I am learning about surface integrals and flux integrals and I have done the following exercise for practice. Since I am still not entirely sure if I have understood correctly I would appreciate some feedback on my solution below (whether it is correct, if/what can be improved), thanks.
Let $S$ be that portion of the plane $x+2y+2z=4$ lying in the first octant, oriented with outward normal pointing upward. Find
(a) the area of $S$; (b) $\int_{S} (x-y+3z)\sigma$; (c) $\int_{S} zdx\wedge dy+ydz\wedge dx+xdy\wedge dz$ where $\sigma=n_1dy\wedge dz+n_2dz\wedge dx+n_3dx\wedge dy$ is the area $2$-form (and $\mathbf{n}=(n_1,n_2,n_3)$ is the outward-pointing unit normal to the surface $S$.)
What I have done:
We parametrize the surface by $\mathbf{g}:\Omega\to\mathbb{R}^3,\ \mathbf{g}\begin{pmatrix}u\\v\end{pmatrix}=\begin{bmatrix} u\\v \\ \frac{4-u-2v}{2} \end{bmatrix}$ and note that $D\mathbf{g}=\begin{bmatrix}1 & 0\\0 & 1\\-\frac{1}{2} & -1\end{bmatrix}$ and that $\mathbf{n}=\frac{\frac{\partial\mathbf{g}}{\partial r}\times\frac{\partial\mathbf{g}}{\partial\theta}}{\lVert\frac{\partial\mathbf{g}}{\partial r}\times\frac{\partial\mathbf{g}}{\partial\theta}\rVert}=\frac{\left(\frac{1}{2},1,1\right)}{\frac{3}{2}}=\begin{bmatrix}\frac{1}{3}\\\frac{2}{3}\\\frac{2}{3}\end{bmatrix}$ is a viable outward pointing unit normal vector so
(a) $\text{area}(S)=\int_{S}\sigma=\int_{\Omega}\mathbf{g}^*\sigma=\int_{\Omega}\left(\frac{1}{2}\cdot \frac{1}{3}+\frac{2}{3}+\frac{2}{3}\right)dudv=\int_{\Omega}\frac{3}{2}dudv\overset{*}{=}\int_{u=0}^{u=4}\left(\int_{v=0}^{v=2-\frac{1}{2}u}dv\right)du=\frac{3}{2}\int_{u=0}^{u=4}\left(2-\frac{1}{2}u\right)du=6.$
($^*$ To be in the first octant it must be $u\geq 0,\ v\geq 0$ and $z=\frac{4-u-2v}{2}\geq 0\Leftrightarrow v\leq 2-\frac{1}{2}u$.)
(b) $\int_{S}(x-y+3z)\sigma=\int_{\Omega}\mathbf{g}^* \left( (x-y+3z)\sigma \right)=\int_{\Omega}\left( \left(u-v+\frac{3}{2}(4-u-2v)\right)\left(\frac{3}{2}\right) \right)dudv=\int_{u=0}^{u=4}\left(\int_{v=0}^{v=4}\left(-\frac{3}{4}u-6v+9\right)dv\right)du=\int_{u=0}^{u=4}\left(6-\frac{3}{8}u^2\right)du=16.$
(c) $\int_{S} zdx\wedge dy+ydz\wedge dx+xdy\wedge dz=\int_{\Omega}\mathbf{g}^*\left(zdx\wedge dy+ydz\wedge dx+xdy\wedge dz\right)=\int_{\Omega}\left( \left(\frac{4-u-2v}{2}\right)+v+u\cdot\frac{1}{2} \right)dudv=\int_{\Omega}2dudv=2\int_{u=0}^{u=4}\left(\int_{v=0}^{v=2-\frac{1}{2}u}dv\right)du=2\int_{u=0}^{u=2}\left(2-\frac{1}{2}u\right)du=8.$
They are all correct. As regards (c), we have $$\begin{align} \iint_S \mathbf{F}\cdot d \mathbf{S}&=\iint_{T} \big\langle (x,y,2-\frac{x}{2}-y),(\frac{1}{2},1,1)\big\rangle\, dx dy\\ &=\iint_{T} 2\, dx dy=2|T|=8. \end{align}$$ where $T$ is the triangle $\{(x,y): x/4+y/2\leq 1,x\geq 0, y\geq 0\}$ and $|T|=4$ is its area.