Filling up spheres

First, we derive an explicit formula for the volume of the segment of a sphere. We use the concept of solids of revolution to do that.

The volume $V$ for any sphere filled up to a height of $x$ is

$V = \large \pi \: \int_0^{2r} \: ( \sqrt { r^{2} - (x-r)^{2} } \:)^{2} dx$

$V = \pi \: (rx^{2} - \frac{x^{3}}{3})$

Then, by virtue of related rates, we get

$\large \frac{dV}{dt} = (2\pi rx - \pi rx^{2}) (\frac{dx}{dt} )$

so

$\large \frac{dx}{dt} = \large \frac { (\frac{dV}{dt} )}{(2\pi rx - \pi rx^{2}) }$

or

$\large \frac{dx}{dt} = \large \frac { (\frac{dV}{dt} )}{\pi (x)(2r - x) }$

now, we substitute $V = \frac {325}{24}\pi$ to the predetermined equation to determine the value of the height which it corresponds to. The equation then simplifies to

$325 = 72x^{2} - 8x^{3}$

or

$(2x-5) (4x^2 - 26x + 65 )=0$

It will be obvious to select the value $x = \frac {5}{2}$ , as it is the only value which fits in the range $0 < x < 6$ which is a necessity.

So, using that value of $x$ , we find $\frac{dx}{dt}$ as

$\large \frac{dx}{dt} = \large \frac{\pi}{\pi (\frac {5}{2})(6 - \frac {5}{2}) }$

$\large \frac{dx}{dt} = \large \frac{4}{35 }$

So then the answer is $39$ .