Lời giải chi tiết:

\(+)\,V={{V}_{\left( N \right)}}-{{V}_{\left( N' \right)}}=\frac{1}{3}\pi {{R}^{2}}SB-\frac{1}{3}\pi R{{'}^{2}}SA\,\left( * \right)\)

+) Ta có:

\(\begin{align} & \frac{SA}{SB}=\frac{R'}{R}\Rightarrow \frac{SA}{SB-SA}=\frac{R'}{R-R'} \\  & \Rightarrow SA=\frac{hR'}{R-R'}\Rightarrow SB=SA+h=\frac{hR}{R-R'} \\ \end{align}\)

+) Thay SA, SB vào phương trình (*) ta có:

\(\begin{align} & V=\frac{1}{3}\pi {{R}^{2}}SB-\frac{1}{3}\pi R{{'}^{2}}SA\, \\ & \Leftrightarrow V=\frac{1}{3}\pi {{R}^{2}}.\frac{hR}{R-R'}-\frac{1}{3}\pi R{{'}^{2}}.\frac{hR'}{R-R'} \\ & \Leftrightarrow V=\frac{\pi h\left( {{R}^{3}}-R{{'}^{3}} \right)}{3\left( R-R' \right)}=\frac{\pi h}{3}\left( {{R}^{2}}+R{{'}^{2}}+RR' \right) \\ \end{align}\)