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Câu hỏi: Cho hình phẳng D giới hạn bởi parabol $y=-\dfrac{1}{2}{{x}^{2}}+2\text{x}$, cung tròn có phương trình $y=\sqrt{16-{{x}^{2}}}$ với $0\le x\le 4$, trục tung (phần tô đậm trong hình vẽ). Diện tích của hình D bằng
A. $8\pi -\dfrac{16}{3}$
B. $2\pi -\dfrac{16}{3}$
C. $4\pi +\dfrac{16}{3}$
D. $4\pi -\dfrac{16}{3}$
A. $8\pi -\dfrac{16}{3}$
B. $2\pi -\dfrac{16}{3}$
C. $4\pi +\dfrac{16}{3}$
D. $4\pi -\dfrac{16}{3}$
Diện tích hình phẳng D là
$S=\int\limits_{0}^{4}{\left( \sqrt{16-{{x}^{2}}}-\left( -\dfrac{1}{2}{{x}^{2}}+2\text{x} \right) \right)d\text{x}}=\int\limits_{0}^{4}{\sqrt{16-{{x}^{2}}}d\text{x}}-\int\limits_{0}^{4}{\left( -\dfrac{1}{2}{{x}^{2}}+2\text{x} \right)d\text{x}}$.
Xét $I=\int\limits_{0}^{4}{\sqrt{16-{{x}^{2}}}d\text{x}}$. Đặt $x=4\sin t\Rightarrow d\text{x}=4\cos tdt$. Đổi cận: $\left\{ \begin{aligned}
& x=0\Rightarrow t=0 \\
& x=4\Rightarrow t=\dfrac{\pi }{2} \\
\end{aligned} \right.$
$I=\int\limits_{0}^{\dfrac{\pi }{2}}{\sqrt{16-16{{\sin }^{2}}t}.4\cos tdt}=16\int\limits_{0}^{\dfrac{\pi }{2}}{{{\cos }^{2}}tdt}=8\int\limits_{0}^{\dfrac{\pi }{2}}{\left( 1+\cos 2t \right)dt}=\left. 8\left( t+\dfrac{1}{2}\sin 2t \right) \right|_{0}^{\dfrac{\pi }{2}}=8\dfrac{\pi }{2}=4\pi $.
Xét $J=\int\limits_{0}^{4}{\left( -\dfrac{1}{2}{{x}^{2}}+2\text{x} \right)d\text{x}}=\left. \left( \dfrac{-{{x}^{3}}}{6}+{{x}^{2}} \right) \right|_{0}^{4}=\dfrac{16}{3}$. Vậy $S=4\pi -\dfrac{16}{3}$.
$S=\int\limits_{0}^{4}{\left( \sqrt{16-{{x}^{2}}}-\left( -\dfrac{1}{2}{{x}^{2}}+2\text{x} \right) \right)d\text{x}}=\int\limits_{0}^{4}{\sqrt{16-{{x}^{2}}}d\text{x}}-\int\limits_{0}^{4}{\left( -\dfrac{1}{2}{{x}^{2}}+2\text{x} \right)d\text{x}}$.
Xét $I=\int\limits_{0}^{4}{\sqrt{16-{{x}^{2}}}d\text{x}}$. Đặt $x=4\sin t\Rightarrow d\text{x}=4\cos tdt$. Đổi cận: $\left\{ \begin{aligned}
& x=0\Rightarrow t=0 \\
& x=4\Rightarrow t=\dfrac{\pi }{2} \\
\end{aligned} \right.$
$I=\int\limits_{0}^{\dfrac{\pi }{2}}{\sqrt{16-16{{\sin }^{2}}t}.4\cos tdt}=16\int\limits_{0}^{\dfrac{\pi }{2}}{{{\cos }^{2}}tdt}=8\int\limits_{0}^{\dfrac{\pi }{2}}{\left( 1+\cos 2t \right)dt}=\left. 8\left( t+\dfrac{1}{2}\sin 2t \right) \right|_{0}^{\dfrac{\pi }{2}}=8\dfrac{\pi }{2}=4\pi $.
Xét $J=\int\limits_{0}^{4}{\left( -\dfrac{1}{2}{{x}^{2}}+2\text{x} \right)d\text{x}}=\left. \left( \dfrac{-{{x}^{3}}}{6}+{{x}^{2}} \right) \right|_{0}^{4}=\dfrac{16}{3}$. Vậy $S=4\pi -\dfrac{16}{3}$.
Đáp án D.