Fourier decomposition and green formula I've got trouble to prove Green Formula with certain assumptions. My goal is to prove the following, with $\Omega \subset \mathbb{R}^3$ is a tore (axially symmetrically disposed around the central axis of the tore) and $d\Omega = dxdydz$, $v$ scalar and $u \in \mathbb{R}^3$,
$$
\int_\Omega curl(u) v d\Omega = \int_\Omega u\nabla v  d\Omega - \int_{\partial\Omega} u\times\mathbf{n} v  d\Omega 
$$
$\mathbf{n}$ being the normal vector pointting outside $\partial\Omega$.
First of all, I switched to cylindrical coordinates $(r, \Phi, z)$ and performed a Fourier decomposition according to the $\Phi$-coordinate, s.t for any vector field $A$, $A(r, \Phi, z) = \sum_\alpha A^\alpha(r,z)e^{i\alpha\Phi}$. We thus have ($\hat{\Omega}$ is the domain cylindrical coordinates, and $d\hat{\Omega} = rdrd\Phi dz$),
$$
\int_\Omega curl(\sum_\alpha u^\alpha(r,z)e^{i\alpha\Phi}) \overline{\sum_\lambda v^\lambda(r,z)e^{i\lambda\Phi}} d\hat{\Omega} = \sum_\alpha\sum_\lambda\int_\Omega curl( u^\alpha(r,z)) v^\lambda(r,z)e^{i(\alpha - \lambda)\Phi} d\hat{\Omega}
$$
Then, we note that when $\alpha \ne \lambda$, the integral vanishes since the integration of $cos$ and $sin$ over $[0, 2\pi]$ is zero. Therefore we only have,
$$
\sum_\alpha\int_\Omega curl( u^\alpha(r,z)) v^\alpha(r,z) d\hat{\Omega}
$$
But now, how to get rid of the sum to have a formulation in terms of $u^\alpha$ and $v^\alpha$ only ?
Thanks in advance
 A: 
$\Omega \subset \mathbb{R}^3$ is a tore. 

The condition for the domain in which Green formula to hold is that $\Omega$ is bounded and has certain smoothness. Being axially symmetric is not the key premise.

$\displaystyle\int_\Omega \mathrm{curl}(u) v d\Omega = \int_\Omega u\nabla v  d\Omega - \int_{\partial\Omega} u\times\mathbf{n} v  \, d\Omega. $

There are two minor typos in the formula: (a) the integral on the surface should be with respect to the surface measure $dS$, not the volume measure $d\Omega$. (b) there should be a cross product sign between $\mathbf{u}$ and $\nabla v$. 
The proof is rather straight forward using Gauss-Green Theorem (Evans's PDE book appendix C.2 uses this name):
$$
\int_{\Omega} \partial_{x_i} \phi\, dx = \int_{\partial \Omega} \phi n_i \,dS.
$$ 
This formula leads us to for the vector field $\mathbf{u} = (u_1,u_2,u_3)$:
$$
\int_{\Omega} (\partial_{x_j} u_i - \partial_{x_i} u_j)d\Omega = \int_{\partial \Omega} (n_j u_i - n_i u_j)dS \implies \int_{\Omega} \nabla \times \mathbf{u} \,d\Omega = \int_{\partial \Omega}  \mathbf{n}\times \mathbf{u} \,dS.\tag{1}
$$
Now use the product rule for curl:
$$
\nabla\times (v\mathbf{u}) = \nabla v\times \mathbf{u} + v\nabla \times \mathbf{u},
$$
and replace the $\mathbf{u}$ with $v\mathbf{u}$, and use the fact that cross product is anti-commutative, we have:
$$
\int_{\Omega} (\nabla v\times \mathbf{u} + v\nabla \times \mathbf{u}) \,d\Omega = \int_{\partial \Omega}  \mathbf{n}\times (v\mathbf{u}) \,dS
\\
\implies  \int_{\Omega} v\nabla \times \mathbf{u}  \,d\Omega = \int_{\Omega} \mathbf{u} \times \nabla v - \int_{\partial \Omega}  (\mathbf{u} \times \mathbf{n} ) v\,dS.
$$
This is the Green formula you wanna prove.
