The symplectic geometry of cold coconuts Consider the open set  $M \subset \mathbb{C}^{2}$ given by the union of the unit ball $|z_1|^2 + |z_{2}|^2 < 1$ (the coconut) and the cylinder $|z_1| < \epsilon$, $0 < \epsilon < \! \!< 1$, (the straw, which in this case pierces the coconut through and through, but this is not important).
Fix a number $r$ strictly between $\epsilon$ and $1$. 
Does there exist a strictly positive number $c$ so that the volume of the intersection of the unit ball with any symplectic image of the ball of radius $r$ lying wholly inside $M$ is greater than $c$? If so, is there a reasonable estimate for $c$ as a function of $r$ and $\epsilon$?
By Gromov's non-squezing theorem we know we cannot symplecticaly move the whole ball of radius $r$ up the straw, but it is not clear to me how much of its volume can we sip out of the coconut.
Motivation.
This problem is related to the comments I got on this question. Both questions arose from trying to understand a comment that V.I. Arnold made in one of his papers (I forgot where, but for some reason the statement came back to my mind after many years) saying or implying (I read this a long time ago ...) that despite the non-squeezing theorem and the symplectic camel theorem people in statistical mechanics still happily interchange regions of phase space that have the same volume. Brett's answer seems to imply that Arnold's criticism should not be taken too seriously and that the physicists are right: even if you cannot exchange the regions by canonical transformations, you can exchange all of their volume up to an arbitrarily small amount. 
On the other hand, it is still possible that there exists a symplectic refinement of Poincaré recurrence. 
 A: Anton's answer is roughly what I meant by my comment above. In what follows, I'll just try to explain the intuition behind why thin straws can pretty much be as wiggly as you like. (You should be able to extend this argument to a full proof.)
It is easiest to visualize bending a symplectic cylinder when you consider it as a product of a small disk with an infinite strip, instead of as a product of a small disk with an unbounded plane. It is also a bit easier to imagine taking up all the volume with a straw with a square cross section, so I'll start with the following model for a thin straw.
$$ -2\epsilon< x_1<0, \ \ \ - \epsilon <y_1< \epsilon,  \ \ \ - \epsilon <y_2<\epsilon  $$
Visualize this straw in the coordinates $x_1$, $x_2$. We will try to bend it around the positive $x_1$-axis.
To bend this straw,  consider a Hamiltonian $H$ which is equal to 
  $x_1 y_2-x_2y_1$  where $y_2<\epsilon$ or $x_1>\epsilon^2$, equal to  0 when $x_1<0$ and which has small derivative close to 0. 
In the region where $y_2<0$ or $x_1>\epsilon^2$,  the effect of the flow of $H$ is to rotate $y_2$ into $y_1$, and $x_2$ into $x_1$. Running this flow will fold the straw back on itself until it looks roughly like a straw that has been bent almost 180 degrees.  Using a further Hamiltonian flow, you can adjust this bent straw away from $0$ to look like the union of the upper half of the original straw with the upper half of the straw
$$ \epsilon^2< x_1<2\epsilon+\epsilon^2, \ \ \ - \epsilon <y_1< \epsilon,  \ \ \ - \epsilon <y_2<\epsilon  $$
How messy this bent straw gets around $0$ where it is bent is essentially controlled by how small you can get the derivatives of $H$ around $0$. 
Using this construction, you can imagine your straw as roughly a Euclidean straw with cubic cross section, which may be bent however you like, so long as you give up a little space for each bend. With that intuition, you should be able to dream up an algorithm for taking up most of the volume with a sufficiently thin straw, and give a rigorous proof that $c=0$.
A: As it is explained in the comment of Brett Parker,
$$c=0.$$ 
It is probably clear for the specialists, but it took some time for me. 
Let me give more details.
Consider a symplectomorphism $s_1$ from a thin straw to the "coconut with the straw" which is standard outside of the coconut. 
Then you can move nearly all the volume in the image of $s_1$ from the coconut
applying a symplectomorphism $h_1$ which moves points only in the embedded straw.
Repeat the operation as many times as you want choosing a sequence of symplectomorphisms $s_n$ and $h_n$ as above. 
With the appropriate choice of symplectomorphisms you can move out nearly all the volume of the coconut.
The needed symplectomorphisms $s_n$ and $h_n$ can be assumed to be Hamiltonian. 
The choices of Hamiltonians are straightforward the support of the Hamiltonian of $s_n$ should be in the coconut 
and the support of the Hamiltonian of $h_n$ should be in the image of $s_n$.
