Suppose the area, volume or hyper volume covered by a vector is
$$ \mathrm{V}\left(\vec{u}\right) = u_x \times u_y \times \ldots $$
And the area, volume or hyper volume covered by a matrix is
$$ \mathrm{V}\left(A \right) = \mathrm{c}_x\left(A\right) \wedge \mathrm{c}_y\left(A\right) \wedge \ldots $$
Note that $\wedge$ is the wedge product which I don't really understand but is similar to the cross product and when fully calculated out gives a volume that the vectors bound.
The area, volume or hyper volume covered by a differential form is: $$ \mathrm{V}\left(\mathrm{d} \vec{s} \right) = \mathrm{d} s_x \wedge \mathrm{d} s_y \wedge \ldots $$
which is just the volume element $\mathrm{d} V = \mathrm{d} x \, \mathrm{d} y \, \mathrm{d} z$ most of the time.
And when fully calculated out I think is something like:
$$ \mathrm{V}\left(\mathrm{d} \vec{s} \right) = \sum^n_{k=0} \left(-1\right)^k \left(\prod^{k-1}_{i=0} s_i \right) \left(\prod^n_{i=k + 1} s_i \right) \, \mathrm{d} s_k $$
Surface normals are:
$$ \hat{n} = \frac{\partial \vec{u}}{\partial v_x} \wedge \frac{\partial \vec{u}}{\partial v_y} \wedge \ldots $$ And if I use $\nabla_{\vec{v}} \otimes \vec{u}$ to denote the Jacobian derivative: $$ \mathrm{V}\left(\nabla_{\vec{v}} \otimes \vec{u} \right) = \mathrm{c}_x\left(\nabla_{\vec{u}} \otimes \vec{s}\right) \wedge \mathrm{c}_y\left(\nabla_{\vec{u}} \otimes \vec{s}\right) \wedge \ldots $$ $$ \mathrm{V}\left(\nabla_{\vec{v}} \otimes \vec{u} \right) = \frac{\partial \vec{u}}{\partial v_x} \wedge \frac{\partial \vec{u}}{\partial v_y} \wedge \ldots $$
I think according to the divergence theorem then:
$$ \int_{\vec{s} \in C} \mathrm{V}\left(\mathrm{d} \vec{s}\right) = \int_{\vec{s} \in C} \frac{\nabla_{\vec{s}}\cdot \vec{s}}{\mathrm{N}} \, \mathrm{V}\left(\mathrm{d} \vec{s}\right) = \oint_{\vec{u} \in \partial C} \frac{\vec{s}}{\mathrm{N}} \cdot \mathrm{V}\left(\nabla_{\vec{u}} \otimes \vec{s} \right) \, \mathrm{V}\left(\mathrm{d} \vec{u} \right) $$
where $\mathrm{N}$ is the dimensionality of the vector. I'm not sure how this works for cases higher than two or three. Also I'm not sure how the equation works for the one-dimensional case.
I can manually verify this for the case of two-dimensions but I have trouble working this out for multiple dimensions and I'm not sure I am right. Also, the dimensionality fudge factor seems off to me somehow.
Am I on the right track here?
What's the proper way to multiply together the wedge product of the differential forms?
How would I prove this special case of the divergence theorem for multiple dimensions?
Is it possible to get to the full divergence theorem from this special case I noted here?