Can the volume of a partial sphere be re-written as $V= \frac{4\pi}{3}r^{2}h$? I am doing a related rate problem where I have a spherical surface given by the spherical equation:
$$V_{sphere} = \frac{4}{3}\pi r^3,$$
$$ρ = 4(1+cos(ϕ)),$$ bounded by the $xy$ plane
I have a rate at which water is flowing in $\frac{dV}{dt} = 25$ $\frac{ft^3}{\min}$

Taking the derivative of the Volume of sphere and setting that equal to the rate of water pumping in:
 $$25 = 4 \pi r^2 \quad \frac{dr}{dt}$$
but as I was looking at the graph, I am concerned that this will not work because the water starts at the $xy$ plane- it is not a full sphere. Can I separate the r's into $r^2h$ for the height? or will this just be taken care of during integration when I set limits? OR am I completely off base?
I am trying to find the amount of time it takes the water to reach $3$ ft.
Thank you!
edit: I also realize that I will have to convert to spherical coordinates. I was just thinking that I could convert to spherical and when I do have my height be equal to $4-ρ$ since we start with a $ρ = 4$ at the floor and the height should increase with that relationship, is that correct?
 A: The 'volume' of a sphere of radius R, (or diameter, $D=2R$) is given by:
$V=\frac{4}{3}\pi R^3$
ie
$V=\frac{\pi}{6}D^3$
If the sphere is 'partially filled' with a liquid to a level $h$, the volume of liquid inside the sphere is given by:
$V_p=\frac{\pi}{6} h^2(3D-2h)$

A: I assume the sphere is a rigid container and that the water is supposed to fill
it up to a certain level, that is, the water will be bounded above by the
plane $z = h(t)$ where $h(t)$ is the height of the water at time $t$.
It is not very useful to try to build a solution around $\frac{dr}{dt}$
in this problem, because it is not a radius $r$ that increases, it is merely
the height of the water that increases. (In fact the only changing radius is the
radius of the disk-shaped surface of the water, which is actually decreasing.)
And while you can solve this in spherical coordinates, to do so you will have to
do one integral to include the cone with the vertex at the origin of coordinates
and base at the top surface of the water, and a separate integral for
the rest of the water.
It is simpler to do this in cylindrical coordinates.
You were correct when you said the limits of integration might take care
of the problem that you started at the $x,y$-plane rather than filling the
whole sphere--in fact, they do.
For cylindrical coordinates, the lower limit of integration would occur at $z = 0.$
