Water in a U-tube I'm currently working on a problem that states:

a U-tube has vertical arms of radii $r$ and $2r$.... The U-tube contains liquid up to a height $h$ in each arm. The liquid is set oscillating and at a given instant the liquid in the norrower arm is a distance $y$ above the equilibrium level. Show that the potential $U=\frac 58g\rho\pi r^2y^2$.

First of all I want to check if I'm understanding how to get the potential energy when the vertical arms have the same radius $r$. If the tube is filled up to $h$, and adittionaly in one of the arms a block of water of height $y$ is put, then $U=\int_0^ydm\,gy'=\int_0^y\pi r^2dy'\rho gy' = \frac12 \pi r^2\rho g y^2$. What I did is to consider a disk of water of height $dy'$ and calculate $dU$, then I integrated over the pile of water of height $y$ to get the total potential energy. I've seen on the internet (here) that this potential energy is actually $\rho Agy^2$ (the $\frac12$ is missing), and don't understand why.
Secondly, I'm not seeing why $\frac 58$ appears in the problem. It has to do with the difference in radius between the arms, but don't know how to write it down.
I appreciate your help. Thanks.
 A: The $\frac{1}{2}$ is missing in their derivation because the potential you are both calculating is different. You integrate from $0$ to $y$. In here is the implicit assumption that at the equilibrium height there is zero potential. 
In the derivation you linked, they use the formula $U = \Delta mgy$. What they are doing is taking a column of fluid of height y, and then raising it by a distance $y$. This makes a lot of sense if both sides of the tube are the same radius because you directly get the potential difference. But using your convention, what they should do is $\int_{-y}^{0}dmg y'=-\frac{1}{2}\pi r^2 g\rho y^2$. With your convention, the original potential of the fluid element they consider has a negative potential energy. The total difference between the two potentials using your convention is still equal to $\pi g\rho r^2y^2$. 
So you're on the right track to the solution. The next step is to realize that if one tube is a distance $y$ above equilibrium, then you know the volume of liquid above the equilibrium line. This is equal to the volume of liquid missing in the larger tube. Thus you can relate the height increase in the smaller tube with the height decrease in the larger tube. With this you can calculate the potential energy associated with the missing column of water in the large tube and find the total potential of the system.
