Rate of Change of Cylindrical Roll's Volume as it Unrolls This is purely a "for-fun" question. I was minding my own business in the washroom this morning when I began to unroll some toilet paper from the roll, and in typical Breaking Bad fashion (sorry if you don't understand the reference) I had a serious moment where I wondered how the rate of change of the volume of the roll of paper changes as it is unrolled. Obviously the circumference of the cylinder keeps decreasing and as a result the ROC changes as well. I thought about it a bit but got stumped when I started writing things out (my math skills are a bit rusty).  
I believe the beginning part of this solution Accounting for changing radius of a paper roll to always unroll the same amount of paper is the correct starting point for this question.
For simplicity, assume that the paper is rolled onto itself - no cardboard piece in the centre and the innermost layer's circumference approaches 0.
Also, assume that the roll is being unravelled at 1 circumference per second ($2\Pi r$/s) of the outermost (initial) layer.
Lastly, assume that there are 100 layers (100 circles as defined in the mentioned question) and each is 1 unit-of-your-liking.
Feel free to adjust any of these parameters to your liking!
 A: Let the Initial radius be $R$
Let the Radius at time $t$ be $r_t$ [$r_t$ is a function of $t$]
At time $t$ let the angular displacement be $\theta$ then the Volume that is reduced in $dt$ is:$$r_t*d\theta*{R\over 100}*h$$
That is:
$${dV\over dt }=r_t*d\theta*{R\over 100}*h$$
This should now be solved by a Double integral over $t$ and $\theta$.
Where:
${R\over 100}$ is the thickness of the roll.
${h}$ is the height of the roll.
A: If the roll is rotating at a constant angular rate, you will remove one layer per second all the way down.  As you started with $100$ layers, it will take $100$ seconds to unroll.  The remaining volume is proportional to the square of the remaining time.  If the starting volume is $V$, at time $t$ (seconds) there is $V\frac {(100-t)^2}{100^2}$ of the initial volume remaining.
Added:  if you want the paper removed at a constant rate, the angular velocity goes up as $\frac 1r$.  This is because the linear velocity is constant and $v=r\omega$.  Then we have $\frac {dr}{dt}=-k\frac 1r, \frac 12r^2=C-kt, r=\sqrt{2(C-kt)}$.  With $r(0)=r_0, r(100)=0$ we have $C-100k=0, C=\frac 12r_0^2,\\ r=r_0\sqrt{1-\frac t{100}}$
