This is an auxiliary question to improve my understanding
Q: Given the integral,
$$I = \iint_\Omega \varphi (x, y) dxdy$$
If the integrand $\varphi(x, y) dxdy$ happens to be the differential of second order for a function, say, $\psi(x, y)$, meaning that
$$ d(d(\psi(x, y))) = \varphi(x, y) dxdy$$
Then the integral becomes
$$\iint_\Omega d^2 \psi = \psi|_\Omega $$
If I am trying to derive or develop some formula which is 2D analogue of that of gradient:
$$\int_{\vec{a}}^{\vec{b}} \nabla U(\vec{r}) d\vec{r} = U(\vec{b}) - U(\vec{a})$$
I guess the idea is to consider something like
$$\int_a^b \int_c^d \frac{\partial^2 \varphi(x, y)}{\partial y \partial x} dxdy $$
In that case, the double application of the fundamental theorem of calculus leads to the desired answer. In fact, when dealing with definite integrals, we worked with the sets $[a, b]$ where $a<b$; while double integrals are evaluated over more complicated sets. Hence, here arises another idea, although the arbitrary multiple integral is still evaluated through fundamental theorem of calculus. However, what to if we are given
$$\iint_\Omega \frac{\partial^2 \varphi(x, y)}{\partial y \partial x} dxdy$$
In some case, we could consider the simple manifold $a \le x \le b \wedge u(x) \le y \le v(x) $ and come to the same conclusion once again. In the way above, the integral was defined to be an iterated one which is not always allowed.
Stokes theorem states that
$$\int_\Omega d\omega = \int_{\partial \Omega} \omega$$
Trivial cases are Green and divergence theorems. They can be formally derived from this generalised Stokes theorem. I need, as you may guess, Green one stated as
$$\iint_\Omega \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dxdy = \int_{\partial \Omega } Pdx+Qdy$$
Putting, for example, $Q = \frac{\partial \varphi}{\partial y}$ and $P = 0$ then I get
$$\iint_\Omega \frac{\partial^2 \varphi}{\partial y \partial x}dxdy = \int_{\partial \Omega} \frac{\partial \varphi}{\partial y} dy$$
The last integral resembles the gradient theorem and if the boundary of $\Omega$ allows Gradient theorem or performs all obligations that are stated in the theorem, then I have virtually arrived at the answer of my question but using Stokes theorem. Then the expression at the beginning is incorrect and the true idea is behind Stokes again?