Q:

Rotational volume and differential equation

A container with the shape of some function y=f(x) is rotated around the y-axis. It's filled with a fluid and has a hole in the bottom where the fluid leaks out. The rate of liquid flowing out should be proportional to the square root of the height of liquid in the container. 
Determine the function f(x) such that the height of liquid in the container has a constant negative slope.  
$v(t)$ is the volume of water in the container at time t
$v'(t)$ is the rate of change in volume at time t
$h(t)$ is the height of the water level in the container.
$h'(t)$ is the rate of change of the water level and its constant. 
$v'(t)=-k*\sqrt{h(t)}$
$v(t)=\pi\int^{h(t)}_{0}f^2(x) dx$
I feel as if maybe the fundamental theorem of calculus would get involved here.
$v'(t)=\pi f^2(h(t))$
But I don't really get anywhere from here. Am I going at this from the wrong direction?
The answer is supposed to be $f(x)=Kx^4$, where K is a constant

A:

Let $x = g(y)$ be the inverse function of $y=f(x)$. The volume of the liquid, w.r.t the height $y$ is given by
$$ V(y) = \pi \int_0^y g(s)^2 ds$$
By the fundamental theorem of calculus:
$$ \frac{dV}{dy} = \pi g(y)^2 $$
By the chain rule
$$ \frac{dV}{dt} = \frac{dV}{dy}\frac{dy}{dt} = -c\pi g(y)^2 $$
where $\frac{dy}{dt}=-c$ is some negative constant
It is also given that
$$ \frac{dV}{dt} = -k\sqrt{y} $$
Therefore
$$ -c\pi g(y)^2 = -k\sqrt{y} \implies x = g(y) = \sqrt{\frac{k}{c\pi}}y^{1/4} $$
Inverting gives
$$ y = f(x) = \left(\frac{c\pi}{k}\right)^2x^4 = Kx^4 $$

