The question is
Let $f:D \rightarrow \mathbb{R}^3$ be $f(u,v)=(u,v,u^2+v^2+1)$, where $D=[0,1]\times [0,1]$. Let $\omega=y\,\mathrm{d}y\wedge \mathrm{d}z+xz\,\mathrm{d}x\wedge \mathrm{d}z$. Calculate $\int_D{f^*\omega }$.
I calculate the integral by two ways, but results are different.
Here is the first way:
$f^*w=v\,\mathrm{d}v\wedge \mathrm{d}(u^2+v^2+1)+u(u^2+v^2+1)\,\mathrm{d}u\wedge \mathrm{d}(u^2+v^2+1)=2vu\,\mathrm{d}v\wedge \mathrm{d}u+u(u^2+v^2+1)2v\,\mathrm{d}u\wedge \mathrm{d}v=2uv(u^2+v^2)\,\mathrm{d}u\wedge \mathrm{d}v$
$\int_D{f^*\omega }=\int_0^1\,\mathrm{d}u\int_0^12uv(u^2+v^2)\,\mathrm{d}v=\frac{1}{2}$
Here is the second way:
Let $M$ be the region by $x=0,x=1,y=0,y=1,z=0,f(D)$. As $\mathrm{d}\omega=0$, we have $\int_M\mathrm{d}\omega=\int_{\partial M}\omega=0$.
$\int_D{f^*\omega }=\int_{f(D)}\omega=\int_{\partial M\cap\{x=0\}}\omega-\int_{\partial M\cap\{x=1\}}\omega+\int_{\partial M\cap\{y=0\}}\omega-\int_{\partial M\cap\{y=1\}}\omega=-\int_0^1y\,\mathrm{d}y\int_{y^2+1}^{y^2+2}\,\mathrm{d}z-\int_0^1x\,\mathrm{d}x\int_{x^2+1}^{x^2+2}z\,\mathrm{d}z=-\frac{3}{2}$
It seems to me that you have made a sign error somewhere in your second method. Double check your orientation for the surfaces $\partial M\cap\{y=0\}$ and $\partial M\cap \{y=1\}$. What you should have is: \begin{align} \int_Df^*\omega & =\int_{f(D)}\omega\\ &= -\int_{\partial M\setminus f(D)}\omega\\ &= -\int_0^1y\;dy\int_{y^2+1}^{y^2+2}\;dz + \int_0^1x\;dx\int_{x^2+1}^{x^2+2}z\;dz\\ &=\frac{1}{2} \end{align}