Integrating over the two form Let $A=(0,1)^2$. Let $\alpha:A\to\Bbb R^3$ be given by the equation
$$\alpha(u,v)=(u,v,u^2+v^2+1)$$
Let $Y$ be the image set of $\alpha$. Evaluate the integral over $Y_\alpha$ of the 2-form 
$x_2dx_2\land dx_3+x_1 x_3 dx_1\land dx_3$.
Can someone please give me a useful hint for solving this?
Thanks in advance!

Given parametrization:
$$\phi (u,v) = (u,v,{u^2} + {v^2} + 1)$$
Setting
$$x = u,y = v,z = {u^2} + {v^2} + 1$$
Then calculate differential resp. to parametrization:
$$\begin{gathered}
  dx = du \hfill \\
  dy = dv \hfill \\
  dz = 2udu + 2vdv \hfill \\ 
\end{gathered} $$
Form in old coordinates:
$$\omega (x,y,z) = ydy \wedge dz + xzdx \wedge dz = (ydy + xzdx) \wedge dz$$
Form in new coordinates resp. surface, the world famous pull-back:
$$\begin{gathered}
  {\phi ^ * }\omega (u,v) = (u({u^2} + {v^2} + 1)du + vdv) \wedge (2udu + 2vdv) \hfill \\
  {\phi ^ * }\omega (u,v) = 2uv({u^2} + {v^2} + 1)du \wedge dv - 2uvdu \wedge dv \hfill \\
  {\phi ^ * }\omega (u,v) = 2({u^3}v + u{v^3})du \wedge dv \hfill \\ 
\end{gathered}$$
Ready to integrate:$$2 \cdot \int\limits_0^1 {\int\limits_0^1 {({u^3}v + u{v^3})dudv} }  = \int\limits_0^1 {(\frac{1}{2}v + {v^3})dv = } \frac{1}{2}$$
Another user (Callus) saw this:
$$\begin{gathered}
  d\omega  = d(xzdx \wedge dz) \hfill \\
   = d(xz) \wedge dx \wedge dz \hfill \\
   = (zdx + xdz) \wedge dx \wedge dz \hfill \\
   = zdx \wedge dx \wedge dz + xdz \wedge dx \wedge dz = 0 \hfill \\ 
\end{gathered}  $$
That means the form is closed. On the other hand: We know formula for exterior
derivative for wedge of forms:
$$d(\alpha  \wedge \beta ) = d\alpha  \wedge \beta  + {( - 1)^k}\alpha  \wedge d\beta$$
Here $\alpha$ is a k-form. Therefore $$\omega  = y{\kern 1pt} dy \wedge dz + xz{\kern 1pt} dx \wedge dz$$
may be written like:
$$\omega  = d(\frac{1}{2}{y^2} \wedge dz) + d(\frac{1}{2}{x^2} \wedge zdz) = d(\frac{1}{2}({y^2}dz + {x^2}z)dz)$$
Or, if we set:
$$\theta  = \frac{1}{2}({x^2}z + {y^2})dz$$
we get for short:
$$\omega  = d\theta $$
At least resp. local coordinates, $\omega$
is exact. It's the exterior derivative from another form $\theta$.
So, by Stoke's theorem, that means:
$$\int\limits_A {d\theta }  = \int\limits_{\partial A} \theta   = \int\limits_{\partial A} {\frac{1}{2}({x^2}z + {y^2})dz}$$
Let's evaluate the RHS:
If $z = {x^2} + {y^2} + 1$, so $dz = 2xdx + 2ydy$ and for RHS we get:
$$\int\limits_{\partial A} {\frac{1}{2}({x^2}z + {y^2})dz}  = \int\limits_{\partial A} {({x^2}({x^2} + {y^2} + 1) + {y^2})xdx}  + \int\limits_{\partial A} {({x^2}({x^2} + {y^2} + 1) + {y^2})ydy}$$
With counterclockwise orientation of boundary $\partial A = A\backslash (0,1) \times (0,1)$, we have to manage following four integrals:
$$\left. {\begin{array}{*{20}{c}}
  {x \in [0,1]} \\ 
  {y = 0} 
\end{array}} \right\}\int\limits_0^1 {({x^5} + {x^3})dx}  = \frac{5}{{12}}$$
$$\left. {\begin{array}{*{20}{c}}
  {x = 1} \\ 
  {y \in [0,1]} 
\end{array}} \right\}2\int\limits_0^1 {({y^3} + y)dy}  = \frac{3}{2}$$
$$\left. {\begin{array}{*{20}{c}}
  {x \in [0,1]} \\ 
  {y = 1} 
\end{array}} \right\} - \int\limits_0^1 {({x^5} + 2{x^3} + x)dx}  =  - \frac{1}{6} - \frac{1}{2} - \frac{1}{2} =  - \frac{7}{6}$$
$$\left. {\begin{array}{*{20}{c}}
  {x = 0} \\ 
  {y \in [0,1]} 
\end{array}} \right\} - \int\limits_0^1 {{y^3}dy}  =  - \frac{1}{4}$$
Last two integrals got minus-sign resp. orientation. Summing up, we have:
$$\int\limits_{\partial A} {\frac{1}{2}({x^2}z + {y^2})dz}  = \frac{5}{{12}} + \frac{3}{2} - \frac{{14}}{{12}} - \frac{3}{{12}} = \frac{6}{{12}} = \frac{1}{2}$$
