Cho (H) là hình phẳng giới hạn bởi Parabol y=căn bậc hai 3 . x^2

* Hướng dẫn giải

Đáp án B

Xét phương trình tương giao:

$\begin{array}{l}\sqrt{3}{x}^2=\sqrt{4-x^2}\iff 3x^4=4-x^2\iff \left[\begin{array}{l}{x}^2=1\Rightarrow x=\pm 1\\ x^2=-\frac{4}{3}\left(L\right)\end{array}\right.\\ S=\underset{0}{\overset{1}{\int }}\sqrt{3}{x}^{2}dx+\underset{1}{\overset{2}{\int }}\sqrt{4-x^2}dx=\frac{\sqrt{3}{x}^3}{3}\left|\begin{array}{l}1\\ 0\end{array}\right.+S_2\\ S_2:x=2\sin t,t\in (-\frac{\pi }{2};\frac{\pi }{2})\Rightarrow dx=2\cos tdt\\ S_2:\underset{\frac{\pi }{6}}{\overset{\frac{\pi }{2}}{\int }}2\cos t.2\cos tdt=\underset{\frac{\pi }{6}}{\overset{\frac{\pi }{2}}{\int }}4{\cos }^{2}tdt=2\underset{\frac{\pi }{6}}{\overset{\frac{\pi }{2}}{\int }}(1+\cos 2t)dt\\ =2[t+\frac{\sin 2t}{2}\text{]}\left|\begin{array}{l}\frac{\pi }{2}\\ \frac{\pi }{6}\end{array}\right.=\frac{2\pi }{3}-\frac{\sqrt{3}}{2}\\ \Rightarrow S=\frac{\sqrt{3}}{3}+\frac{2\pi }{3}-\frac{\sqrt{3}}{2}\end{array}$