2. Given the curve $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a \geqslant b>0)$ passes through the point $(\sqrt{3}, 1)$. All these curves that satisfy $y \geqslant 1$ are $\qquad$ .
2. $\frac{4 \pi}{3}-\sqrt{3}$.
From the problem, we have $\frac{3}{a^{2}}+\frac{1}{b^{2}}=1$,
$$
\begin{array}{l}
\therefore b^{2}=\frac{a^{2}}{a^{2}-3} . \\
\because a \geqslant b, \therefore a^{2} \geqslant \frac{a^{2}}{a^{2}-3}, \text { so } a^{2} \geqslant 4 . \\
\because \frac{x^{2}}{a^{2}}+\frac{\left(a^{2}-3\right) y^{2}}{a^{2}}=1, \\
\therefore a^{2}\left(y^{2}-1\right)=3 y^{2}-x^{2} . \\
\text { Also, } \because y \geqslant 1, \\
\therefore 4\left(y^{2}-1\right) \leqslant 3 y^{2}-x^{2} . \\
\therefore x^{2}+y^{2} \leqslant 4 .
\end{array}
$$

The points that satisfy $x^{2}+y^{2} \leqslant 4$ and
$y \geqslant 1$ are shown in the shaded
area of Figure 2. The area is
$$
\begin{array}{l} 
\frac{1}{3} \cdot \pi \cdot 2^{2}-\frac{1}{2} \times \\
2^{2} \sin 120^{\circ}=\frac{4 \pi}{3}-\sqrt{3} .
\end{array}
$$