8. 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 the area of the figure formed by all points on these curves satisfying $y \geqslant 1$ is $\qquad$
8. $\frac{4 \pi}{3}-\sqrt{3}$.

Solution: From the problem, we have $\frac{3}{a^{2}}+\frac{1}{b^{2}}=1, b^{2}=\frac{a^{2}}{a^{2}-3}$,
From $a^{2} \geqslant b^{2}$, we get $a^{2} \geqslant 4$.
$$
\because \frac{x^{2}}{a^{2}}+\frac{\left(a^{2}-3\right) y^{2}}{a^{2}}=1, \therefore a^{2}\left(y^{2}-1\right)=3 y^{2}-
$$
$x^{2} \geqslant 4\left(y^{2}-1\right)(\because y \geqslant 1)$, thus, $x^{2}+y^{2} \leqslant 4$. The area of the figure that satisfies $x^{2}+y^{2} \leqslant 4, y \geqslant 1$ is $\frac{1}{3} \cdot \pi \cdot 2^{2}-\frac{1}{2} \times$
$$
2^{2} \sin 120^{\circ}=\frac{4 \pi}{3}-\sqrt{3}
$$