I'm reading through some short notes found here https://faculty.math.illinois.edu/~kapovich/481-07/int.pdf and got confused by something I saw.
Let $M \subset \mathbb{R}^3$ be an oriented surface with an outward unit normal $\vec{n} = (n^1,n^2,n^3)$. The definition for the area form $dA$ which I've seen is the pullback along the inclusion $M \to \mathbb{R}^3$ of the form $i_{\vec{n}} dV$, where $dV = dx \wedge dy \wedge dz$ and $i_{\vec{n}}$ is the interior derivative along $\vec{n}$. So the area form $dA$ is the pullback of $n^1 dy \wedge dz - n^2 dx \wedge dz + n^3 dx \wedge dy$, which agrees with the reference. But the reference also says $n^1 dA = dy \wedge dz$, $n^2 dA = dz \wedge dx$, and $n^3 dA = dx \wedge dy$, and I'm confused how they drew these conclusions. Aren't the $n^i$ just smooth real-valued functions so that $(n^1(p),n^2(p),n^3(p))$ is orthogonal to $T_p M$ at every $p \in M$? So how does $$ n^1 dA = (n^1)^2 dy \wedge dz - n^1 n^2 dx \wedge dz + n^1 n^3 dx \wedge dy$$ become $dy \wedge dz$? I'm sure I must be missing something obvious..
Indeed, the general result you should prove is that if $\vec F=(F^1,F^2,F^3)$, then $$(\vec F\cdot \vec n)dA = F^1\,dy\wedge dz + F^2\,dz\wedge dx + F^3\,dx\wedge dy.$$ Note, in your case, that since $\vec n$ is normal to the surface, the $1$-form $n^1\,dx + n^2\,dy+ n^3\,dz$ vanishes on the surface. This means that (on the surface) $n^1\,dx = -(n^2\,dy + n^3\,dz)$, from which you see that $$-n^1n^2\,dx\wedge dz + n^1n^3\,dx\wedge dy = (n^2){}^2 dy\wedge dz - (n^3){}^2 dz\wedge dy,$$ and since $\vec n$ is a unit vector, you're done.