A hollow glass bowl in the shape of a sphere of radius 4” is filled with water to a depth of 3”. If the water flows out through a small hole in the bottom so that the level drops 2”, how much water has escaped in in^3?

$V = \dfrac{\pi h^2}{3}(3R - h)$
 

Original volume of water (depth = h₁):
$V_o = \dfrac{\pi(3^2)}{3}\left[ 3(4) - 3 \right]$

$V_o = 27\pi ~ \text{in}^3$
 

Volume of remaining water (depth = h₂):
$V_r = \dfrac{\pi(1^2)}{3}\left[ 3(4) - 1 \right]$

$V_r = \frac{11}{3}\pi ~ \text{in}^3$
 

Required volume (Volume of water that escaped from the bowl):
$V = V_o - V_r = 27\pi - \frac{11}{3}\pi$

$V = \frac{70}{3}\pi ~ \text{in}^3 = 73.304 ~ \text{in}^3$