Use Stokes's theorem to calculate the integral $$I= \int_\Gamma (x^2+2y)dx+(y+z)dy+(z^2+x^2)dz$$ where $\Gamma$ is the boundary of $$\gamma=\left\{ (x,y,z):3x+y+3z=3,x\ge0,y\ge0,z\ge0\right\} $$
Using Stokes's theorem I got $$\int_\Gamma (x^2+2y)dx+(y+z)dy+(z^2+x^2)dz=\int _\gamma -dy\wedge dz-2xdz\wedge dx-2dx\wedge dy$$but I have problem finding parameric representation for $\gamma$ since I don't know the limits of $(x,y,z)$ (I know that the parametric representation for $\gamma $ will be $(u,v,w)\mapsto(u,v,1-u-\frac{v}{3})$ but what are the limits of $u,v,w$ that's a mystery for me).
EDIT: To achieve that I used the version for $F=(P,Q,R)$ of Green theorem, i.e $$\int_{\partial\Omega} Pdx+Qdy+Rdz=\iint_{\Omega}\left(\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z}\right)dy\wedge dz-\left(\frac{\partial P}{\partial z}-\frac{\partial R}{\partial x}\right)dz\wedge dx+\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dx\wedge dy.$$ In my case $$\int_\Gamma Pdx+Qdy+Rdz=\int_\gamma (0-1)dydz-(0-2x)dzdx+(0-2)dxdy.$$ What am I doing wrong?
Consider $3x+y+3z=3$ intersects $z=0$ along the line $3x+y=3$ on the $xy$-plane. Or, $y = 3-3x$. This line connects $(0,3,0)$ on the $y$-axis to $(1,0,0)$ on the $x$-axis. Hence, $0 \leq x \leq 1$ and $0 \leq y \leq 3-3x$ describes the shadow on the $xy$-plane which provides the domain for your parametrization of the triangular part of the plane $3x+y+3z=3$ bounded by the coordinate planes and in the totally positive octant. More to the point, $\int_{0}^1 \int_{0}^{3-3x} ... dy \, dx$ is what you want. That is how the two-form integral in your post is calculated.
Oh, also, you should use $dz = -dx-\frac{1}{3}dy$ to replace $dz$ with $dx$ and $dy$ then all that remains is $dx \wedge dy$ at the end of it. Or, you could work with $du$ and $dv$ if you prefer.