Torricelli's law modelling Sphere Struggling connecting Water-volume to time. Could anyone be kind and guide me in the right direction? 

A spherical tank is filled with liquid. A tap in the bottom of the
  tank opens, and the liquid starts to drain. After one hour is The tank
  emptied halfway. How long does it take before it is completely empty?
  (The flow is assumed to follow Torricelli's law: its is proportional
  to the square root of the surface 

So far I'm mostly grasping at straws. 
I've come up with 
$V'(x) = -k\sqrt{h}$
Assuming r=1 =>
$V(h) = \int_0^h\pi(1-y^2)dy $
$V(h)$ is a primitive to  $V'(h) = \pi (1-h^2)$
$V(h=1)= \frac{2\pi}{3}$
$V(h=\frac{1}{2})= \frac{2\pi}{6}$
$ t = 0 \iff h = 1$,
$ t = 1 \iff h = \frac{1}{2}$
Not sure how to connect this to time further to figure out when 
$\int_0^tV'(t)dt$ = 0
 A: Hint:
For a sphere of radius $1$,
$$V(h)=\pi\int_{-1}^h(1-h^2)\,dh$$ so that
$$\frac{dV}{dt}=\frac{dV}{dh}\frac{dh}{dt}=\pi(1-h^2)\frac{dh}{dt}=-k\sqrt h.$$
Then
$$(h^{-1/2}-h^{3/2})\,dh=-k\,dt$$ and by integration
$$-2h-\frac25h^{5/2}=C-kt.$$
From the given conditions, you can draw $k$ and $C$.
A: Toricelli's law states that $$\frac{dV}{dt}(t) = -k\sqrt{h(t)}$$
To find the dependence $V(h)$, recall the formula for the volume of a spherical cap of height $h$:
$$V(h) =\pi\left(h^2-\frac{h^3}3\right)$$ 
To connect the two, use the chain rule:
$$-k\sqrt{h} = \frac{dV}{dt} = \frac{dV}{dh}\frac{dh}{dt} = \pi(2h-h^2)\dot{h}$$
Hence $$T = \int_0^T dt = \int_2^{H} \frac{\pi(2h-h^2)\,dh}{-k\sqrt{h}} = \frac{\pi}{15k}\left(3H^{5/2}-10H^{3/2} + 16\sqrt2\right)$$
because at the beginning $H = 2$.
If $T = 1$ then half the volume has flown out, i.e. $H = 1$ so $$1 = \frac{\pi}{15k}\left(3-10+16\sqrt2\right) = \frac{\pi}{15k}(-7+16\sqrt2)$$
which gives you $k$.
The entire water has flown out when $H = 0$ so
$$T = \frac{\pi}{15k} 16\sqrt{2} = \frac{16\sqrt2}{-7+16\sqrt2}$$
