As suggested by fourierwho in their comment, perhaps the most the natural domains for which the divergence (also called Gauss-Green) theorem holds are the sets of finite perimeter, i.e. Caccioppoli sets, so let's precisely see why.
Definition 1 ([1], §3.3 p. 143). Let $\Omega$ a Lebesgue measurable set in $\mathbb{R}^n$. For any open subset $G\subseteq\mathbb{R}^n$ the perimeter of $\Omega$ in $G$, denoted as $P(\Omega,G)$, is the variation of $\chi_\Omega$ in $\Omega$ i.e.
$$
\begin{split}
P(\Omega,G)&=\sup\left\{\int_\Omega \nabla\cdot\varphi\,\mathrm{d}x\,:\,\varphi\in [C_c^1(G)]^n, \|\varphi\|_\infty\leq1\right\}\\
& =| \nabla \chi_{\Omega\cap G}|=TV(\Omega,G)
\end{split}\tag{1}\label{1}
$$
where $[C_c^1(G)]^n$ is the set of compact support continuously differentiable vector functions in $G$ and $TV$ is the total variation of the set function $\nabla \chi_{\Omega\cap G}$.
The set $\Omega$ is a set of finite perimeter (a Caccioppoli set) in $G\subseteq\mathbb{R}^n$ if $P(\Omega,G)<\infty$.
- If $G=\mathbb{R}^n$, then we can speak of perimeter of $\Omega$ tout court, and denote it as $P(\Omega)$.
- If $P(\Omega,G^\prime)<\infty$ for every bounded open set $G^\prime\Subset\mathbb{R}^n$, $\Omega$ is a set of locally finite perimeter.
Why definition \eqref{1} implies a natural extension of the classical divergence (Gauss-Green) theorem? For simplicity lets consider sets of finite perimeter: $P(\Omega)<\infty$ implies that the distributional derivative of the characteristic function of $\Omega$ is a vector Radon measure whose total variation is the perimeter defined by \eqref{1}, i.e.
$$
\nabla\chi_\Omega(\varphi)=\int_\Omega\nabla\cdot\varphi\,\mathrm{d}x=\int_\Omega \varphi\,\mathrm{d}\nabla\chi_\Omega\quad
\varphi\in [C_c^1(\mathbb{R}^n)]^n\tag{2}\label{2}
$$
Now the support in the sense of distributions of $\nabla\chi_\Omega$ is $\subseteq\partial\Omega$ ([2], §1.8 pp. 6-7): to see this note that if $x\notin\partial\Omega$, it should belong to an open set $A\Subset\mathbb{R}^n$ such that either $A\Subset\Omega$ or $A\Subset\mathbb{R}^n\setminus\Omega$:
- if $A\Subset\Omega$, then $\chi_\Omega=1$ on $A$ and hence \eqref{2} is equal to zero for each $\varphi\in [C_c^1(A)]^n$
- if $A\Subset\mathbb{R}^n\setminus\Omega$, then $\chi_\Omega=0$ on $A$ and hence \eqref{2} is again equal to zero for each $\varphi\in [C_c^1(A)]^n$
Also, as a general corollary of (one of the versions of) Radon-Nikodym theorem ([1], §1.1 p. 14) we can apply a polar decomposition to $\nabla\chi_\Omega$ and obtain
$$
\nabla\chi_\Omega=\nu_\Omega|\nabla\chi_\Omega|_{TV}\equiv\nu_\Omega|\nabla\chi_\Omega|\tag{3}\label{3}
$$
where $\nu_\Omega$ is a $L^1$ function taking values on the unit sphere $\mathbf{S}^{n-1}\Subset\mathbb{R}^n$, and rewriting \eqref{2} by using \eqref{3} we obtain the sought for general divergence (Gauss-Green) theorem
$$
\int_\Omega\!\nabla\cdot \varphi\, \mathrm{d}x =\int_{\partial\Omega} \!\varphi\,\cdot\nu_\Omega\, \mathrm{d}|\nabla\chi_\Omega|\quad\forall\varphi\in [C_c^1(\mathbb{R}^n)]^n\tag{4}\label{4}
$$
Note that this result is an almost direct consequence of definition 1 above, with minimal differentiability requirement imposed on the data $\varphi$: it seems to follow directly from the given definition of perimeter \eqref{2} through the application of general (apparently unrelated) theorems on the structure of measures and distributions, and in this sense it is the most "natural form" of the divergence/Gauss-Green theorem.
Further notes
- When $\Omega$ is a smooth bounded domain, \eqref{4} "reduces" the standard divergence (Gauss-Green) theorem.
- There are more general statement of the theorem, relaxing further both the conditions on $\Omega$ and on $\varphi$: however they require further, more technical, assumptions and therefore are in some sense "less natural".
- The notion of perimeter \eqref{1} was introduced by Ennio De Giorgi by using a gaussian kernel in order to "mollify" the set $\Omega$. By using De Giorgi's ideas, Calogero Vinti and Emilio Bajada further generalized the notion of perimeter: however I am not aware of a corresponding generalization of the divergence theorem.
[1] Ambrosio, Luigi; Fusco, Nicola; Pallara, Diego (2000), Functions of bounded variation and free discontinuity problems. Oxford Mathematical Monographs, New York and Oxford: The Clarendon Press/Oxford University Press, New York, pp. xviii+434, ISBN 0-19-850245-1, MR1857292, Zbl 0957.49001.
