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If a sphere is cut by a plane, what will be the volume of the smaller segment?
Take a sphere radius  
\[r\]
, centre the origin, and let the sphere be cut by a horizontal plane  
\[y=a\]
. We can consider the sphere segment to be generated by rotating an arc of the circle about the  
\[y\]
  axis. We can write  
\[x^2+y^2=r^2 \]
  for a circle in the  
\[xy\]
  plane, then  
\[y^2=r^2-x^2\]
. The volume of revolution is
\[\begin{equation} \begin{aligned} V &= \pi \int^r_a (r^2-x^2)dx \\ &= \pi \int^r_a r^2 dx - \pi \int^r_a x^2dx \\ &= \pi [r^2x]^r_a - \pi [\frac{x^3}{3} ]^r_a \\ &=\pi (r^3-r^2a)- \pi ( \frac{r^3}{3} - \frac{a^3}{3} ) \\ &= \pi (r-a)(r^2- \frac{r^2}{3}- \frac{ar}{3}- \frac{a^2}{3} ) \\ &= \frac{\pi (r-a)}{3}(2r^2-ar-a^2) \end{aligned} \end{equation}\]