Wedge product of a one- form and a Kähler form

Let $x$, $y$, $v$, $w$ be coordinates on $R^{4}$ and $g$ be the Riemannian metric whose matrix with respect to these coordinates is $$g=\left ( \begin{array} {cccc} 1 & 0 & -kx & 0\\ 0 & \alpha & ky\alpha & kx\\ -kx & ky\alpha & \alpha\beta & k^{2}xy\\ 0 & kx & k^{2}xy & 1 \end{array}\right ),$$ where $\alpha = 1 + k^{2}x^{2}$ and $\beta = 1 + k^{2}y^{2}$.

We consider on $R^{4}$ the $(1,1)$ tensor $J$:

$$J\frac{\partial}{\partial x} = \frac{\partial}{\partial w},\; J\frac{\partial}{\partial y} = -ky\frac{\partial}{\partial y} + \frac{\partial}{\partial v},\; J\frac{\partial}{\partial v} = -\beta \frac{\partial}{\partial y} + ky\frac{\partial}{\partial v},\; J\frac{\partial}{\partial w} = - \frac{\partial}{\partial x}.$$

If $\Omega$ is the Kähler 2-form, that is $\Omega (X,Y) = g(X,JY)$, and $\omega = kdv$, then $d\Omega = \omega \wedge \Omega$ holds.

I wanted to check that $d\Omega = \omega \wedge \Omega$ is valid. First, I calculated $d\Omega$ as:

$$d\Omega = -X^{1}Y^{2}kdx - X^{1}Y^{3}k^{2}xdy - X^{1}Y^{3}k^{2}ydx + X^{2}Y^{1}kdx - 2X^{2}Y^{3}k^{2}xdx + X^{3}Y^{1}k^{2}ydx + X^{3}Y^{1}k^{2}xdy + 2X^{3}Y^{2}k^{2}xdx + X^{3}Y^{4}kdx - X^{4}Y^{3}kdx,$$

where $X=X^{1}\frac{\partial}{\partial x} + X^{2}\frac{\partial}{\partial y} + X^{3}\frac{\partial}{\partial v} + X^{4}\frac{\partial}{\partial w}$ and $Y=Y^{1}\frac{\partial}{\partial x} + Y^{2}\frac{\partial}{\partial y} + Y^{3}\frac{\partial}{\partial v} + Y^{4}\frac{\partial}{\partial w}$.

So, in the expression for $d\Omega$ there is no $dv$. I don't know how will I calculate $\omega \wedge \Omega$. Did I make a mistake in calculation of $d\Omega$, should it be calculated as:

$$d\Omega (X,Y,Z) = X\Omega (Y,Z) - Y\Omega (X,Z) + Z\Omega (X,Y) - \Omega ([X,Y],Z) + \Omega ([X,Z],Y) - \Omega ([Y,Z],X)?$$