Variational formula for volume element Suppose $S$ is a $k$-dimensional smooth orientable surface in $\mathbb{R}^{d}$.  Assuming at each point $x \in S$ that $[n^{1}(x),\dots,n^{d - k}(x), v^{1}(x),\dots,v^{k}(x)]$ is positively oriented (where $n^{j}(x)$ is a normal vector and $v^{k}(x)$ is tangent), then the volume form on $S$ is defined by
$$dA_{x}(u_{1},\dots,u_{k}) = \det \left( n^{1}(x), \dots, n^{d - k}(x), v^{1}(x), \dots, v^{k}(x)\right).$$
My question is: if $f : \mathbb{R}^{d} \to \mathbb{R}$ is smooth, is the following formula true?
$$\int_{S} |f| dA = \sup \left\{ \int_{S} |f| \omega \, \mid \, \omega \in \Lambda^{k}(S), \, \, \forall x \in S, \, \, \|\omega_{x}\| \leq 1 \right\},$$
where 
$$\|\omega_{x}\| = \sup \left\{ |\omega_{x}(u_{1},\dots,u_{k})| \, \mid \, |u_{1}|, \dots, |u_{k}| \leq 1 \right\}$$
and $|\cdot|$ is the Euclidean norm on $\mathbb{R}^{d}$.
I know this is true when $k = d -1$.  This follows because to each $\omega \in \Lambda^{k}(S)$ there is a vector field $a_{\omega}$ such that 
$$\omega_{x}(u_{1},\dots,u_{d - 1}) = \det \left(a_{\omega}(x), u_{1},\dots,u_{d - 1}\right)$$
from which we obtain (without too much fuss)
$$\int_{S} |f| \omega = \int_{S} |f(x)| \langle a_{\omega}(x), n(x) \rangle dA_{x}.$$
However, it's unclear to me there's an analogous trick when the codimension is higher.
 A: Let $(\theta_{i},U_{i})_{i \in \mathbb{N}}$ be a partition of unity on $S$ (i.e. $\theta_{i}$ are smooth functions taking values in $[0,1]$ that vanish outside of the coordinate patches $U_{i}$, the open sets cover $S$, and $\sum_{i = 1}^{\infty} \theta_{i} = 1$).  Let $\varphi_{i} : V_{i} \to U_{i}$ denote parametrizations of $S$.  Define the norm on alternating $k$-multilinear maps by
$$\|\omega\| = \left( \sum_{1 \leq i_{1} < i_{2} < \dots < i_{k} \leq d} \omega(e_{i_{1}},\dots,e_{i_{k}})^{2} \right)^{\frac{1}{2}}.$$
(I'm not convinced this norm equals the one introduced in the original question, although it does agree with that norm on $k$-vectors.)
Let $x_{1},\dots,x_{d}$ denote the standard coordinates on $\mathbb{R}^{d}$.  
Fix a smooth integrable function $f$ on $S$ and a $k$-form $\omega$ with $\|\omega_{x}\| \leq 1$ pointwise everywhere on $S$.  Observe that if $u_{1},\dots,u_{k} \in \mathbb{R}^{d}$, then
$$|\omega_{x}(u_{1},\dots,u_{k})| \leq \|\omega_{x}\| \|u_{1} \wedge \dots \wedge u_{k}\|$$
by the Cauchy-Schwarz inequality and the Binet-Cauchy formula.  Thus,
\begin{align*}
\int_{S} |f| \omega &= \sum_{i = 1}^{\infty} \int_{V_{i}} |f(\varphi(u))| \omega_{\varphi(u))} \left( \frac{\partial \varphi}{\partial u_{1}}(u),\dots,\frac{\partial \varphi}{\partial u_{k}}(u) \right) \theta_{i}(\varphi(u)) du \\
&\leq \sum_{i = 1}^{\infty} \int_{V_{i}} |f(\varphi(u))| \left\| \frac{\partial \varphi}{\partial u_{1}}(u) \wedge \dots \wedge \frac{\partial \varphi}{\partial u_{k}}(u)\right\| \theta_{i}(\varphi(u)) \, du \\
&= \sum_{i = 1}^{\infty} \int_{U_{i}} |f| \theta_{i} dA \\
&= \int_{S} |f| dA.
\end{align*}
This implies
$$\sup \left\{ \int_{S} |f| \omega \, \mid \, \omega \in \Lambda^{k}(\mathbb{R}^{d}), \, \, \forall x \in S \quad \|\omega_{x}\| \leq 1 \right\} \leq \int_{S} |f| dA.$$
On the other hand, we can extend $dA$ to a smooth $k$-form on $\mathbb{R}^{d}$.  Evidently, on $S$, we have $\|dA_{x}\| = 1$ and we can carry out the extension (e.g. using a tubular neighborhood of $S$) in such a way that this inequality holds on all of $\mathbb{R}^{d}$.  Therefore, 
$$\int_{S} |f| dA = \max \left\{ \int_{S} |f| \omega \, \mid \, \omega \in \Lambda^{k}(\mathbb{R}^{d}), \, \, \forall x \in S \quad \|\omega_{x}\| \leq 1 \right\}.$$
