Lời giải

\( -x^2 + 2x = 0  \Leftrightarrow \left[ \begin{array}{l} x = 0 \\ x = 2 \end{array} \right. \)

a) \( V = \pi \int_0^2 (-x^2 + 2x)^2 \, dx = \frac{16\pi}{15} \, \) (đvtt)

b) \( y = -x^2 + 2x = 1 - (x - 1)^2 \iff (x - 1)^2 = 1 - y \)

\( \Leftrightarrow x - 1 = \pm \sqrt{1 - y} \Leftrightarrow \left[ \begin{array}{l} x = 1 + \sqrt{1 - y} \\ x = 1 - \sqrt{1 - y} \end{array} \right. \)

\( V = \pi \int_0^1 \Big[ (1 + \sqrt{1 - y})^2 - (1 - \sqrt{1 - y})^2 \Big] \, dy \)

\( = \pi \int_0^1 \Big( 4\sqrt{1 - y} \Big) \, dy = 4\pi  (1 - y)^{\frac{3}{2}}(\frac{-2}{3}) \bigg|_0^1 = \frac{8\pi}{3}\) (đvtt)