Visible Portion of the Earth's Surface 
EDIT: I need help converting the right side to a function of h
Let $A_h$ be the area of the zone corresponding to height h. If we set up a rectangular co-ordinate syustem with the origin at the center A of the spherical Earth with radius R, and if the surface of the earth is obtained by rotating the curve $x = g(y), y_B \le y \le y_E$ about the y-axis, then the surface area is given by $$A_h = 2\pi \int_{yb}^{ye} g(y) \sqrt{1+[g'(y)]^2} dy$$ 1. Derive a formula for the observable area $A_h$ as a function of the altitude h above the Earth's surface.
Okay, so I've been looking at this problem for a few days now and I'm having trouble deriving this equation based on the pictue. I know I need to revolve the curve $CE$ around the y axis but I'm having a hard time figuring out what the equation will be. I know this has to do with horizon and such, and the equation for line $$CD = \sqrt{h(2R+h)}$$ I also know the $$\sqrt{1+[g'(y)]^2}$$is an arclength
I'm just very confused becuase I know once I plug all these numbers in I will get a constant and integrating a constant is just adding (in this case) a y the result and then plugging in the bounds. Once I find this equation I have the answer for the rest of these problems.
(First post, I'm sorry if this isn't super clear, all help is greatly appreciated)

Adding up the sections of $A_h$ is a lot like stacking up incrementally smaller onion rings and adding up the surface areas of their exteriors.


a very dextrous sketch of a stack of onion rings in the shape of the Earth’s crust
We get the surface area of an onion ring by taking the length of its slant $\ell$ and multiplying it by $2\pi r$, just as if we ‘unwound’ it and calculated the area of the resultant rectangle.
You were right to recognize the arc length formula—that’s exactly what $\ell$ is: a length of arc parametrized by $x$ and $y$. We could also see this by breaking up the crust of Earth into teeny-tiny right triangles, where the hypotenuse measures length $d\ell =\sqrt{dx^2+dy^2}$. However, $g(y)$ is the horizontal piece that we will break into tiny increments, not $x$.
Since you already know the variations of the arc length formula, let’s skip right to $$d\ell=\sqrt{1+\left(\frac{dg}{dy}\right)^2}\,dy =\sqrt{1+\left[g’(y)\right]^2}\,dy$$
Next, to get the area of the exterior surface of each ring, multiply by $2\pi r$, just as if we were dealing with a rectangle of height $d\ell$ and length $2\pi r$. The radius of each ring is given by $g(y)$. This gets us to $$dA_h=2\pi\,g(y)\,\sqrt{1+\left[g’(y)\right]^2}\,dy$$
Finally, we’ll want to add this incremental areas up for all the tiny rings between $y_B$ and $y_E$. Luckily, our variable of integration is already $y$!
$$\begin{align}
\int dA_h &= \int_{y_B}^{y_E} 2\pi \,g(y)\,\sqrt{1+\left[g’(y)\right]^2}\,dy \\
A_h &= 2\pi\int_{y_B}^{y_E} g(y)\,\sqrt{1+\left[g’(y)\right]^2}\,dy \\
\end{align}$$
Voilà!
Addendum: I would also recommend researching the calculus-based derivation of the formula for the surface area of a circle and/or the formula for the surface area of a cone.
