The Complete Works of Archimedes

...composed of parallelograms, is greater than the surface of the cylinder excluding the base.

13

The surface of any right cylinder, excluding the bases, is equal to a circle whose radius is a mean proportional between the side of the cylinder and the diameter of the base of the cylinder.

Let the base of some right cylinder be circle $A$, and let $ΓΔ$ be equal to the diameter of circle $A$, and let $EZ$ be equal to the side of the cylinder; and let $K$ be a mean proportional between $ΓΔ$ and $EZ$. Let circle $B$ be set out, whose radius is equal to $K$. I say that circle $B$ is equal to the surface of the cylinder, excluding the bases.

For if it is not equal, it is either greater or less. First, if possible, let it be less. Since there are two unequal magnitudes, namely the surface of the cylinder and circle $B$, it is possible to inscribe an equilateral polygon in circle $B$, and to circumscribe another, such that the circumscribed figure has to the inscribed figure a ratio less than that which the surface of the cylinder has to circle $B$. Let the circumscribed and inscribed figures be conceived. And about circle $A$ let a rectilinear figure be circumscribed similar to the one circumscribed about circle $B$. And let a rectilinear prism be erected from it; this will then be circumscribed about the cylinder.

Now, let $KΛ$ be equal to the perimeter of the rectilinear figure about circle $A$, and let $Λ Z$ be equal to $KΔ$. And let $Γ T$ be half of $ΓΔ$. Triangle $KΔ T$ will then be equal to the rectilinear figure about circle $A$.

Since the base has a perimeter equal to $KΛ$, and its height is the radius of circle $A$, and the surface composed of the parallelograms of the prism circumscribed about the cylinder is contained by the side of the cylinder and a line equal to the perimeter of the base of the prism; let $EP$ be set equal to $EZ$, so that triangle $PKΛ$ is also equal to the surface composed of the parallelograms, and thus to the surface of the prism.

And since the rectilinear figures circumscribed about circles $A$ and $B$ are similar, the rectilinear figures will have the same ratio as the squares of their radii. Therefore, triangle $KTΔ$ will have to the rectilinear figure about circle $B$ the ratio which the side of $A$, namely $KΔ$, has in square [to the radius of $B$]. For $TΔ$ and $K$ are equal to the radii. But the ratio which the side $KΔ$ has to $K$ in square, this same ratio the side $KΔ$ has [to the other] in length. For $K$ is a mean proportional between $TΔ$ and $PZ$, because it is also between $ΓΔ$ and $EZ$.

Geometric diagrams illustrating proposition 13 of Archimedes. At the top, two circles (labeled α and β) are shown with circumscribed rectilinear polygons. Below them is a large, complex geometric construction featuring a right-angled triangle with vertices and intersection points labeled with Greek letters (including κ, λ, ρ, ε, ζ, τ, δ). This diagram supports the mathematical proof concerning the surface area of a cylinder by relating it to properties of triangles and circles.

But how are they equal? For since $Λ T$ is equal to $TΓ$, and $PE$ is equal to $EZ$, therefore $ΓΔ$ is double $Δ T$, and $PZ$ is double $PE$. As, therefore, $Δ T$ is to $ΓΔ$, so is $PE$ to $PZ$. And therefore the rectangle contained by $ΓΔ$ and $EZ$ is equal to the rectangle contained by $Δ T$ and $PZ$.

But the rectangle contained by $ΓΔ$ and $EZ$ is equal to the square on $K$. And the rectangle contained by $Δ T$ and $PZ$ is therefore equal to the square on $K$. Therefore, as the side $Δ T$ is to $K$, so $K$ is to $PZ$. Thus, as the side $Δ T$ is to $PZ$, so is the square on $Δ T$ to the square on $K$. For if three straight lines be proportional, as the first is to the third, so is the figure on the first to the similar and similarly situated figure on the second.

But what ratio $Δ T$ has to $PZ$ in length, this ratio triangle $KTΔ$ has to triangle $PΛ Z$, since $KΔ$ and $Λ Z$ are equal. Therefore, triangle $KTΔ$ has the same ratio to the rectilinear figure circumscribed about circle $B$ as triangle $KTΔ$ has to triangle $PZΛ$. Therefore, triangle $PΛ Z$ is equal to the rectilinear figure circumscribed about circle $B$.