ and whose height is the same as the cylinder, and the triangles $AEB, CZD$ has as its boundary the plane of the parallelogram $ACDB$; and the one encloses the other, and both are concave in the same direction—therefore the cylindrical surface contained by the straight lines $AC, BD$ and the plane segments $AEB, CZD$ is greater than the composite surface consisting of the parallelograms whose bases are $AE, EB$ and whose height is that of the cylinder itself, and the triangles $AEB, CZD$. Let the common triangles $AEB, CZD$ be subtracted. Therefore the remaining cylindrical surface contained by the straight lines $AC, BD$ and the plane segments $AE, EB, CZ, ZD$ is greater than the composite surface consisting of the parallelograms whose bases are $AE, EB$ and whose height is the same as the cylinder. But the parallelograms whose bases are $AE, EB$ and whose height is the same as the cylinder are greater than the parallelogram $ACDB$ and the area $H$. Therefore, the remaining cylindrical surface contained by $AC, BD$ is greater than the parallelogram $ACDB$.

But now let the area $H$ be less than the plane segments $AE, EB, CZ, ZD$. And let each of the arcs $AE, EB, CZ, ZD$ be bisected at points $Th, K, L, M$; and let $A Th, Th E, EK, KB, CL, LZ, ZM, MD$ be joined. Since more than half of the plane segments $AE, EB, CZ, ZD$ is taken away by the triangles $A Th E, EKB, CLZ, ZMD$, and this is done continually, certain segments will be left which shall be less than the area $H$. Let them be left, and let them be $A Th, Th E, EK, KB, CL, LZ, ZM, MD$. Similarly, we shall show that the parallelograms whose bases are $A Th, Th E, EK, KB$, and whose height is the same as the cylinder, are greater than the parallelograms whose bases are $AE, EB$, and whose height is the same as the cylinder. And since the cylindrical surface contained by the straight lines $AC, BD$ and the plane segments $AEB, CZD$ has as its boundary the plane of the parallelogram $ACDB$; and also the composite surface consisting of the parallelograms whose bases are $A Th, Th E, EK, KB$ and whose height is the same as the cylinder, and the rectilinear [areas] $A Th E, EKB, CLZ, ZMD$—let the common rectilinear [areas] $A Th E K B, CLZMD$ be subtracted. Therefore the remaining cylindrical surface contained by the straight lines $AC, BD$ and the plane segments $A Th, Th E, EK, KB, CL, LZ, ZM, MD$ is greater than the composite surface consisting of the parallelograms whose bases are $A Th, Th E, EK, KB$ and whose height is the same as the cylinder.

But the parallelograms whose bases are $A Th, Th E, EK, KB$ and whose height is the same as the cylinder are greater than the parallelograms whose bases are $AE, EB$ and whose height is the same as the cylinder. And therefore the cylindrical surface contained by the straight lines $AC, BD$ and the plane segments $A Th, Th E, EK, KB, CL, LZ, ZM, MD$ is greater than the parallelograms whose bases are $AE, EB$ and whose height is the same as the cylinder. But the parallelograms whose bases are $AE, EB$ and whose height is the same as the cylinder are greater than the parallelogram $ACDB$ and the area $H$. Therefore the cylindrical surface contained by the straight lines $AC, BD$ and the plane segments $A Th, Th E, EK, KB, CL, LZ, ZM, MD$ is greater than the parallelogram $ACDB$ and the area $H$. And the segments $A Th, Th E, EK, KB, CL, LZ, ZM, MD$ being subtracted, they are less than the area $H$.