Archimedes' derivation of the spherical cap area formula

Archimedes derived a formula for the area of a spherical cap.

so Archimedes says that the curved surface area of a spherical cap is equal to the area of a circle with radius equal to the distance between the vertex at the curved surface and the base of the spherical cap. $$A = \pi(h^2+a^2)$$
I want to know how Archimedes derived this formula. I have searched on the net and only found solutions using integration. Is there a method to do this without using integration?

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According to Morris Kline in Mathematical Thought from Ancient to Modern Times, Volume I, page 109, he used the method of exhaustion. See this for a description of the method (phrased in modern terms) applied to this problem. – David Mitra Jun 28 '13 at 15:05

Enclose the sphere inside a cylinder of radius $r$ and height $2r$ just touching at a great circle. The projection of the sphere onto the cylinder preserves area.

That is the way Archimedes derived that the area of the sphere is same as lateral surface area of the cylinder which is $= (2 \pi r) (2r)=4\pi r^2$. The projection of the cap on the cylinder has area $(2 \pi r)h$. And since $a^2=r^2-(r-h)^2=2rh-h^2 \Rightarrow 2rh=h^2+a^2$, the area of the cap is $\pi (2rh) = \pi (h^2+a^2).$

Edit: corrected grammar

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To convince yourself that the projection preserves area, note that although the radius of circle after projection increases(by a factor of $\cos \theta$) the width of the shell decreases(by a factor of $\cos \theta$) as the width is at an angle on the sphere and vertical on the cylinder, where $\theta$ is the angle between the circle and the great circle touching the cylinder. The two effects cancel out. – nsoum Jun 28 '13 at 14:18
what is the great circle? could you show me a diagram? – udiboy1209 Jun 28 '13 at 14:21
Just imagine a cylinder of radius $r$ and height $2r$, put it on a table and drop the sphere inside it. It just touches the cylinder along a circle. That is the great circle I was talking about. – nsoum Jun 28 '13 at 14:43

In Archimedes codex, or the Palimpset, they go into great detail how he arrived at this. He used a parabolic segment, the area intercepted between a parabola and a straight line crossing through it. I see the formula posted but thought you were looking for the history or thought behind it.

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Welcome to Math.SE! This is interesting information; are you able/willing to add mathematical details, perhaps including diagrams? – Andrew D. Hwang Nov 26 '15 at 2:08