Given the curve $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a \geqslant b>0)$ passes through the point $(\sqrt{3}, 1)$. Then, when $a, b$ vary, the area of the figure formed by all points on these curves satisfying $y \geqslant 1$ is $\qquad$ .
8. $\frac{4 \pi}{3}-\sqrt{3}$.

From the problem, we have $\frac{3}{a^{2}}+\frac{1}{b^{2}}=1$, so $b^{2}=\frac{a^{2}}{a^{2}-3}$. Since $a \geqslant b$, we have $a^{2} \geqslant \frac{a^{2}}{a^{2}-3}$, hence $a^{2} \geqslant 4$. Because
$$
\begin{array}{c}
\frac{x^{2}}{a^{2}}+\frac{(a-3) y^{2}}{a^{2}}=1, \\
a^{2}\left(y^{2}-1\right)=3 y^{2}-x^{2} .
\end{array}
$$

Since $y \geqslant 1$, we have $4\left(y^{2}-1\right) \leqslant 3 y^{2}-x^{2}$. Therefore, $x^{2}+y^{2} \leqslant 4$.
$$
\frac{1}{3} \cdot \pi \cdot 2^{2}-\frac{1}{2} \times 2^{2} \sin 120^{\circ}=\frac{4 \pi}{3}-\sqrt{3} .
$$