geometry
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?
Let $C(A, B)$ denote a truncated cylinder whose left edge has length $A$ and whose right edge has length $B$. (I'm assuming that it's placed so that its longest and shortest edges are to the left and right).
The volume of $C(A, 0)$ is $\frac{1}{2} \pi r^2 A$. Reason: Stack up $C(A, 0)$ atop $C(0, A)$ to get a cylinder of height $A$. The volume of $C(A, 0)$ is clearly half the volume of this cylinder.
Now for a cylinder C(A, B), where $A > B$, consider the stack
C(0, Q) C(A, B) C(B, A) C(Q, 0)
where $Q = A - B$. THat's a cylinder of height $A + B + (A-B) = 2A$. If we subtract from it the volumes of the top and bottom pieces (which we computed in part 1), we get $$ \pi r^2 (2A) - \pi r^2 (Q) = \pi r^2 (2A - (A-B)) = \pi r^2 (A + B) $$ But the volume we've just computed is twice the volume of the sliced cylinder $C(A,B)$ that we're interested in. So the volume of $C(A,B)$ isjust $$ \pi r^2 \frac{A + B}{2} $$
A rather similar proof works for surface area.
Note that the proof given above works for a cylinder cut by ANY two planes, as long as the longest and shortest "line" of the truncated cylinder are diametrically opposite each other.
I'm not sure to well understand your question, so I add a figure.
The figure is a plane section of your sphere passing from $P$. If I well understand the distance that you want is the length of the arc $PB$.( If this is wrong than my answer is wrong)
You know the radius of the cap, that is the arc $AB=\beta$. In this case the arc $PB$ is simply $\dfrac{\pi}{2}-\alpha-\beta$ , where $\alpha$ is the arc that fix the position of $P$ with respect to the equatorial plane of the sphere(its latitude).
If you know as radius of the cap the distance $CB$ than you can find $\beta=\arcsin (CB)$.
Best Answer
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