1. The volume of liquid in a hemispherical bowl is given by the formula

      $V={\frac{1}{3} \pi h}^{2}(3R-h)$

where $R$ is the radius of the bowl and $h$ is the depth of liquid.

(ie the height between the bottom of the bowl and the level of the liquid).

In a particular case, liquid is leaking through a small hole in the bottom of a bowl at a rate directly proportional to the depth of liquid.

When the bowl is full, the rate of volume loss is equal to $\pi$.

Show that the rate of change of the depth of the liquid is inversely proportional to $R(h-2R)$