Area form for $M^2 \subseteq \Bbb R^4$ We know that in general, given a orientable hypersurface $M^{n-1} \subseteq \Bbb R^n$, the volume form on $M$ is given by $$dM = \sum_{i=1}^n(-1)^{i-1}n_i\,dx^1 \wedge\cdots\wedge \widehat{dx^i}\wedge \cdots \wedge dx^n,$$where $(n_1,\cdots,n_n)$ is the unit normal to $M$. Moreover, we have $$n_i\,dM = (-1)^{i-1}dx^1 \wedge\cdots\wedge \widehat{dx^i}\wedge \cdots \wedge dx^n$$for tangent vectors. Another general case is for bi-dimensional surfaces in $\Bbb R^n$, for which in coordinates we can write using $\sqrt{EG-F^2}$. 
But I wanted to write something like the first expression for surfaces $M^2 \subseteq \Bbb R^4$, specifically. There is no standard choice of orthonormal basis for the normal space as far as I know. Is there a way to get around this, at least in this case? If there is, I'd guess it would adapt for submanifolds of codimension $2$, but that would be the cherry on the cake.
 A: I pulled up my sleeves and did the computation for $M^2 \subseteq \Bbb R^4$. I'll post it here, but I'll leave the question open for a while in case someone sees a non-medieval way to do this.
First of all, take $\{e_1,e_2\}$ an orthonormal positive basis of $T_xM$, and pick any orthonormal positive basis $\{n,\nu\}$ of $T_xM^\perp$ with $\{e_1,e_2,n,\nu\}$ being again an orthonormal basis of $T_x(\Bbb R^4)$. We note that $\det(e_1,e_2,n,\nu) = 1$, so that $$dM(v,w) = \det(v,w,n,\nu)$$is the area form in $M$. Let's do a quick reality check and see that this does not depend on our choice of $n$ and $\nu$. If $n',\nu'$ is another choice with the properties above, write $$n' = (\cos t) n + (\sin t) \nu, \quad \nu' = (-\sin t) n + (\cos t) \nu$$for some convenient $t$. Using that $\det$ is multilinear, it is easy to see that $$\det(v,w,n,\nu)=\det(v,w,n',\nu')$$for all $v$ and $w$. Now write $n = (n^1,\cdots,n^4)$ and $\nu = (\nu^1,\cdots,\nu^4)$, and similarly for $v$ and $w$. Recall that $$dx^i \wedge dx^j(v,w) = \begin{vmatrix} v^i & v^j \\ w^i & w^j\end{vmatrix}.$$ Then you take $$dM(v,w) = \begin{vmatrix} v^1 & v^2 & v^3 & v^4 \\   w^1 & w^2 & w^3 & w^4 \\ n^1 & n^2 & n^3 & n^4 \\ \nu^1 & \nu^2 & \nu^3 & \nu^4 \end{vmatrix}$$and expand that on the bottom row to get a combination of $3 \times 3$ subdeterminants with some signs and coefficientes $\nu_i$. Then you expand these subdeterminants by their bottom rows again, and after some pain and suffering it yields $$dM= \sum_{1 \leq i<j\leq 4}(-1)^{i+j-1}\begin{vmatrix} n_{i'} & n_{j'} \\ \nu_{i'} &\nu_{j'}\end{vmatrix}dx^i \wedge dx^j,$$where $i'<j'$ are the remaining indices (is there a way to write that neatly?).
In practice, if you have a parametrization $\varphi\colon U \subseteq \Bbb R^2 \to M \subseteq \Bbb R^4$, one of the vectors of the standard basis will not be tangent to $M$, say, $e_4$. Then take $n = \varphi_u\times \varphi_v \times e_4$ normalized and $\nu = \varphi_u \times \varphi_v \times n$ normalized, and fix some signs if needed.
One can imagine the general case for $M^k \subseteq \Bbb R^n$, picking $n-k$ orthonormal normal directions, and we'd have some combination like $$dM = \sum_{1 \leq i_1 <i_2 < \cdots < i_k \leq n}(-1)^{-1+\sum i_j} \det(\text{something }(n-k)\times (n-k))\,dx^{i_1}\wedge\cdots \wedge dx^{i_k},$$but take that with a grain of salt.
