1. For a 2-dimensional phase space, they are the same.

2. More generally, for a $2n$-dimensional [symplectic manifold](s://en.wikipedia.org/wiki/Symplectic_manifold) $(M;\omega)$ with symplectic two-form 
$$\tag{1} \omega~=~\frac{1}{2} dz^I ~\omega_{IJ} \wedge dz^J, \qquad \omega_{IJ}~=~-\omega_{JI}, $$
the Poisson bracket is given by
$$ \tag{2} \{f,g\}_{PB}~=~\frac{\partial f}{\partial z^I}\pi^{IJ} \frac{\partial g}{\partial z^J}, \qquad \pi^{IJ} ~:=~(\omega^{-1})^{IJ} . $$
See also e.g. [this](http://physics.stackexchange.com/q/302288/2451) and [this](http://physics.stackexchange.com/q/201620/2451) Phys.SE posts. 


3. The canonical volume form on $(M;\omega)$ 
$$\tag{3} \Omega~=~\omega^n~=~\rho~ dz^1\wedge \ldots \wedge dz^{2n},$$
is the $n$'th exterior power of the symplectic two-form $\omega$, with volume density 
$$\tag{4} \rho~=~{\rm Pf}(\omega_{IJ}) $$
given by the [Pfaffian](http://en.wikipedia.org/wiki/Pfaffian), which is a square root of the determinant. This is closely related to [Liouville's theorem](http://en.wikipedia.org/wiki/Liouville's_theorem_%28Hamiltonian%29).