6.1. Find the area of the figure defined on the coordinate plane by the inequality $\sqrt{\arcsin \frac{x}{3}} \leq \sqrt{\arccos \frac{y}{3}} \cdot$ In your answer, specify the integer closest to the found value of the area.
Answer: 16. Solution. The domain of admissible values of the inequality is determined by the system $\left\{\begin{array}{l}\arcsin \frac{x}{3} \geq 0, \\ \arccos \frac{y}{3} \geq 0\end{array} \Leftrightarrow\left\{\begin{array}{c}0 \leq x \leq 3, \\ -3 \leq y \leq 3 .\end{array}\right.\right.$ If $-3 \leq y \leq 0$, then $\arccos \frac{y}{3} \geq \frac{\pi}{2} \geq \arcsin \frac{x}{3}$. Therefore, the entire square $\{0 \leq x \leq 3,-3 \leq y \leq 0\}$ with an area of 9 is included in the desired set.

On the remaining set $\{0 \leq x \leq 3,0 \leq y \leq 3\}$, the inequalities $0 \leq \arcsin \frac{x}{3} \leq \frac{\pi}{2}, 0 \leq \arccos \frac{y}{3} \leq \frac{\pi}{2}$ hold, and since the function $\sin t$ is increasing for $t \in\left[0 ; \frac{\pi}{2}\right]$, we obtain equivalent inequalities $\sin \left(\arcsin \frac{x}{3}\right) \leq \sin \left(\arccos \frac{y}{3}\right) \Leftrightarrow \frac{x}{3} \leq \sqrt{1-\left(\frac{y}{3}\right)^{2}} \Leftrightarrow x^{2}+y^{2} \leq 9$. Therefore, here the desired set includes a quarter circle, the area of which is $\frac{\pi \cdot 3^{2}}{4}$.

In total, the desired area is $9\left(1+\frac{\pi}{4}\right) \approx 16.069 \ldots$