How to calculate radius of the surface of water at different points in time when it is poured into a spherical container? So I have a sphere with radius $1\ cm$, and I'm pouring in water at $0.5\ cm^3/s$. 
How would I find a the radius of the surface of the water at any given time? So in the end this would look like a quadratic equation relating time and radius, where the maxima would be where radius $= 1\ cm$. How would I go about using differentiation in solving for this equation?
 A: Notice, let $r$ be the radius of the circular surface of water at any time $t$ then the volume of water poured (in a sphere of radius $R=1\ cm$ ) at the same time $t$ is given by using geometry $$\color{red}{V=\frac{\pi}{3}\left(2R^3-(2R^2+r^2)\sqrt{R^2-r^2}\right)}$$
Differentiating w.r.t. time $t$, we get
$$\frac{dV}{dt}=\frac{\pi}{3}\left(\frac{(2R^2+r^2)r}{\sqrt{R^2-r^2}}-2r\sqrt{R^2-r^2}\right)\frac{dr}{dt}$$
$$\frac{dV}{dt}=\frac{\pi r^3}{\sqrt{R^2-r^2}}\frac{dr}{dt}$$$$\implies \frac{dr}{dt}=\frac{\sqrt{R^2-r^2}}{\pi r^3}\frac{dV}{dt}$$
Now, setting the values $R=1\ cm$ & constant rate of pouring water $\frac{dV}{dt}=0.5\ cm^3/s$, we get
$$\frac{dr}{dt}=\frac{\sqrt{1-r^2}}{\pi r^3}(0.5)$$
$$\frac{2\pi r^3\ dr}{\sqrt{1-r^2}}=dt$$$$\implies \int \frac{2\pi r^3\ dr}{\sqrt{1-r^2}}=\int dt$$
$$\frac{-2\pi(2+r^2)\sqrt{1-r^2}}{3}=t+C$$
$\color{blue}{\text{Condition}}$: At initial time $t=0$, the radius of water surface $r=0$, hence, we get
$$\frac{-2\pi(2+0)\sqrt{1-0}}{3}=0+C\implies C=\frac{-4\pi}{3}$$
Now, setting the value of $C$, we get
$$\frac{-2\pi(2+r^2)\sqrt{1-r^2}}{3}=t-\frac{4\pi}{3}$$
$$\frac{2\pi(2+r^2)\sqrt{1-r^2}}{3}=\frac{4\pi}{3}-t$$
$$\color{red}{(1-r^2)(2+r^2)^2=\left(\frac{4\pi-3t}{2\pi}\right)^2}$$

Comment :
Setting maximum value of radius, $r=1\ cm$, we get
$$t=\frac{4\pi}{3}$$
We conclude that the maximum radius $\color{red}{r=1\ cm}$ of water surface will occur at the time $\color{red}{t=\frac{4\pi}{3} \ sec}$

