How else can we describe the volume of a lagrangian submanifold in a Kahler manifold? Suppose $(V^{2g}, g, \omega, J)$ is an almost Kahler manifold. ie. $(V,\omega)$ is a symplectic manifold with $\omega$-compatible almost complex structure $J$ ($J$ is a symplectomorphism) and such that $\omega(\cdot, J\cdot)$ coincides with a riemannian metric $g$ on $V$. 
Suppose further that $L$ is a lagrangian submanifold of $V$. Then $L$ has a volume, as defined with respect to the riemannian metric $g$. 
Nonetheless I want to believe that there still is an alternative expression of the metric quantity $vol(L,g)$ which exploits both the (almost) Kahler structure and the fact that the submanifold $L$ is lagrangian (ie. totally isotropic) and that we have a canonical decompostion $TM|L=TL \oplus T JL$, ie. the normal bundle of $L$ is precisely $JL$. 
Can anybody testify or provide some evidence or enlightenment on this matter? 
Alternatively, an even simpler question: if we suppose $V$ is closed, then can we give an expression to the volume of $V$ (wrt g) in terms of $vol(L,g)$ which depends essentially on the circumstances of $L$ being lagrangian and $g=\omega(\cdot, J\cdot)$?
 A: This is probably closer in spirit to what you're looking for than what you've received in the comments. If $(V^{2m}, J, \omega, g)$ is Calabi-Yau (which for me means that $J$ is integrable, and the first Chern class $c_1(V) = 0$), then one can say much more. In this case there exists a holomorphic nowhere vanishing $(m,0)$ form $\Omega$, called the "holomorphic volume form." The form $\Omega$ is unique up to multiplication by a nowhere vanishing holomorphic function. We don't need to assume that $\Omega$ is parallel, but Yau's theorem does tell us (if $V$ is compact) that we can change the metric, keeping the Kaehler class $[\omega]$ unchanged, to make $\Omega$ parallel (and consequently also the new metric will be Ricci-flat.) But I have digressed.
In such a situation, if $L$ is Lagrangian, then it is well known that the restriction of $\Omega$ to $L$ is equal to $e^{i \theta} \mathrm{Vol_L}$, where $\mathrm{Vol}_L$ is the volume form of $L$ (with the induced metric) and $e^{i \theta}$ is the "phase" of the Lagrangian, where $\theta : L \to \mathbb R/ (2 \pi \mathbb Z)$ is a smooth, multivalued function on $L$. In addition, the mean curvature $H$ of $L$ in $V$ is given by $H = J \nabla \theta$. So the minimal Lagrangian submanifolds (vanishing mean curvature) correspond to those with constant phase function $\theta = \theta_0$. In this case, if $L$ is compact, then the volume of $L$ is given by
$\mathrm{Vol} (L) = \int_L \mathrm{Vol}_L = \int_L e^{-i \theta_0} \Omega = e^{- i \theta_0} [\Omega] \cdot [L],$
which is topological. (It looks complex, but it's actually real, because $[\Omega]$ is a class in $H^g(V, \mathbb C)$.) If you prefer, you can just replace $\Omega$ by $e^{- i \theta_0} \Omega$ to get rid of the phase factor.
Such "minimal Lagrangian" submanifolds, whose volume is purely topological, are also called special Lagrangian submanifolds, and are widely studied in calibrated geometry and differential geometric approaches to mirror symmetry. The best place to start looking is the text "Riemannian Holonomy Groups and Calibrated Geometry" by Dominic Joyce and its multiple references.
I'm not sure how much of this will extend to the case of $J$ non-integrable and $c_1(V) \neq 0$. I'd have to think about it.
A: Take an open rectangle in $\mathbb R^2$ with side lengths $a$ and $b$. This is a Riemannian manifold, thus symplectic, and has an almost complex structure. The center line is a Lagrangian submanifold and has volume $a$.
Two versions of this whole ensemble are symplectomorphic if $a_1b_1=a_2b_2$. So the volume of the center line is not invariant under symplectomorphisms.
