5. In the Cartesian coordinate system $x o y$, the area of the figure formed by points whose coordinates satisfy the condition $\left(x^{2}+y^{2}+2 x+2 y\right)\left(4-x^{2}-y^{2}\right) \geqslant 0$ is

保留源文本的换行和格式，直接输出翻译结果如下：

5. In the Cartesian coordinate system $x o y$, the area of the figure formed by points whose coordinates satisfy the condition $\left(x^{2}+y^{2}+2 x+2 y\right)\left(4-x^{2}-y^{2}\right) \geqslant 0$ is
A circle minus their common
part. Let the first circle be
$$
(x+1)^{2}+(y+1)^{2}
$$
$-2 \leqslant 0$ with area $S_{1}$,
the second circle $x^{2}+y^{2}-4$
$\leqslant 0$ with area $S_{2}$, the area of the required region is $S$, and the common area of the two circles is
5. $2 \pi+4$.
Rewrite the inequality as $\left[(x+1)^{2}+(y+1)^{2}-\right.$ 2] $\left[x^{2}+y^{2}-4\right] \leqslant 0$. The required region is shown in the figure as the two

It is not difficult to see that the center of the first circle lies on the chord connecting $(-2,0)$ and $(0,-2)$ of the second circle. The first circle passes through the points $(0,0),(-2,0),(0,-2)$, so
$$
\begin{array}{l}
=\pi+\frac{\pi \times 2^{2}}{4}-2=2 \pi-2 \text {. } \\
\end{array}
$$

and $S_{1}=2 \pi, S_{2}=4 \pi$,
$$
\therefore S=4 \pi+2 \pi-2 \times(2 \pi-2)=2 \pi+4 \text {. }
$$