Applying differential forms to non-vectors $\omega(X,Y)$ $\omega=2xdx\wedge dy + y^2dx\wedge dz$
$X=x^2y\frac{\partial}{\partial y} + x\frac{\partial}{\partial z}$
$Y=x\frac{\partial}{\partial y}$
Calculate $\omega(X,Y)$
I understand how to apply the differential form onto parameters given vectors $(v1,v2)$, where you take  the determinant. Would I perform the same for this operation?
I am trying this, but I'm not sure where to go from here. Should I be taking the partials as each parameter?
$2xdet\begin{bmatrix}
0&0\\ x^2y&x\\ \end{bmatrix} + y^2det\begin{bmatrix}
0&0\\ x&0\\ \end{bmatrix} = 0$
I have looked through Do Carmo Differential Forms, and Faris' Vector fields and differential forms, but have not been able to find an explanation or example.
 A: We wish to compute: $$\omega(X,Y) = (2xdx\wedge dy+y^2dx\wedge dz)(x^2y\partial_y+x\partial_z,x\partial_y).$$
For the first term,
$$2x[dx(x^2y\partial_y + x\partial_z)dy(x\partial_y)-dy(x^2y\partial_y + x\partial_z)dx(x\partial_y)] = 2x(0-0)=0.$$
So the final answer is just the second term, which is
$$y^2[dx(x^2y\partial_y + x\partial_z)dz(x\partial_y)- dz(x^2y\partial_y + x\partial_z)dx(x\partial_y)] = y^2(0-x\cdot0) = 0.$$
So the whole thing is zero (unless I've made a mistake, in which case please comment). I hope this makes the method clear - with practice, you should be able to do these computations fast (skipping the LHS and just writing down the middle and final terms).
A: Your phrase "apply the differential form onto parameters" suggests you're not aware of the isomorphism $T_p(\Bbb R^n) \cong \Bbb R^n$. This explicit correspondence means, for instance, that $$T_{(x,y,z)}(\Bbb R^3) \ni x^ 2y \partial_y + x\partial_z \stackrel{\cong}{\mapsto} (0, x^2y, x) \in \Bbb R^3,$$and similarly for $Y$. In other words, if you know how to evaluate $\omega$ at $(0,x^2y,x)$ and $(0,x,0)$, then you know how to evaluate it at $x^2y\partial_y+x\partial_z$ and $x\partial_y$, there's no excuse. Namely, $\{{\rm d}x, {\rm d}y, {\rm d}z\}$ is dual to $\{\partial_x,\partial_y,\partial_z\}$, so that you have ${\rm d}x(\partial_x) = 1$, ${\rm d}x(\partial_y) = 0$, etc. So $$\begin{align} \omega(X,Y) &= 2x({\rm d}x\wedge {\rm d}y)(X,Y) + y^2({\rm d}x\wedge {\rm d}z) \\ &=2x \begin{vmatrix}{\rm d}x(X) & {\rm d}x(Y) \\ {\rm d}y(X) & {\rm d}y(Y) \end{vmatrix}+ y^2\begin{vmatrix} {\rm d}x(X) & {\rm d}x(Y) \\  {\rm d}z(X) & {\rm d}z(Y) \end{vmatrix} \\ &= 2x \begin{vmatrix} 0 & 0 
 \\ x^2y & x\end{vmatrix} + y^2 \begin{vmatrix} 0 & 0 \\ x & 0\end{vmatrix}  \\ &= 0.\end{align}$$
