Voss Weyl formula and divergence theorem in curvilinear coordinates I am unable to reconcile the divergence theorem in curvilinear coordinates, and what I get by an application of the Voss Weyl formula and the divergence theorem in $\mathbb{R}^2$. Could someone help identify what I am doing wrong?

With reference to the above figure, and notation, $S_{\alpha\beta}$ = the metric tensor, $S$ = its determinant, $\mathbf{S}_\alpha$ is the covariant basis for the tangent space, and $\mathbf{S}^\alpha$ is the contravariant basis.
Divergence theorem:
$\int\limits_\Omega \nabla_\alpha v^\alpha \text{d}\Omega 
= \oint\limits_\Gamma v^\alpha n_\alpha\text{d}\Gamma$
My derivation:
First, starting from the left hand side,
$
\begin{aligned}
    \int\limits_\Omega \nabla_\alpha v^\alpha \text{d}\Omega
 &= \int\limits_A \frac{1}{\sqrt{S}}\frac{\partial}{\partial S^\alpha}
                  \left(v^\alpha\sqrt{S}\right) \sqrt{S}\;\text{d}A
                  \quad \text{by the Voss-Weyl formula} \\
 &= \int\limits_A \frac{\partial}{\partial S^\alpha}
                  \left(v^\alpha\sqrt{S}\right) \text{d}A \\
 &= \oint\limits_C v^\alpha n_\alpha \sqrt{S}\; \text{d}C
                  \quad \text{by the divergence theorem in $\mathbb{R}^2$}
\end{aligned}
$
On the other hand, starting from the right hand side,
$
  \oint\limits_\Gamma v^\alpha n_\alpha\text{d}\Gamma
= \oint\limits_C v^\alpha n_\alpha \sqrt{S_{\beta\gamma}t^\beta t^\gamma}\; \text{d}C
$
which is different.
I am clearly not understanding something right. Any pointers would be much appreciated.
 A: 
Thanks to Prof. Pavel Grinfeld for pointing out the flaw in the derivation in the question (see comment here). The key is to distinguish between the normal to $\Gamma$ and the normal to $C$. (This type of distinction is also made between normals in the context of the divergence theorem applied to volumes on page 242 of Prof. Grinfeld's book).
Here is the normal, $\mathbf{n}$, to $\Gamma$.
\begin{equation}
\begin{aligned}
    \mathbf{n} &= \frac{t^\alpha\mathbf{S}_\alpha}
                       {\sqrt{S_{\beta\gamma}t^\beta t^\gamma}}
                  \times \nu \\
               &= \frac{t^\alpha}
                       {\sqrt{S_{\beta\gamma}t^\beta t^\gamma}}
                  \left(\mathbf{S}_\alpha \times \nu\right) \\
               &= \frac{t^\alpha}
                       {\sqrt{S_{\beta\gamma}t^\beta t^\gamma}}
                  \epsilon_{\delta\alpha}\mathbf{S}^\delta \\
               &= n_\delta\mathbf{S}^\delta
\end{aligned}
\end{equation}
Thus,
\begin{equation}
    n_\alpha = \frac{\epsilon_{\alpha\delta}t^\delta}
                    {\sqrt{S_{\beta\gamma}t^\beta t^\gamma}}
\end{equation}
By analogy, the coefficients of the normal, $\bar{\mathbf{n}}$, to $C$, are given by
\begin{equation}
    \bar{n}_\alpha = \frac{\bar{\epsilon}_{\alpha\delta}t^\delta}
                          {\sqrt{\bar{S}_{\beta\gamma}t^\beta t^\gamma}}
\end{equation}
Now, $\bar{\epsilon}_{\alpha\delta} = \frac{\epsilon_{\alpha\delta}}{\sqrt{S}}$
and $\bar{S}_{\beta\gamma} = \delta_{\beta\gamma}$, so that
\begin{equation}
    \bar{n}_\alpha = \frac{1}{\sqrt{S}}
                         \frac{\sqrt{S_{\beta\gamma}t^\beta t^\gamma}}
                              {\sqrt{\delta_{\mu\upsilon}t^\mu t^\upsilon}}
                     n_\alpha
\end{equation}
Now, the derivation in the question looks like this:
\begin{equation}
\begin{aligned}
    \int\limits_\Omega \nabla_\alpha v^\alpha \text{d}\Omega
 &= \int\limits_A \frac{1}{\sqrt{S}}\frac{\partial}{\partial S^\alpha}
                  \left(v^\alpha\sqrt{S}\right) \sqrt{S}\;\text{d}A
                  \quad \text{by the Voss-Weyl formula} \\
 &= \int\limits_A \frac{\partial}{\partial S^\alpha}
                  \left(v^\alpha\sqrt{S}\right) \text{d}A \\
 &= \oint\limits_C v^\alpha \bar{n}_\alpha \sqrt{S}\; \text{d}C
                  \quad \text{by the divergence theorem in $\mathbb{R}^2$} \\
 &= \oint\limits_C v^\alpha n_\alpha 
                         \frac{\sqrt{S_{\beta\gamma}t^\beta t^\gamma}}
                              {\sqrt{\delta_{\mu\upsilon}t^\mu t^\upsilon}}\; \text{d}C \\
 &= \oint\limits_\Gamma v^\alpha n_\alpha \text{d}\Gamma
\end{aligned}
\end{equation}
and all is well!
