Calculating area of sphere with constraint on zenith Let $\mathbb S_R$ be the sphere of radius $R$ centered about the origin. Consider $$A_R= \left\{ (x,y,z)\in \mathbb S_R \mid x^2+y^2+(z-R)^2\leq 1 \right\}.$$
I want to calculate the area of this region of the sphere with radius $R$ about the origin. I already calculated the area of the region of $\mathbb S_R$ of vectors with zenith $\leq \alpha$ is given by $2\pi R^2(1-\cos \alpha)$.
For $A_R$, here's my idea. I want to find the zenith $\alpha$ which satisfies $R\sin \alpha=\sin \alpha+R$, since this is the zenith of points of both $\mathbb S_R$ and the unit sphere translated up $R$ units at points which are of the same height. Then, I just want to integrate the zenith $0\leq \phi\leq \alpha$.
Solving gives $\phi=\arcsin \frac R{R-1}$ and things get a little messy.
On the other hand, another student posted a solution which makes sense. He just writes $\alpha$ should satisfy $2R^2-2R^2\cos\alpha=1$ and then obtains the area of $A_R$ is $\pi$, independently of the radius.
Why is my method wrong?
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I think @David K has addressed your point. Here is how I would proceed:
Parametrize the surface as follows:
\begin{cases}
x=x\\
y=y\\
z=\sqrt{R^2-x^2-y^2}
\end{cases}
with $(x,y)\in D$, where $D$ is projection of the intersection of both spheres in the $xy$-plane. When both spheres intersect, we have
\begin{cases}
x^2+y^2+z^2 = R^2 \\
x^2+y^2+(z-R)^2 = 1
\end{cases}
which leads to $x^2+y^2=1-\frac{1}{4R^2}$. In other words, $D$ is a disc with radius $\sqrt{1-\frac{1}{4R^2}}=\frac{\sqrt{4R^2-1}}{2R}$ (assuming $R\ge \frac{1}{{2}}$). This parametrization only works when the radius of the bottom sphere is smaller than (or equal to) the radius of $D$, i.e. when $R\le \frac{\sqrt{4R^2-1}}{2R}$, or $R\ge 1/\sqrt{2}$.
We need the norm of the normal vector to $S$, i.e., the norm of  $$\pmatrix{1\\0\\\frac{-x}{\sqrt{R^2-x^2-y^2}}}\times\pmatrix{0\\1\\\frac{-y}{\sqrt{R^2-x^2-y^2}}}=\pmatrix{\frac{x}{\sqrt{R^2-x^2-y^2}} \\ \frac{y}{\sqrt{R^2-x^2-y^2}} \\ 1} $$
which is 
$$
\sqrt{1+\frac{x^2+y^2}{R^2-x^2-y^2}}=\frac{R}{\sqrt{R^2-x^2-y^2}}
$$
 It follows that the wanted area equals
$$
\iint_D \frac{R}{\sqrt{R^2-x^2-y^2}}\; dA = \int_0^{2\pi}\int_0^{\frac{\sqrt{4R^2-1}}{2R}}  \frac{R}{\sqrt{R^2-r^2}}\; r dr d\theta = 2\pi R^2-\pi |2R^2-1|\\
$$
Remember that we assumed that $R\ge 1/\sqrt{2}$, so this expression equals $\pi$.
If $0<R\le\frac{1}{2}$ the  intersection is at most a point, and it is the entire surface of the sphere of radius $R$ that we want, i.e., $4\pi R^2$. And for the other values of $R$ ($1/2\le R\le 1/\sqrt{2}$), we are better off in spherical coordinates and the same reasoning also leads us to $\pi$: you should get
$$
\int_0^{2\pi} \int_0^{\cos^{-1}(1-\frac{1}{2R^2})} R^2\sin\phi\; d\phi d\theta = \pi\quad (1)
$$
So in summary the area equals
\begin{cases}
4\pi R^2 \quad &\mbox{if }0<R\le \frac{1}{2}\\
\pi \quad &\mbox{if } R\ge 1/2
\end{cases}
Note: spherical coordinates are much more interesting here. Equation $(1)$ is simpler to find, to solve, and it is valid for all $R\ge 1/2$.
