Proposition 13 – Sphere and the Cylinder I

Here’s a version of proposition 13 that fills in a few details. I have tried to write this in a way that makes it easy to follow.

Proposition 13: The area of a cylinder (excluding the ends) is equal to a circle whose radius is a mean proportional between the height of the cylinder and the diameter of the base.

Comment: Call the radius of the cylinder ‘r’ and the radius of the circle mentioned in the theorem ‘R’. And call the height of the cylinder h. Then the theorem talks of R as being a mean proportional between 2r and h. Hence (2r)/R = R/h or R²=2rh . Since πR² is the area of the circle, we have that the area of the cylinder is being claimed to be 2πrh. I.e. we really have the modern formula. It is important to understand that the Greeks had to express newly proven areas (like the cylinder) in terms of areas they already knew (like the circle in this example).

Proof: The proof will use the fact that if a is not greater than b and a is not less than b then a must equal b.

We will use the notation introduced in the comment and add to it. Let A be the base of the cylinder (and also use A to denote its area). Let B be the circle whose radius R has the property that R²=2rh (and also use B to denote the area).

Let S denote the area of the cylinder. We consider the two cases: B<S and B>S. If we can get contradictions from both of these we’ll have shown that B=S.

Suppose that B<S. We can find P_CB and P_IB , two polygons circumscribed and inscribed around B such that the ratio of the outer polygon to the inner polygon is less than S/B. (Note: two different uses of ‘B’ in the same sentence.) Now construct around A a polygon similar to the polygon around B and use this to construct a prism of height h which surrounds the cylinder. Call this polygon P_CA (which we can read as polygon circumscribed around A).

By taking a triangle with one side equal to r and the other side equal to the perimeter of P_CA we will have a triangle whose area equals the area of P_CA. Call this T_PCA standing for “triangle with area equal to P_CA”. Note that:

(area T_PCA ) = r(perimeter P_CA). Also let R_PCA denote the rectangle whose side is the perimeter of P_CA and whose base is h. Then the area of R_PCA equals the area of the prism around the cylinder. From R_PCA we can construct a triangle whose area is the same as R_PCA; the triangle will have the same height as R_PCA but twice the base. I’ll call this last triangle T_prism standing for triangle with same area as the prism and note:

(area T_prism ) = (perimeter of P_CA )h

Now: (area of P_CA ) / (area of P_CB ) = r² / R² and:

(area T_PCA ) / (area of P_CB )= r² / R²

= r² / rh

= r/h

= (r(perimeter P_CA) )/(h(perimeter P_CA))

= (area T_PCA ) / (area T_prism )

Therefore: (area of P_CB ) = (area T_prism )

But (area P_CB )/( area P_IB ) < S/B which implies:

(area T_prism )/( area P_IB ) < S/B or

(area T_prism )/ S < ( area P_IB ) / B

But this is impossible, LHS > 1 and RHS<1. Students: How do we know that LHS > 1 ? I.e. How do we know that (area T_prism ) > S ?

Thus B is not less than S. I.e. B ≥ S.

Now suppose B>S. (This part is left to you.)

QED