TL;DR: Is there a nice expression for the surface integral of a vector field?
The surface integral of a scalar field $\phi$ over a surface parameterised by $\mathbf{r}(u,v)$ is $$ \iint_T \phi(\mathbf{r}(u, v)) \ \Bigg \lVert \frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v} \Bigg \rVert \ du \ dv, $$ which can be rewritten using the vector identity $\lVert \mathbf{x} \times \mathbf{y} \rVert^2 = \lVert \mathbf{x} \rVert^2 \lVert \mathbf{y} \rVert^2 - (\mathbf{x} \cdot \mathbf{y})^2$ into the much nicer and more intuitive expression $$ \iint_T \phi(\mathbf{r}(u, v)) \sqrt{g} \ du \ dv, $$ where $g$ is the determinant of the metric $\mathbf g = J^{\text T} J$, and $J = \begin{bmatrix} \mathbf r_u & \mathbf r_v \end{bmatrix}$ is the Jacobian.
Now that made me think: is there a similar expression for the surface integral of a vector field?
I started with the usual definition $$ \iint_T \mathbf F(\mathbf{r}(u, v)) \cdot \bigg ( \frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v} \bigg ) \ du \ dv $$ and worked my way through some identities involing dot products and scalar products, determinants of block matrices, and a bit of matrix algebra to rewrite the integral into $$ \iint _T \lVert \mathbf F \rVert \sqrt{\det \Bigg(J^{\text T} \bigg(I- \frac{\mathbf F \mathbf F^{\text T}}{\mathbf F^{\text T}\mathbf F}\bigg)J \Bigg)} \ du \ dv. $$ This is about as far as I could get. I have a feeling that there should be some nice geometric intuition behind this, but I can't quite get hold of it. First of all, the matrix sandwiched between the Jacobians is a projection matrix, right? Its job is to project vectors onto the orthgonal complement of $\mathbf F$, i.e., to zero out anything that points along the vector field at a given point. The determinant of such a matrix is zero, so won't it be zero after transforming by the Jacobian too? Is there something I'm missing here? Any insights would be greatly appreciated.