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)
Best Answer
Geometric Approach
In this answer, it is shown that the area of the green strip on the sphere is the same as the area of the red projection onto the cylinder circumscribing the sphere and sharing its axis with the green strip.
We can compute the height of the cap using similar triangles:
Thus, the area of the cap is $$ 2\pi R\frac{Rh}{R+h}=\frac{2\pi R^2h}{R+h} $$
Calculus Approach
Because $\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac xy$, we have
$$ \begin{align} \int_0^{\frac{R}{R+h}\sqrt{2Rh+h^2}}2\pi x\sqrt{1+x^2/y^2}\,\mathrm{d}x &=\int_0^{\frac{R}{R+h}\sqrt{2Rh+h^2}}2\pi R \frac{x}{\sqrt{R^2-x^2}}\,\mathrm{d}x\\ &=-2\pi R\left[\sqrt{R^2-x^2}\right]_0^{\frac{R}{R+h}\sqrt{2Rh+h^2}}\\[6pt] &=\frac{2\pi R^2h}{R+h} \end{align} $$