(8) Given a moving point $P(x, y)$ satisfies $\left\{\begin{array}{l}2 x+y \leqslant 2, \\ x \geqslant 0, \\ \left(x+\sqrt{x^{2}+1}\right)\left(y+\sqrt{y^{2}+1}\right) \geqslant 1,\end{array}\right.$ then the area of the figure formed by the moving point $P(x, y)$ is $\qquad$
82 Prompt: From $\left(x+\sqrt{x^{2}+1}\right)\left(y+\sqrt{y^{2}+1}\right) \geqslant 1$ we get
$$
x+\sqrt{x^{2}+1} \geqslant \sqrt{y^{2}+1}-y,
$$

Therefore,
$$
\ln \left(x+\sqrt{x^{2}+1}\right) \geqslant \ln \left(\sqrt{y^{2}+1}-y\right) .
$$

Let $f(x)=\ln \left(x+\sqrt{x^{2}+1}\right)$, then it is easy to see that $f(x)$ is an increasing and odd function.
The condition essentially means $f(x) \geqslant f(-y)$, so $x \geqslant -y, x+y \geqslant 0$. From $\left\{\begin{array}{l}2 x+y \leqslant 2, \\ x+y \geqslant 0\end{array}\right.$ we get the intersection point coordinates as $(2,-2)$. Therefore, the area is $\frac{1}{2} \times 2 \times 2=2$.