Q:

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?
Picture

A:

The problem seems to be what you mean by
"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."
The quantity $R \sin \alpha$ is the distance from the $z$ axis of points whose spherical coordinates are $(R, \theta, \alpha)$ for any $\theta,$
that is, the points on the sphere $\mathbb S_R$ that are at a zenith angle of $\alpha$ relative to the center of that sphere.
So far, this makes sense.
The quantity $\sin \alpha$ is the distance from the $z$ axis of points on the unit sphere about the origin at a zenith angle of $\alpha.$
This does not change when you translate that sphere upward $R$ units,
so it's unclear why you would add $R.$
In fact, since $0 < \sin\alpha \leq 1$ whenver $0 < \alpha < \pi,$
for any such $\alpha$ we would find that 
$R \sin\alpha \leq R < R + \sin\alpha,$
so it is impossible to have $R \sin\alpha = R + \sin\alpha.$
And that equation is just as impossible for $\alpha = 0$ or $\alpha = \pi$
as long as $R > 0.$
In short, the equation you proposed just does not make sense.
Now there may be some sense in equating the heights ($z$ coordinates)
of points at the intersection of $\mathbb S_R$ and the unit sphere around $(0,0,R).$ For that, we use the cosine of the azimuth angle relative to each sphere, and we note that the azimuth angle of these points relative to the sphere around $(0,0,R)$ is greater than their azimuth angle relative to  $\mathbb S_R.$
There is a relationship between the two angles, but it's not what your equation says.

As for the claim that $2R^2-2R^2\cos\alpha=1,$ which the other student presumably did not explain, let $P$ be a point where the two spheres intersect. Then the points $(0,0),$ $(0,R),$ and $P$ form an isoceles triangle with two legs of length $R$ (each a radius of $\mathbb S_R$) and a base of $1$ (the segment from $(0,R)$ to $P,$ which is a radius of the unit sphere around $(0,R).$)
If we drop a perpendicular from either $P$ or $(0,R)$ to the opposite leg, we divide the isoceles triangle into two right triangles, one with hypotenuse $R$ and angle $\alpha$ and one with hypotenuse $1$ and angle $\frac\alpha2.$
If we drop a perpendicular from $(0,0)$ to the base of the isoceles triangle, we get two congruent right triangles with hypotenuse $R$ and angle $\frac\alpha2,$ with leg $\frac12$ opposite the angle $\frac\alpha2.$
In either case we end up having to deal with a right triangle with angle $\frac\alpha2.$
Let's try working with the second pair of right triangles. 
The construction tells us that $R \sin\frac\alpha2 = \frac12.$
Square both sides and multiply by $2$: 
$$2R^2 \sin^2\frac\alpha2 = 2\left(\frac14\right) = \frac12.$$
Use the trigonometry identity 
$2 \sin^2\frac\alpha2 = 1 - \cos\alpha$
(which you can get from either the half-angle sine formula or the double-angle cosine formula)
to substitute $1 -\cos\alpha$ for $2 \sin^2\frac\alpha2$:
$$R^2(1 - \cos\alpha) = \frac12.$$
You can then distribute the $R^2$ over $1-\cos\alpha$ and multiply both sides by $2$ to get the other student's formula if you want, although I think the formula above is more convenient for just plugging into your known area formula.

