Integrating 2-form 

In $\mathbb{R}^3$ I consider the compact 2-dimensional manifold
    $$
M=\left\{(x,y,z)\in\mathbb{R}^2: z=xy\right\}
$$
    which is orientated by the (global) map $\phi\colon\mathbb{R}^2\to\mathbb{R}^3, (x,y)\mapsto (x,y,xy)$.
    Furthermore a 2-form is given by
    $$
\omega:=3zdy\wedge dz+(x^2+y^2)dz\wedge dx+xzdx\wedge dy.
$$
    Consider
    $$
A:=\left\{(x,y,z)\in M: \lvert x\rvert\leq 1,\lvert y\rvert\leq 1\right\}
$$
    and my task now is to calculate the integral
    $$
\int_{A}\omega.
$$


First of all i calculated the pullback $\phi^{\star}\omega$, which is to my calculation
$$
\phi^{\star}\omega=x^2ydx\wedge dy-(x^2+y^2)dx\wedge y dx+xdy+3xydy\wedge ydx+xdy\\
=x^2ydx\wedge dy-(x^2+y^2)xdx\wedge dy-2xy^2dx\wedge dy\\
=(-x^3+x^2y-4xy^2)dx\wedge dy
$$
To my opinion it is $\phi^{-1}(A)=[-1,1]\times [-1,1]$.
So if my calculations are correct I have to calculate the following integral:
$$
\int\limits_{[-1,1]\times [-1,1]}(-x^3+x^2y-4xy^2)dx\wedge dy
$$
1.) Is this right?
2.) If yes, how can I calculate this?
Greetings!
 A: In short, your computation of $\phi^\ast \omega$ is correct, and you were on the right track. If you want to be completely pedantic, you have the following ingredients:


*

*your given map $\phi : \mathbb{R}^2 \to M$,

*the inclusion $\iota : M \to \mathbb{R}^3$,

*the $2$-form $\omega = 3z dy \wedge dz + (x^2+y^2)dz \wedge dx + xz dx \wedge dy$ on $\mathbb{R}^3$,

*the closed subset $A = \phi([-1,1]^2)$,


and you want to compute
$$
 \int_A \iota^\ast\omega = \int_{[-1,1]^2} (\iota \circ \phi)^\ast\omega = \int_{[-1,1]^2} \psi^\ast \omega,
$$
where, for convenience,
$$
 (\psi^1(x,y),\psi^2(x,y),\psi^3(x,y)) = \psi(x,y) := (\iota \circ \phi)(x,y) = (x,y,xy).
$$
First, observe that
$$
 \psi^\ast dx = d \psi^1 = \frac{\partial \psi^1}{\partial x} dx + \frac{\partial \psi^1}{\partial y} dy = dx,\\
 \psi^\ast dy = d \psi^2 = \frac{\partial \psi^2}{\partial x} dx + \frac{\partial \psi^2}{\partial y} dy = dy,\\
 \psi^\ast dz = d \psi^3 = \frac{\partial \psi^3}{\partial x} dx + \frac{\partial \psi^3}{\partial y} dy = ydx + xdy.
$$
Hence,
$$
 \psi^\ast \omega = \psi^\ast(3z dy \wedge dz + (x^2+y^2)dz \wedge dx + xz dx \wedge dy)\\
 = 3\psi^3 \psi^\ast dy \wedge \psi^\ast dz + ((\psi^1)^2+(\psi^2)^2)\psi^\ast dz \wedge \psi^\ast dx + \psi^1 \psi^2 \psi^\ast dx \wedge \psi^\ast dy\\
 = 3xy dy \wedge (ydx+xdy) + (x^2+y^2)(ydx+xdy)\wedge dx + x(xy)dx\wedge dy\\
 = -3xy^2 dx \wedge dy -x(x^2+y^2)dx \wedge dy + x^2 y dx \wedge dy\\
 = (-x^3+x^2y-4xy^2)dx \wedge dy,
$$
exactly as you computed.
Finally, since $dx \wedge dy$ is just the volume form on $\mathbb{R}^2$, it follows that, by definition,
$$
 \int_A \iota^\ast \omega = \int_{[-1,1]^2} \psi^\ast \omega = \int_{[-1,1]^2} (-x^3+x^2y-4xy^2)dx \wedge dy := \int_{[-1,1]^2} (-x^3+x^2y-4xy^2)dx dy,
$$
which is just an honest-to-goodness double integral over a rectangle in the plane.
