The area of the region \ (x, y): y ≤ π - |x|, y ≤ |x x|, y ≥ 0\ is:

The given region is defined by the inequalities $y \leq \pi - |x|$, $y \leq |x \sin x|$, and $y \geq 0$.

Since replacing $x$ with $-x$ leaves the inequalities unchanged, the region is symmetric with respect to the y-axis. We can find the area of the region in the first quadrant ($x \geq 0$) and multiply it by $2$.

For $x \geq 0$, the inequalities become:
$y \leq \pi - x$
$y \leq x \sin x$
$y \geq 0$

The condition $y \geq 0$ and $y \leq \pi - x$ restricts $x$ to the interval $[0, \pi]$. In this interval, $\sin x \geq 0$, so $|x \sin x| = x \sin x$.

To find the intersection of the curves $y = \pi - x$ and $y = x \sin x$, we equate them:
$x \sin x = \pi - x \Rightarrow x(1 + \sin x) = \pi$

By inspection, $x = \dfrac{\pi}{2}$ is a solution since $\dfrac{\pi}{2}(1 + 1) = \pi$.

For $x \in \left[0, \dfrac{\pi}{2}\right]$, $x \sin x \leq \pi - x$.
For $x \in \left[\dfrac{\pi}{2}, \pi\right]$, $x \sin x \geq \pi - x$.

The area in the first quadrant, $A_1$, is given by:
$A_1 = \int_{0}^{\pi/2} x \sin x \, dx + \int_{\pi/2}^{\pi} (\pi - x) \, dx$

Evaluating the first integral using integration by parts:
$\int_{0}^{\pi/2} x \sin x \, dx = [-x \cos x]_{0}^{\pi/2} - \int_{0}^{\pi/2} (-\cos x) \, dx$
$= 0 + [\sin x]_{0}^{\pi/2} = 1$

Evaluating the second integral:
$\int_{\pi/2}^{\pi} (\pi - x) \, dx = \left[ -\dfrac{(\pi - x)^2}{2} \right]_{\pi/2}^{\pi} = 0 - \left( -\dfrac{\pi^2}{8} \right) = \dfrac{\pi^2}{8}$

Thus, the area in the first quadrant is:
$A_1 = 1 + \dfrac{\pi^2}{8}$

The total area of the region is twice the area in the first quadrant:
$\text{Total Area} = 2 \times A_1 = 2 \left( 1 + \dfrac{\pi^2}{8} \right) = 2 + \dfrac{\pi^2}{4}$

Answer: $2 + \dfrac{\pi^2}{4}$