Evo uradio sam ja drugi dio dobijem [inlmath]1680\pi\mbox{ cm}^3[/inlmath]
[dispmath]V=V_1+V_2[/dispmath][dispmath]V_1=BH=r_1^2\pi H=144\pi\cdot 10=1440\pi\mbox{ cm}^3[/dispmath][dispmath]V_2=\frac{1}{3}BH=\frac{1}{3}144\pi\cdot 5=240\pi\mbox{ cm}^3[/dispmath][dispmath]V=240\pi+1440\pi=1680\pi\mbox{ cm}^3[/dispmath]
I odnos se dobije [inlmath]\frac{95}{84}[/inlmath]