A hemispherical and a conical hole are scooped out of a solid wooden cylinder. The height of the solid cylinder is $7 \; cm$, radius of each of hemisphere, cone and cylinder is $3 \; cm$. The height of the cone is $3 \; cm$. Find the volume of the remaining solid. Give your answer correct to the nearest whole number. Take $\pi = 3.142$
Height of cylinder $= H = 7 \; cm$
Radius of cylinder $= R = 3 \; cm$
Radius of hemisphere $= R_S = 3 \; cm$
Radius of cone $= r = 3 \; cm$
Height of cone $= h = 3 \; cm$
Volume of cylinder $= V = \pi R^2 H$
Volume of hemisphere $= V_S = \dfrac{2}{3} \pi R_S^3$
Volume of cone $= V_C = \dfrac{1}{3} \pi r^2 h$
Volume of remaining solid $= V - \left(V_S + V_C\right)$
$= \pi R^2 H - \left(\dfrac{2}{3} \pi R_S^3 + \dfrac{1}{3} \pi r^2 h\right)$
$= \pi \left[3^2 \times 7 - \left(\dfrac{2}{3} \times 3^3 + \dfrac{1}{3} \times 3^2 \times 3\right)\right]$
$= \pi \left[63 - \left(18 + 9\right)\right] = 3.142 \times 36 = 113.112$
$\therefore \;$ Volume of the remaining solid $= 113 \; cm^3$ $\;\;\;$ [correct to the nearest whole number]