I have the upper half of the ball $\ x^2 + y^2 + z^2 = R , \ (z \ge 0)$ and the plane $\ z = h $ where $\ 0 < h < R $ .
How do I compute that surface of the ball above the plane? and show it is $\ 2 \pi R(R-h) $
My attempt:
The intersection projection of the plane $\ z = h $ and the ball is $\ x^2 + y^2 = R - h^2 $ hence the area I want to compute is a circle with radius $\ R $ and then subtract from it the circle with radius $\ h $ . Though the answer I get is $\ 2\pi(R-h) $ and not $\ 2\pi *R*(R-h) $
$$\ \int_0^{2\pi} \int_0^{cos^{-1}(\frac{h}{R})} R^2 \sin \theta = R^2 \int \sin \theta [cos^{-1}(\frac{h}{R})] = R^2 \cos^{-1}(\frac{h}{R}) \cdot [- \cos \theta]_0^{2\pi} = R^2 \cos^{-1}(\frac{h}{R}) \cdot (-1 +1) = 0 $$
If you were asked to compute the surface area of a sphere of radius $R$; i.e of the sphere $x^2 + y^2 + z^2 = R^2 $, the typical approach would be to use spherical coordinates \begin{equation} \begin{cases} x &= R \sin \theta \cos \phi \\ y &= R \sin \theta \sin \phi \\ z &= R \cos \theta \end{cases} \end{equation} with $0 \leq \theta \leq \pi$, and $0 \leq \phi \leq 2\pi$. But now you don't want the entire sphere, so how would the limits of integration of integration change? (draw a picture; it's very simple trigonometry).
Edit:
After looking at your work, it seems you are slightly unclear about which variables are being held fixed in integrals, and which are varying. So the integral you want to evaluate is \begin{equation} \int_0^{2\pi} \left(\int_0^{\arccos(h/R)} R^2 \sin \theta \, d\theta \right) \, d\phi \end{equation} I added the () to make it clear in what order to do things. Evaluate the inner integral first; you should get something which depends only on $R$ and $h$. Then, you can pull those constants out and solve the outer integral