[2] Giusti, Enrico (1984), Minimal surfaces and functions of bounded variations, Monographs in Mathematics, 80, Basel–Boston–Stuttgart: Birkhäuser Verlag, pp. XII+240, ISBN 978-0-8176-3153-6, MR 0775682, Zbl 0545.49018
I am not sure what you mean by the coordinates being independent of eachother, but in any case we of course have $\partial_x x = 1,$ and $\partial_x y = \partial_x z = 0$, and analogously for the other partial derivatives simply directly from the definition.
On the other hamd, knowing that
$$\vec u = \hat i (\omega_y z - \omega _z y) + \hat j (\omega_z x - \omega _x z) + \hat k (\omega _x y - \omega _y x),$$
and assuming that $\vec\omega = \vec\omega(x,y,z)$ depends on $x,y$ and $z$, we get
\begin{align}
\nabla \times \vec u &=
\begin{vmatrix}
\hat i & \hat j & \hat z\\
\partial_x & \partial_y & \partial_z\\
\omega_y z - \omega _z y & \omega_z x - \omega _x z & \omega _x y - \omega _y x
\end{vmatrix}\\[.2cm]
&= \begin{pmatrix}
\partial_y(\omega _x y - \omega _y x) - \partial_z(\omega_z x - \omega _x z)\\
\partial_z(\omega_y z - \omega _z y) - \partial_x(\omega _x y - \omega _y x)\\
\partial_x (\omega_z x - \omega _x z) - \partial_y(\omega_y z - \omega _z y)
\end{pmatrix}\\[0.2cm]
&= \begin{pmatrix}
\big( (\partial_y\omega _x) y + \omega_x - (\partial_y\omega _y) x - 0\big) - \big((\partial_z\omega_z) x + 0 - (\partial_z\omega _x) z - \omega_x\big)\\
\big( (\partial_z\omega_y) z + \omega_y - (\partial_z\omega _z) y - 0 \big) - \big( (\partial_x\omega _x) y + 0 - (\partial_x\omega _y) x - \omega_y)\\
\big( (\partial_x\omega_z) x + \omega_z - (\partial_x\omega _x) z - 0 \big) - \big( (\partial_y\omega_y) z + 0 - (\partial_y\omega _z) y - \omega_z)
\end{pmatrix}\\[0.2cm]
&= 2\vec \omega +
\begin{pmatrix}
(\partial_y\omega_x)y-(\partial_y\omega_y)x-(\partial_z\omega_z)x+(\partial_z\omega_x)z\\
(\partial_z\omega_y)z-(\partial_z\omega_z)y-(\partial_x\omega_x)y + (\partial_x\omega_y)x\\
(\partial_x\omega_z)x - (\partial_x\omega_x)z
-(\partial_y\omega_y)z + (\partial_y\omega_z)y\end{pmatrix}.
\end{align}
The latter vector is generally not equal to $0$ unless $\vec\omega$ is constant (take e.g. $\vec\omega(x,y,z) =(x,y,z)$ ), so the constancy condition was probably implicitly included in the question.
Best Answer
In inviscid flow, the fluid velocity satisfies the no-flux condition $\mathbf{u} \cdot \mathbf{n} = 0$ on the boundary $\partial D$, but the tangential component of velocity need not vanish as in viscous flow.
Let $\Sigma(t)$ be a smooth material surface element, bounded by the simple, smooth closed curve $C(t)$. As $\Sigma(t)$ evolves in time, it contains the same fluid particles present at time $t= 0$. If $\Sigma(0) \subset \partial D$, then no flux condition ensures that $\Sigma(t) \subset \partial D$ for all $t > 0$.
By Stokes' theorem we have
$$\oint_{C(t)}\mathbf{u} \cdot dl = \int_{\Sigma(t)}(\nabla \times \mathbf{u}) \cdot \mathbf{n} \, dS = \int_{\Sigma(t)} \mathbf{\omega} \cdot \mathbf{n} \, dS $$
In an inviscid flow with the absence of body forces it follows from Kelvin's circulation theorem that
$$\frac{d}{dt}\oint_{C(t)}\mathbf{u} \cdot dl = \frac{d}{dt}\int_{\Sigma(t)} \mathbf{\omega} \cdot \mathbf{n} \, dS =0$$
By a generalization of the Leibniz integral rule we can pass the derivative under the integral on the RHS to obtain
$$\frac{d}{dt}\int_{\Sigma(t)} \mathbf{\omega} \cdot \mathbf{n} \, dS = \int_{\Sigma(t)} \left[\frac{\partial\mathbf{\omega}}{\partial t}+ (\nabla \cdot \mathbf{\omega})\mathbf{u} \right]\cdot \mathbf{n} \, dS- \oint_{C(t)}(\mathbf{u}\times \mathbf{\omega}) \cdot dl$$
Note that $(\nabla \cdot \mathbf{\omega})\, \mathbf{u} \cdot \mathbf{n} = 0$ and $\mathbf{u}\times \mathbf{\omega} = \mathbf{u}\times (\nabla \times \mathbf{u}) = \mathbf{u} \cdot \nabla\mathbf{u} - \mathbf{u} \cdot \nabla\mathbf{u} = \mathbf{0}$.
Thus,
$$\int_{\Sigma(t)} \frac{\partial\mathbf{\omega}}{\partial t}\cdot \mathbf{n} \, dS = 0$$
Under the usual smoothness assumption for the velocity and vorticity fields, since this holds for any $\Sigma(t) \subset \partial D$, we get $\frac{\partial\mathbf{\omega}}{\partial t}\cdot \mathbf{n} = 0$ for all $t > 0$.