Consider a smooth enough function $u:\Omega\subset\mathbb{R}^2\to\mathbb{R}^3$, where $\Omega$ is an open bounded domain. For any $x\in\Omega$, $\nabla u(x)\in\mathbb{R}^{3\times2}$. Now, define $A\in\mathbb{R}^{3\times2}$ such that $A_{ij}=\int_{\Omega}\partial_ju_i(x)dx$ for $i=1,2,3$ and $j=1,2$.
Question: What assumptions of $u$ can guarantee that $A$ has full rank?
So formally, $A=\int_{\Omega}\nabla u(x)dx$ is the integral of the Jacobian. One guess of mine is: if $\nabla u(x)$ is of full rank at any point $x\in\Omega$ then $A$ has full rank.
$\nabla u(x)$ has full rank means that two vectors $\partial_1u(x)$ and $\partial_2u(x)$ are linearly independent at $x$. When $u$ is a parametrization of a surface, $\partial_1u(x)$ and $\partial_2u(x)$ then form a basis of the tangent plane of the surface at the point $u(x)$. The $i$-th column of $A$ should be $\int_{\Omega}\partial_i u(x)dx$ formally, and it looks like an "average" of the vector field $\partial_i u(x)$ over $\Omega$ if we further divide it by constant $|\Omega|$. I think in general two arbitrary vectors field $\mathbf{f}(x)$ and $\mathbf{g}(x)$ linearly independent at any point $x$ doesn't imply that their averages over the entire domain $\frac{1}{|\Omega|}\int_{\Omega}\mathbf{f}(x)$ and $\frac{1}{|\Omega|}\int_{\Omega}\mathbf{g}(x)$ are still linearly independent. However, in this case, the considered two vector fields are not arbitrary, and they have to be the two columns of the Jacobian matrix.
To sum up, my guess can be rephrase as: if $\partial_1u(x)$ and $\partial_2u(x)$ are linearly independent at any $x\in\Omega$, do we have $\int_{\Omega}\partial_1 u(x)dx$ and $\int_{\Omega}\partial_2 u(x)dx$ are linearly independent? If not, what is the counterexample, and is there any possible conditions on $u$ that makes $A$ has full rank?
Updated: My guess is not correct, as pointed out in the comment there is a counterexample. Then what other conditions on $u$ can make $A$ has full rank?