Segmented area between circles The following is a geometry problem that I came across with in the course of a research project.
Consider a ray starting at some initial point $t$. Place point $s_1$ at distance $r$ from $t$ on the ray and draw a circle centered at $s_1$ that passes through $t$. Likewise, centered at $t$, an arc with radius $r$ goes through $s_1$. Let $\mathcal{A_1}$ be the area enclosed between the intersecting arcs.
Next, arbitrarily place another point somewhere on the free end of the ray and call it $s_2$ such that $|s_1 - t| < |s_2 - t|$, where $|.|$ denotes the Euclidean distance. A circle with radius $r$ is centered at $s_2$ and another arc centered at $t$ goes through $s_2$. The area enclosed between these intersecting arcs we call $\mathcal{A}_2$. It is easy to show that $\mathcal{A}_1 < \mathcal{A}_2 < \lim_{|s_2 - t| \to \infty} \mathcal{A}_2 = \frac{1}{2} \pi r^2$.
Now, assume that we mark the segments of the ray within the enclosed areas in the middle and arcs centered at $t$ pass through the marks segmenting $\mathcal{A}_1$ and $\mathcal{A}_2$. We call these segmented areas $\mathcal{A}_{11}$ and $\mathcal{A}_{12}$ and $\mathcal{A}_{21}$ and $\mathcal{A}_{22}$ as depicted below (dashed lines are the arcs centered at $t$).

Question: How does $\mathcal{A}_{22}$ change as $s_2$ gets farther from $t$? (i.e., does it increase or decrease?) What can we say about $\mathcal{A}_{22}$ in comparison with $\mathcal{A}_{12}$?
Any idea or comment is much appreciated.
EDIT: The question has been edited in a way that makes the comments incomprehensible. Please see the edit history if you want to make sense of the comments.
EDIT: Here is the link to the same question at mathoverflow.net
 A: The shape of $A_1$ is always the same, so we can calculate its area as the sum of two circular segments: $A_1 = r^2(2\pi/3 - \sqrt{3}/2)$.
Let's first let $t$ be the origin (why the hell would you name the origin of a ray $t$?!). Let's set $r=1$, keeping in mind all areas will be scaled by $r^2$ later.
The coordinates of the vertices of $A_{11}$ are $(1/8,\pm \sqrt{15}/8)$, and the subtended angles are $2\tan^{-1} \sqrt{15}$ and $2\tan^{-1} \sqrt{15}/7$.
This then gives $A_{11} \approx 0.350767 (r^2)$, $A_{1} \approx 1.22837$, and $A_{12} \approx 0.877603$.
Note that all these numbers can be made precise; they're just huge ugly expressions, and remember they are multiplied by $r^2$.
Now let $s_2$ be located at coordinates $(R,0)$ where $R > 1$ according to our assumptions.
By similar reasoning, the vertices of $A_2$ are $(\frac{2R^2-1}{2R},\pm \sqrt{1-\frac{1}{4R^2}})$.
Similarly now, we can compute $A_{22}$. The expression for $A_{22}/A_{12}$ is horrendously large, so I will just have Mathematica plot it as a function of $R$:

The limit according to Mathematica is $1.09003$.
Edit: I have corrected a number of mistakes. Now, the area ratio is always greater than unity for $R>1$.
