Answer
$8160~\pi \dfrac{in^3}{s}$
Work Step by Step
Apply the chain rule: $\dfrac{dV}{dt}=(\dfrac{\partial V}{\partial r})(\dfrac{dr}{ dt})+(\dfrac{\partial V}{\partial h})(\dfrac{dh}{dt})$
Now, $\dfrac{\partial V}{ \partial r} \approx \dfrac{\partial }{ \partial r} (\dfrac{1}{3} \pi r^2 h)=\dfrac{2}{3}\pi r^2 h$
and
$\dfrac{\partial V}{ \partial h} \approx \dfrac{\partial }{ \partial h} (\dfrac{1}{3} \pi r^2 h)=\dfrac{1}{3}\pi r^2 $
Thus, we have $\dfrac{dV}{dt}=(11200 \pi)(1.8)+(4800 \pi)(-2.5)= 8160 \pi \dfrac{in^3}{s}$
