A solid, consisting of a right circular cone standing on a hemisphere, is placed upright in a right circular cylinder, full of water, and touches the bottom. Find the volume of water left in the cylinder, having given that the radius of the cylinder is 3 cm and its height is 6 cm; the radius of the hemisphere is 2 cm and the height of cone is 4 cm. Give your answer to the nearest cubic centimeter.

Given,

Height of cylinder (H) = 6 cm

Radius of cylinder (R) = 3 cm

Height of cone (h) = 4 cm

Radius of cone = Radius of hemisphere = r = 2 cm

Volume of water left in cylinder = Volume of cylinder - Volume of cone - Volume of hemisphere

$= πR^2H - \dfrac{1}{3}πr^2h - \dfrac{2}{3}πr^3 \\[1em] = π\Big(R^2H - \dfrac{1}{3}r^2h - \dfrac{2}{3}r^3\Big) \\[1em] = \dfrac{22}{7} \times \Big(3^2 \times 6 - \dfrac{1}{3} \times 2^2 \times 4 - \dfrac{2}{3} \times 2^3\Big) \\[1em] = \dfrac{22}{7} \times \Big(54 - \dfrac{16}{3} - \dfrac{16}{3}\Big) \\[1em] = \dfrac{22}{7} \times \Big(54 - \dfrac{32}{3}\Big) \\[1em] = \dfrac{22}{7} \times \Big(\dfrac{162 - 32}{3}\Big) \\[1em] = \dfrac{22}{7} \times \dfrac{130}{3} \\[1em] = \dfrac{2860}{21} \\[1em] = 136 \text{ cm}^3.$

Hence, volume of water left in cylinder = 136 cm³.