Differential equation word problem water leaking $y=x^2$ A tank has the shape of a parabola $y=x^2$ revolved around the y-axis. Water leaks from a hole area $B= .0005 m^2$ at the bottom, let $y(t)$ be the water level at time $t$. How long does it take for the tank to empty if it is initially filled tom height $y_0 = 1 m$
I can tell this is an initial condition differential equation, but I am having trouble setting it up. Is my cross section $\pi x^4$?
I know the basic formula is $\frac{dy}{dt} = \frac{Bv(y)}{A(y)}$
 A: There are two principles at work here: 1) Bernoulli's Principle, which states that
$$2 g y + v(y)^2 = v(0)^2$$
where $v=dy/dt = \dot{y}$ is the rate at which the fluid is sinking, and $y$ is the height of the fuid.  Also, $g$ is the acceleration due to gravity $\approx 9.8 \text{m}/\text{sec}^2$.  $v(0)$ is the speed of the fluid exiting the hole at the bottom.
2) $A(y) v(y) = B v(0)$ - this is a statement that the amount of fluid exiting the container is uniform throughout - sort of a conservation principle.  This allows us to get a diffrential equation for the height of the fluid at any time:
$$\dot{y}^2 = \frac{2 g y}{\pi^2 y^2/B^2 - 1}$$
Note that I used the fact that the area of the fluid $A(y)$ at height $y$ is $\pi y$ for this container.  
In principle, we have a simple ODE which may be expressed in terms of an integral over $y$; the integral, however, is pretty horrid (I get elliptic integrals over imaginary arguments). Nevertheless, we may exploit the fact that the area of the hole at the bottom is very small so that the area of the fluid above the hole at any time is much greater than $B$.  We may therefore neglect the $1$ in the denominator and get the approximate DE:
$$\dot{y} = \pm \frac{B}{\pi}\sqrt{\frac{2 g}{y}}$$
We choose the negative sign because $y$ is decreasing.  The solution to this equation takes the form
$$y^{3/2} = y(0)^{3/2} - \frac{3 B}{2 \pi} \sqrt{2 g} t$$
You may then find the approximate time at which the container is empty by setting the above to zero.
A: If $V$ is the volume of water inside the tank, then $\frac{dV}{dt} = -B$. Observe that $V=\int_{0}^{2 \pi} d\phi \int_0^{\sqrt y} r^2 r dr$. Integrate the latter and substitute the result into the former to get an ODE for $\frac{dy}{dt}$. Solve it using $y(0) = 1$ and $y(t) = 0$ to find the desired $t$.
