Surface Integral of a 2-form Let $\alpha=x\,dx+y\,dy+z\,dz, \gamma=xy\,dz$
Let $D$ be the square $0 \leq x \leq 1, 0 \leq y \leq 1, z=1$ oriented with the upward normal. Calculate $\iint_D \alpha \wedge \gamma$.
My professor have a totally different approach then using the formula that is commonly found over the internet, which seems to work on all cases.
My attempt: 
parameterize the surface $\sigma(u,v)=(u,v,1)$  such that it is a transformation from $\mathbb{R}^2$ to $\mathbb{R}^3$ 
$\alpha \wedge \gamma=-x^2y\,dz\,dx+xy^2\,dy\,dz$ 
$dX(u,v)=\frac{dX}{du}du+\frac{dX}{dv}dv=1du$ 
$dY(u,v)=\frac{dY}{du}du+\frac{dY}{dv}dv=1dv$ 
$dZ(u,v)=\frac{dZ}{du}du+\frac{dZ}{dv}dv=0$
$$\begin{align}
\iint_D \alpha \wedge \gamma & = \iint_{D} -x^2y\,dz\,dx+xy^2\,dy\,dz \\
& = \int_0^1\int_0^1 -u^2v\,dZ\,dX+uv^2\,dY\,dZ\\
&= \int_0^1\int_0^1-u^2v(0)(1du)+uv^2(1dv)(0)\\
&= \int_0^1\int_0^1 0 \\
&= 0
\end{align}$$
Is it correct?
 A: Let's start at the beginning. We need to compute $\alpha\wedge\gamma,$ which you've almost done correctly: the answer should be
$$\alpha\wedge\gamma = x^{2}y\,dx\,dz+xy^{2}\,dy\,dz,$$
using the fact that $dz\wedge dz=0.$ So we want to compute the surface integral
$$\iint_{D}\alpha\wedge\gamma = \iint_{\{0\leq x\leq 1,\,0\leq y\leq 1, \,z=1\}}x^{2}y\,dx\,dz+xy^{2}\,dy\,dz.$$
The answer is plainly $0$, since $z$ is constant over the domain of integration. More explicitly, we can parameterise $D$ via
$$\sigma(u,v)=(u,v,1),\qquad u,v\in[0,1].$$
Now we have
$$\frac{\partial{\sigma}}{\partial{u}}=(1,0,0)\quad\text{ and }\quad\frac{\partial{\sigma}}{\partial{v}}=(0,1,0).$$
By definition, we have
$$(\alpha\wedge\gamma)\left(\frac{\partial{\sigma}}{\partial{u}},\frac{\partial{\sigma}}{\partial{v}}\right)=u^{2}v\det{\begin{pmatrix}1 & 0 \\ 0 & 0\end{pmatrix}}+uv^{2}\det{\begin{pmatrix}0 & 0 \\ 1 & 0\end{pmatrix}}=0,$$
so the integral is
$$\iint_{D}\alpha\wedge\gamma = \int_{0}^{1}\int_{0}^{1}(\alpha\wedge\gamma)\left(\frac{\partial{\sigma}}{\partial{u}},\frac{\partial{\sigma}}{\partial{v}}\right)\,du\,dv=0.$$
