Let $F = (f,g,h)$ be any vector field on an open set $U \subset \mathbb{R}^3$. Suppose that T is a smooth oriented surface in U, and let n be the outward unit normal vector to T that defines its orientation. Show that $$ \int_T F \cdot n dA = \int_T \phi(F)$$ where dA is the standard area form on T, and where $$ \phi(F) = \phi(f,g,h) = f dy \wedge dz - g dx \wedge dz + h dx \wedge dy$$
I am probably way overthinking this problem, but without an actual parametrization of T, I am not sure how to proceed. Also, what is the "standard area form" dA anyway? With F and n both vector fields, $F \cdot n \in C^{\infty}(U)$, so then dA should be some 2-form, but I don't know how to represent that without actually knowing T or without just using the differential form that is already defined to be $\phi(F)$. Any help would be much appreciated!
The standard area $2$-form on $T$ is the $2$-form $\sigma$ with the property that $\sigma(x)(u,v)$ gives the signed area of the parallelogram spanned by two vectors $u$ and $v$ tangent to $T$ at $x$. You can check that this signed area is likewise given by the determinant $\big|n\ u\ v\big|$. Now check that (at $x$) $$(F\cdot n) \,\sigma (u,v) = \big| (F\cdot n)n\ u\ v \big| = \big|F\ u\ v\big| = \phi(F)(u,v).$$