$0\le x\le 2$ thì $y=2x-{{x}^{2}}\Leftrightarrow {{x}^{2}}-2x=0$

Phương trình bậc hai theo y. Ta có $\Delta \prime =1-y,y\le 1.$

$\Rightarrow \left[ \begin{align}  & {{x}_{1}}=1-\sqrt{1-y},x\in \left[ 0;1 \right] \\  & {{x}_{2}}=1+\sqrt{1-y},x\in \left[ 1;2 \right] \\ \end{align} \right.$

${{V}_{y}}=\pi \int\limits_{0}^{1}{\left[ {{\left( 1+\sqrt{1-y} \right)}^{2}}-{{\left( 1-\sqrt{1-y} \right)}^{2}} \right]}dy=4\pi \int\limits_{0}^{1}{\sqrt{1-y}dy}$

Đặt $u=\sqrt{1-y}\Rightarrow {{u}^{2}}=1-y\Rightarrow 2udu=-dy$

Đổi cận $\left[ \begin{align}  & y=1 \\  & y=0 \\ \end{align} \right.$ $\Rightarrow \left[ \begin{align}  & u=0 \\  & u=1 \\ \end{align} \right.$

${{V}_{y}}=4\pi \int\limits_{0}^{1}{\sqrt{1-y}dy}=4\pi \int\limits_{1}^{0}{u\left( -2udu \right)=8\pi \int\limits_{0}^{1}{{{u}^{2}}du=8\pi \left[ \frac{{{u}^{3}}}{3} \right]}_{0}^{1}=\frac{8\pi }{3}}$ (đvtt)