3. Find the area of the figure defined on the coordinate plane by the inequality

$$
\sqrt{\arcsin y} \leqslant \sqrt{\arccos x}
$$
# Problem 3.

Answer: $1+(\pi / 4)$ (in all versions).

Solution (for versions I-1 and I-2). The domain of admissible values of the inequality is determined by the system $\left\{\begin{array}{l}\arcsin x \geqslant 0, \\ \arccos y \geqslant 0\end{array} \Longleftrightarrow\left\{\begin{array}{l}0 \leqslant x \leqslant 1, \\ -1 \leqslant y \leqslant 1 .\end{array}\right.\right.$

If $-1 \leqslant y \leqslant 0$, then $\arccos y \geqslant \frac{\pi}{2} \geqslant \arccos x$. Therefore, the entire square $\{0 \leqslant x \leqslant 1, -1 \leqslant y \leqslant 0\}$ of area 1 is included in the set of solutions.

On the remaining set $\{0 \leqslant x \leqslant 1, 0 \leqslant y \leqslant 1\}$, the inequalities $0 \leqslant \arcsin x \leqslant \frac{\pi}{2}, 0 \leqslant \arccos y \leqslant \frac{\pi}{2}$ hold, and since the function $\sin t$ is increasing for $t \in\left[0, \frac{\pi}{2}\right]$, we obtain equivalent inequalities

$$
\sin (\arcsin x) \leqslant \sin (\arccos y) \Longleftrightarrow x \leqslant \sqrt{1-y^{2}} \Longleftrightarrow x^{2}+y^{2} \leqslant 1
$$

Therefore, the solution here will be a quarter of the unit circle, the area of which is $\frac{\pi}{4}$.