Suppose that a cylindrical tank has a radius of 3 ft and it has a hole in the bottom with radius 1 in. How long will it take for the water, initially 9 ft deep, to drain completely?

My working:

$\displaystyle A(y) \frac{dy}{dt}=-a\sqrt{2gy}$

$\displaystyle \pi*3^2 \frac{dy}{dt}=-\pi*(1/12)^2*\sqrt{2*32y}$

$\displaystyle 1296\frac{dy}{\sqrt{y}} =-8dt$

$\displaystyle 162*2\sqrt{y}=-t+C$

$\displaystyle 324\sqrt{y}=-t+C$

$\displaystyle y(0)=9$

$\displaystyle 324*\sqrt{9}=C$

$\displaystyle C=972$

$\displaystyle 324\sqrt{y}=-t+972$

Substituting $\displaystyle y = 0$,

$\displaystyle 0=-t+972$

$\displaystyle t=972s$

Never mind! I figured it out.