Xét phương trình phần đường bao:
\(\left(x+3\right)^2+\left(y+1\right)^2=1\Leftrightarrow\left(y+1\right)^2=1-\left(x+3\right)^2\)
\(\Leftrightarrow y+1=\pm\sqrt{1-\left(x+3\right)^2}\) (với \(-4\le x\le-2\))
\(\Leftrightarrow y=-1\pm\sqrt{1-\left(x+3\right)^2}\)
\(V=\pi\int\limits^{-2}_{-4}\left[\left(-1-\sqrt{1-\left(x+3\right)^2}\right)^2-\left(-1+\sqrt{1-\left(x+3\right)^2}\right)^2\right]dx\)
\(=\pi\int\limits^{-2}_{-4}4\sqrt{1-\left(x+3\right)^2}dx\)
Đặt \(x+3=sint\Rightarrow dx=cost.dt\) ; \(\left\{{}\begin{matrix}x=-4\Rightarrow t=-\dfrac{\pi}{2}\\x=-2\Rightarrow t=\dfrac{\pi}{2}\end{matrix}\right.\)
\(V=\pi\int\limits^{\dfrac{\pi}{2}}_{-\dfrac{\pi}{2}}4cost.cost.dt=2\pi\int\limits^{\dfrac{\pi}{2}}_{-\dfrac{\pi}{2}}\left(1+cos2t\right)=\pi\left(t+\dfrac{1}{2}sin2t\right)|^{\dfrac{\pi}{2}}_{-\dfrac{\pi}{2}}=2\pi^2\)
Có vẻ cả 4 đáp án đều không chính xác
