Cho (H) là hình phẳng giới hạn bởi parabol $y=\sqrt{3}{{x}^{2}}$...

Phương trình hoành độ giao điểm: $\sqrt{3}{{x}^{2}}=\sqrt{4-{{x}^{2}}}$, Đk: $-2\le x\le 2$
$\Leftrightarrow 3{{\text{x}}^{4}}+{{x}^{2}}-4=0\Leftrightarrow {{x}^{2}}=1\Leftrightarrow x=\pm 1$.
Hình (H) giới hạn bởi: $\left\{ \begin{aligned}
& (P):y=\sqrt{3}{{x}^{2}} \\
& (C):y=\sqrt{4-{{x}^{2}}} \\
& x=-1;x=1 \\
\end{aligned} \right.$ có diện tích là:
$S=\int\limits_{-1}^{1}{(\sqrt{4-{{x}^{2}}}}-\sqrt{3}{{x}^{2}})d\text{x}=\underbrace{\int\limits_{-1}^{1}{\sqrt{4-{{x}^{2}}}}d\text{x}}_{{{I}_{1}}}-\underbrace{\int\limits_{-1}^{1}{\sqrt{3}}{{x}^{2}}d\text{x}}_{{{I}_{2}}}$
.
* Ta có: ${{I}_{2}}={{\left. \dfrac{\sqrt{3}}{3}{{x}^{3}} \right|}_{-1}}^{1}=\dfrac{2\sqrt{3}}{3}$
* Xét ${{I}_{1}}=\int\limits_{-1}^{1}{\sqrt{4-{{x}^{2}}}}d\text{x}$ :Đặt $\text{x=2}\sin t,t\in \left[ -\dfrac{\pi }{2};\dfrac{\pi }{2} \right];d\text{x}=2\cos tdt$.
Khi $x=-1\Rightarrow t=-\dfrac{\pi }{6}$ và $x=1\Rightarrow t=\dfrac{\pi }{6}$.
Ta có: ${{I}_{1}}=\int\limits_{-\dfrac{\pi }{6}}^{\dfrac{\pi }{6}}{\sqrt{4(1-{{\sin }^{2}}x)}}\cos tdt=4\int\limits_{-\dfrac{\pi }{6}}^{\dfrac{\pi }{6}}{{{\cos }^{2}}tdt}$ (Do $\cos t\ge 0$ khi $t\in \left[ -\dfrac{\pi }{2};\dfrac{\pi }{2} \right]$ )
$2\int\limits_{-\dfrac{\pi }{6}}^{\dfrac{\pi }{6}}{(1+\cos 2t)dt}=\left. 2(t+\dfrac{1}{2}\sin 2t) \right|_{-\dfrac{\pi }{6}}^{\dfrac{\pi }{6}}=2(\dfrac{\pi }{3}+\dfrac{\sqrt{3}}{2})$.
Vậy $S=2(\dfrac{\pi }{3}+\dfrac{\sqrt{3}}{2})-\dfrac{2\sqrt{3}}{3}=\dfrac{2\pi +\sqrt{3}}{3}$.