7. Given the elliptical region $D_{1}: \frac{x^{2}}{3}+\frac{y^{2}}{2} \leqslant 1$ and the circular region $D_{2}$ : $x^{2}+y^{2} \leqslant 2.5$. Then the area of the common region between $D_{1}$ and $D_{2}$ is $\qquad$ (accurate to 0.01).
Note: The following result can be used: In the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, $B$ is the upper vertex, and $P\left(x_{0}, y_{0}\right)$ is a point on the first quadrant part of the ellipse, then $S_{\text {sector } O B P}=\frac{a b}{2} \arcsin \frac{x_{0}}{a}$.
7.7. 27 .
Let the intersection point of regions $D_{1}$ and $D_{2}$ be $P$.
Thus, $P\left(\frac{\sqrt{6}}{2}, 1\right)$.
By symmetry, we only need to consider the area of the common region in the first quadrant, $S_{\text {flat } B O P} + S_{\text {sector } O A P}$.
$$
\begin{array}{l}
\text { Also, } S_{\text {sector } B O P}=\frac{\sqrt{3} \times \sqrt{2}}{2} \arcsin \frac{\frac{\sqrt{6}}{2}}{\sqrt{3}} \times \frac{\pi}{360} \\
=\frac{\sqrt{6} \pi}{8} \approx 0.9615, \\
S_{\text {flat } A P P}=\frac{5 \pi}{2} \cdot \frac{\arctan \frac{1}{\sqrt{6}}}{360} \approx 0.8555,
\end{array}
$$

Therefore, the required area is $4\left(S_{\text {flat } B O P}+S_{\text {flat } O A P}\right) \approx 7.27$.