I am trying to figure out what the next calculation of the "area" (or "volume" in higher dimensional analogues) using Stokes' theorem really means. Here is my thought process: 


**$2$-dimensional case:** given a closed simple piece-wise smooth curve $C$ in $\mathbb{R}^2$ you can find out the area enclosed by $C$ using Green's theorem by choosing an orientation on $C$ and calculating 
$$\text{Area} = \left\vert \dfrac{1}{2} \oint\_C (x dy - y dx) \right\vert .$$ 
**$2$-dimensional case in $\mathbb{R}^n$:** When you have a piece-wise smooth non self intersecting map $\phi : S^1 \rightarrow \mathbb{R}^n$ such that $\phi (S^1)$ is contained in a $2$-dimensional plane of $\mathbb{R}^n$, you can again find the area enclosed by $\phi (S^1)$ by choosing an orthonormal coordinate system $(x\_1,...,x\_n)$, choosing an orientation on $S^1$ and calculating using Stokes' theorem
$$\text{Area}^2 = \sum\_{1 \leq i < j \leq n} \left( \dfrac{1}{2} \oint\_{\phi (S^1)} (x\_i dx\_j - x\_j dx\_i) \right)^2 .$$ 


Now my question is what is the meaning of calculating the above quantity when $\phi (S^1)$ is not contained in a $2$-dimensional plane. I want to use the same computation as above, but since I want the number I get to be independent on the coordinate system, I'll have to average with respect to all coordinate systems. To be explicit: let $\phi : S^1 \rightarrow \mathbb{R}^n$ be a piece-wise smooth non self intersecting map and fix an orientation on $S^1$. Let $SO\_n (\mathbb{R})$ be the special orthogonal group and let $\mu$ be the normalized Haar measure on $SO\_n (\mathbb{R})$ (i.e., $\mu (SO\_n (\mathbb{R}) ) =1$). Define the "area" (or the "Stokes area" in lack of a better name) bounded by $\phi (S\_1)$ as
 $$\text{"Stokes area"}^2 = \int\_{SO\_n (\mathbb{R})} \left( \sum\_{1 \leq i < j \leq n} \left( \dfrac{1}{2} \oint\_{\gamma.\phi (S^1)} (x\_i dx\_j - x\_j dx\_i) \right)^2 \right) d \mu (\gamma) .$$
It is not hard to give analogues in any dimension to get "Stokes volume" (note that the analogue in dimension $1$, i.e., for $S^0$, comes out just the usual Euclidean distance between points).


My questions are: 


1. Is this quantity well-known / studied ? If so, I would very much appreciate a reference.
2. What does this "Stokes area" represent geometrically (I have no intuition about it)?
3. Is there a connection (say by some inequality) to the area of a minimal surface enclosed in $\phi (S^1)$?


**Edit:** by the answer of Will Sawin below, it is clear that we can consider only 
$$\text{"Stokes area"}^2 = \sum\_{1 \leq i < j \leq n} \left( \dfrac{1}{2} \oint\_{\gamma.\phi (S^1)} (x\_i dx\_j - x\_j dx\_i) \right)^2 .$$
and that is quantity is always less or equal than the area of a surface enclosed by the curve. The question remains is there is a connection to the infimum of the areas of all the surfaces enclosed by the curve. In the case the curve is contained in a plane, the formula computes this minimum. But what about the general case: is there a constant (maybe depending on the dimension of $\mathbb{R}^n$) $c(n)$ such that 
$$ \text{"Stokes area"} \geq c(n) (\text{infimum of the areas of surfaces enclosed by the curve}) ?$$



Using Stokes' theorem to define "area" enclosed by a curve
Yes, this is a lower bound on the area of a surface bounded by the curve. Parameterize the surface and apply Stokes' theorem. Your squared area is:


$$\sum\_{1 \leq i < j \leq n} \left( \int\_S dx\_i dx\_j \right)^2 = \int\_S \int\_S \sum\_{1\leq i < j \leq n} dx\_i dx\_j dy\_i dy\_j$$


whereas the actual squared area is:


$$ \left( \int\_S \sqrt{ \sum\_{1 \leq i < j \leq n} \left( dx\_i dx\_j \right)^2} \right)^2= \int\_S \int\_S \sqrt{\sum\_{1 \leq i < j \leq n} \left( dx\_i dx\_j \right)^2} \sqrt{\sum\_{1 \leq i < j \leq n} \left( dy\_i dy\_j \right)^2}$$


which is at least as large by Cauchy-Schwartz. Equality occurs exactly when the area element of the surface is always pointed in the same direction in $\wedge^2 \mathbb R^n$ - that is, when the surface is flat.


This also shows that your definition is coordinate-invariant even before you average over $SO\_n$.


