Hemispherical bowl

Volume of the hemisphere,

$V_h =$ $\dfrac{2}{3}$ $\pi r^3$

$V_h =$ $\dfrac{2}{3}$ $\pi 9^3 = 486\pi$

Volume of the cylinder,

$V_c = \pi r^2h$

$V_c = \pi 3^2 \times 6 = 54\pi$

Therefore, the number of bottles required $\implies$ $\dfrac{486\pi}{54\pi}$ $=\boxed{9}$

The volume of a sphere is $v=\dfrac{4}{3} \pi r^3$ , but a hemisphere is half of a sphere so the volume of a hemisphere is $\dfrac{1}{2}\left(\dfrac{4}{3}\right)(\pi r^3)=\dfrac{1}{2}\left(\dfrac{4}{3}\right)(\pi)(9^3)=486 \pi$ . The volume of a cylinder is $\pi r^2 h=(3^2)(6)=54 \pi$ . So the number of bottles required is $\dfrac{486 \pi}{54 \pi} =$ $\color{#D61F06}\large \boxed{9}$