Edit 1:
According to this PDF, second page. The differential equation for a hemisphere becomes $$(r^2h-h^2) h' = -a^2\sqrt{2gh}$$
This differs from the equation I got, which looks like:
$$\pi(r^2-h^2)h'=-a^2\sqrt{2gh}$$
Are they equivalent for some reason, or did I do something wrong?
Edit 2:
I tried out my equation, and I noticed that while $y(0) = 0.05\,m$ as I specified, I got that $y(1) \approx 0.15\,m$ which makes no sense, as the water level is supposed to decrease, not increase.
I want to calculate the volume of water inside a half-sphere (hemisphere) with a hole in the bottom of the sphere with an area $a = 5\times10^{-4}\,m^2$. The sphere has a radius $r = 0.05\,m$.
We have that $\frac{dV}{dt}=-a\sqrt{2g\cdot h(t)}$, where $V$ is volume, $g = 10\,ms^{-2}$, $h(t)$ is the height of the water level above the hole in the bottom of the half-sphere.
Using the second fundamental theorem of calculus, we have that
$$\frac{dV}{dh} = A(h) \implies A(h)\cdot h' = -a\sqrt{2g\cdot h}$$
Using the radius $r$ and the equation for circles we have:
$$x^2+h^2 = r^2 \iff x^2=r^2-h^2$$
As such, the area $A(t) = \pi(r^2-h^2)$. Our equation becomes:
$$\pi(r^2-(h(t))^2)h'(t) = -a\sqrt{2g\cdot h(t)}$$
And since it is an hemisphere, and we assume the water level starts at the top, we get the initial value $h(0) = r$.
Using Wolfram|Mathematica I tried entering the following:
$\text{DSolve}\left[\left\{\pi \left(0.05^2-y(t)^2\right) y'(t)=-0.0005 \sqrt{20 y(t)},y(0)=0.05\right\},y(t),t\right]$
DSolve[{Pi (0.05^2 - (y[t])^2) y'[t] == -0.0005 Sqrt[20 y[t]], y[0] == 0.05}, y[t], t]
Output:
$\left\{y(t)\to \text{Root}\left[967062060875971906790400 \text{$\#$1}^5-24176551521899297669760 \text{$\#$1}^3+151103447011870610436 \text{$\#$1}-3061996020736000000 t^2-7.69564\times 10^{18} t-4.83531\times 10^{18}\&,1\right]\right\}$
My question is, what does this solution even mean? There were three more that looked identical. They make no sense when they have "$\#$" and the "$\&$" symbols attached randomly, what do they mean? And most of all, have I even done this correctly?
How would I use one of these solutions to figure out the volume at $t=120 \textbf{ seconds}$
In Mathematica, $f(\#1, \#2, \dots)\&$ is an abbreviation of the function $(x_1, x_2, ...) \to f(x_1, x_2, \dots)$, and Root[$f$, $k$] returns the $k$th zero of a polynomial function $f(x)$. (I'm not sure how they're ordered.) Thus as a function of time, $y(t)$ satisifes some fifth-degree polynomial with coefficients depending on $t$. To get an explicit value at a given time $t_0$, just substitute $t_0$ for $t$ in the expression above and solve for $y$ via Root.