Let $\phi : D^2 \rightarrow \mathbb{R}^3$ be a smooth map from $D^2 = \{ x \in \mathbb{R}^2 : ||x|| \leq 1 \}$ to $\mathbb{R}^3$.
I think there are two special $2$-forms, $f dx dy$ and $-f dx dy$ $D^2$ that this map induces, corresponding to a choice of normal vector at each point in the image of $D^2$ (it could self intersect, so this isn't quite right). Then $$ \int_{D^2} f dx dy $$ should be the surface area of the image of $D^2$ (suppose $\phi$ is injective so that we do not have it folding on itself).
Question: Can someone give an explicit formula for these two forms in terms of $\phi$?
Now, I suspect that, whenever we have a map into $\mathbb{R}^n$ from an $n-1$ dimensional manifold and given an orientability assumption, we have an induced form which will give rise to some generalization of surface area. If this is indeed true, could you explain a bit about how this might be generalized (I can work out the details myself)?
$f$ is the area element on the embedded surface, which is the length of the cross product of the two partial derivatives of $\phi$ in $\mathbb{R}^3$: $$ f = |\partial_x\phi\times \partial_y\phi |. $$
For the general case in $\mathbb{R}^n$, the set of partial derivatives of $\phi$ give a set of vectors in $\mathbb{R}^n$. $f$ would be the hyper-volume of the parallelogram made by these vectors.