Cho hình trụ có hai đáy là hai hình tròn $\left( O;R...

Gọi là trung điểm$$ $AB\overset\frown{OHO'}=60{}^\circ AB=xO'ABO'H=\dfrac{x\sqrt{3}}{2}OH=\sqrt{O{{A}^{2}}-{{\left( \dfrac{AB}{24} \right)}^{2}}}=\sqrt{{{R}^{2}}-\dfrac{{{x}^{2}}}{4}}O'OHO\cos 60{}^\circ =\dfrac{OH}{{O}'H}\Leftrightarrow \dfrac{1}{2}=\dfrac{\sqrt{{{R}^{2}}-\dfrac{{{x}^{2}}}{4}}}{\dfrac{x\sqrt{3}}{2}}\Leftrightarrow \dfrac{x\sqrt{3}}{4}=\sqrt{{{R}^{2}}-\dfrac{{{x}^{2}}}{4}}\Leftrightarrow \dfrac{3{{x}^{2}}}{16}={{R}^{2}}-\dfrac{{{x}^{2}}}{4}\Leftrightarrow \dfrac{7{{x}^{2}}}{16}={{R}^{2}}\Rightarrow x=\dfrac{4R\sqrt{7}}{7}OO'=O'H\sin 60{}^\circ =\dfrac{x\sqrt{3}}{2}.\dfrac{\sqrt{3}}{2}=\dfrac{3}{4}.\dfrac{4R\sqrt{7}}{7}=\dfrac{3R\sqrt{7}}{7}V=\pi {{R}^{2}}h=\pi {{R}^{2}}.\dfrac{3R\sqrt{7}}{7}=\dfrac{3\pi {{R}^{3}}\sqrt{7}}{7}$.