From a solid right circular cylinder with height 10 cm and radius of the base 6 cm, a right circular cone of the same height and same base are removed. Find the volume of the remaining solid.

Given,

Height of the cylinder (H) = 10 cm

Radius of the base of cylinder (R) = 6 cm

Height of the cone (h) = 10 cm

Radius of the base of cone (r) = 6 cm

Volume of remaining part (V) = Volume of cylinder - Volume of cone

$V = πR^2H - \dfrac{1}{3}πr^2h \\[1em] = \dfrac{22}{7} \times 6^2 \times 10 - \dfrac{1}{3} \times \dfrac{22}{7} \times 6^2 \times 10 \\[1em] = \dfrac{22}{7} \times 6^2 \times 10 \times \Big(1 - \dfrac{1}{3}\Big) \\[1em] = \dfrac{22}{7} \times 36 \times 10 \times \dfrac{2}{3} \\[1em] = \dfrac{22 \times 12 \times 10 \times 2}{7} \\[1em] = \dfrac{5280}{7} \\[1em] = 754\dfrac{2}{7} \text{ cm}^3.$

Hence, volume of remaining solid = $754\dfrac{2}{7}$ cm³.