On an Kähler manifold $(M,g)$ of complex dimensional $n$, equipped with a Kähler form related from the metric as
$$ \omega = \frac{i}{2} g_{\mu \bar{\nu}} dz^{\mu} \wedge d\bar{z}^{\nu},$$
we can construct the corresponding volume form
$$ \omega^n \triangleq \omega \wedge \cdots \wedge \omega = \left(\frac{i}{2}\right)^n n! \det (g_{\mu \bar{\nu}})dz^1 \wedge d\bar{z}^1 \cdots dz^n \wedge d\bar{z}^n$$
Using the identity $dz^j \wedge d\bar{z}^j = (-2i) dx^j \wedge dy^j$, and the fact that the determinant of the Hermitian part of the metric, $g_{\mu \bar{\nu}}$ is the square root of the determinant of the associated Riemannian metric $g$, this becomes:
$$ \omega^n = n! \sqrt{\det(g)}dx^1 \wedge dy^1 \cdots dx^n \wedge dy^n $$
which is a volume form over a real $2n$ dimensional manifold.
My question is if the $\sqrt{\det(g)} dz^1 \wedge d\bar{z}^1 \cdots dz^n \wedge d\bar{z}^n$ factor can be directly treated as a Riemannian volume form, and used to compute volume integrals, instead of using $\sqrt{\det(g)}dx^1 \wedge dy^1 \cdots dx^n \wedge dy^n$
For context, I am trying to numerically evaluate volume integrals using Monte Carlo integration, for example, to compute the volume of $\mathbb{CP}^n$ say, I would like to sample from $\mathbb{CP}^n = S^{2n+1}/S^1 $ (just a $(2n+1)$-dim sphere with the necessary identificatons, and do the following integral:
$$ \textrm{vol}(\mathbb{CP}^n) = \int_{\mathbb{CP}^n} \omega^n = ? $$
which just involves samples from the sphere, instead of doing the real $2n$-dim integral, which needs samples over an infinite domain, which is less nice to do computationally
$$\textrm{vol}(\mathbb{CP}^n) = \int_{\mathbb{R}^{2n}} \sqrt{\det(g)}dx^1 \wedge dy^1 \cdots dx^n \wedge dy^n$$
Just doing the simple example with $n=1$, we expect $\textrm{vol}(\mathbb{CP}^n) = \pi$, and I would like to approximate this just by Monte-Carlo sampling from the sphere $S^3$ instead of sampling from $\mathbb{R}^2$, i.e.
$$ \textrm{vol}(\mathbb{CP}^1) = \int_{\mathbb{CP}^1} \textrm{integrand?} = \int_{\mathbb{R}^2} \frac{ dx \wedge dy}{(1 +x + y)^2} $$
but I don;t know what the integrand should be for the former case...
