A tank is in the form of a cone with the point downward, and the height and diameter are each 10 feet. How fast is the water pouring in at the moment when it is 5 feet deep and the surface is rising at the rate of 4 feet per minute?
The volume of a cone is $$V_c = \frac{1}{12} \pi D^2 H.$$
Given that the diameter of the cone is equal to its height, the volume of the cone may be written $$V_c = \frac{1}{12} \pi H^3.$$
So, $$\frac{dV}{dt} = \frac{1}{4} \pi H^2 \frac{dH}{dt} = \frac{1}{4} \pi (5)^2\times 4 = 25 \pi \textrm{ ft^3/min}$$
So, my answer: $25 \pi \textrm{ ft^3/min}.$
The book's answer: $\frac{25 \pi}{12} \textrm{ ft^3/min}.$
It might be a typo in the book. Your calculations look legit.