Suppose that an initially full hemispherical water tank of radius 1 m has its flat side as its bottom. It has a bottom hole of radius 1 cm. If this bottom hole is opened at 1 P.M., when will the tank be empty?

My working:

$\displaystyle A(y)=\pi r^2=\pi(1-y^2)$

$\displaystyle \pi(1-y^2) \frac{dy}{dt}=-\pi\left(\frac{1}{100}\right)^2 \sqrt{2*9.8y}$

$\displaystyle \int(y^{-1/2}-y^{3/2})dy=-\int{\frac{\sqrt{19.6}}{10^4} dt$

$\displaystyle 2y^{1/2}-\frac{2}{5}y^{5/2}=-\frac{\sqrt{19.6}}{10^4} t+C$

$\displaystyle y(0)=1$

$\displaystyle C=2-\frac{2}{5}=\frac{8}{5}$

$\displaystyle 2y^{1/2}-\frac{2}{5}y^{5/2}=-\frac{\sqrt{19.6}}{10^4} t+\frac{8}{5}$

When $\displaystyle y=0$,

$\displaystyle t=\frac{80000}{5*\sqrt{19.6}}\approx 3614\ s$

The tank is empty about 14 seconds after 2:00 P.M.