Cho hình phẳng D giới hạn bởi parabol y=-1/2x^2+2x

Diện tích hình phẳng D là

$S=\underset{0}{\overset{4}{\int }}\left(\sqrt{16-x^2}-\left(-\frac{1}{2}{x}^2+2\text{x}\right)\right)d\text{x}=\underset{0}{\overset{4}{\int }}\sqrt{16-x^2}d\text{x}-\underset{0}{\overset{4}{\int }}\left(-\frac{1}{2}{x}^2+2\text{x}\right)d\text{x}$.

Xét $I=\underset{0}{\overset{4}{\int }}\sqrt{16-x^2}d\text{x}$ . Đặt$x=4\sin t\Rightarrow d\text{x}=4\cos tdt$ . Đổi cận:$\left\{\begin{array}{l}x=0\Rightarrow t=0\\ x=4\Rightarrow t=\frac{\pi }{2}\end{array}\right.$

$I=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\sqrt{16-16{\sin }^{2}t}.4\cos tdt=16\underset{0}{\overset{\frac{\pi }{2}}{\int }}{\cos }^{2}tdt=8\underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(1+\cos 2t\right)dt={\left.8\left(t+\frac{1}{2}\sin 2t\right)\right|}_0^{\frac{\pi }{2}}=8\frac{\pi }{2}=4\pi$.

Xét$J=\underset{0}{\overset{4}{\int }}\left(-\frac{1}{2}{x}^2+2\text{x}\right)d\text{x}={\left.\left(\frac{-x^3}{6}+x^2\right)\right|}_0^4=\frac{16}{3}$ . Vậy $S=4\pi -\frac{16}{3}$ .