That, from my perspective, resolves the conflict alluded to in my question.
However, I also want to share another calculation, which does not use the tensor notation formally, that guided me in the above derivation (I seem to have to switch between the two ways of thinking, each guiding the other). The following is a demonstration of the divergence theorem on the surface $\Omega$, knowing the divergence theorem in the coordinate space.
Let $\hat{\Xi}$ be the coordinate map, $\tilde{S}$ be a parametrization of $C$, and $\tilde{\Xi}$, the corresponding parametrization of $\Gamma$ as shown in the figure.
\begin{equation}
    \begin{aligned}
        & \oint_\limits\Gamma \mathbf{v}\cdot\mathbf{n}\text{d}\Gamma \\
      = & \int_{t_0}^{t_1} \mathbf{v}\cdot
                           \left(\frac{\tilde{\Xi}'(t)}
                                      {\left\Vert\tilde{\Xi}'(t)\right\Vert}
                                 \times\nu \right)
                           \left\Vert\tilde{\Xi}'(t)\right\Vert \text{d}t \\
      = & \int_{t_0}^{t_1} \mathbf{v}\cdot
                           \left(\tilde{\Xi}'(t) \times\nu \right) \text{d}t \\
      = & \int_{t_0}^{t_1} \mathbf{v}\cdot
                           \left(
                               \left[\begin{array}{c|c}
                                   \frac{\partial\hat{\Xi}}{\partial S^1} &
                                   \frac{\partial\hat{\Xi}}{\partial S^2}
                               \end{array}\right] \tilde{S}'(t)
                               \times\nu 
                           \right) \text{d}t \\
      = & \int_{t_0}^{t_1} \mathbf{v}\cdot
                           \left(
                               \left[\begin{array}{c|c}
                                   \frac{\partial\hat{\Xi}}{\partial S^1} \times \nu &
                                   \frac{\partial\hat{\Xi}}{\partial S^2} \times \nu
                               \end{array}\right] \tilde{S}'(t)
                           \right) \text{d}t \\
      = & \int_{t_0}^{t_1} \left(
                               \left[\begin{array}{c|c}
                                   \frac{\partial\hat{\Xi}}{\partial S^1} &
                                   \frac{\partial\hat{\Xi}}{\partial S^2}
                               \end{array}\right] 
                               \begin{pmatrix} v^1 \\ v^2 \end{pmatrix}
                           \right)
                           \cdot
                           \left(
                               \left[\begin{array}{c|c}
                                   \frac{\partial\hat{\Xi}}{\partial S^1} \times \nu &
                                   \frac{\partial\hat{\Xi}}{\partial S^2} \times \nu
                               \end{array}\right] \tilde{S}'(t)
                           \right) \text{d}t \\
      = & \int_{t_0}^{t_1} \begin{pmatrix} v^1 \\ v^2 \end{pmatrix}^\top
                           \left[\begin{array}{c|c}
                               \frac{\partial\hat{\Xi}}{\partial S^1} &
                               \frac{\partial\hat{\Xi}}{\partial S^2}
                           \end{array}\right]^\top                               
                           \left[\begin{array}{c|c}
                               \frac{\partial\hat{\Xi}}{\partial S^1} \times \nu &
                               \frac{\partial\hat{\Xi}}{\partial S^2} \times \nu
                           \end{array}\right] \tilde{S}'(t)
                           \text{d}t \\
        & \quad \text{the integrand here can be recognized as}\ 
                                 (v^1\mathbf{S}_1 + v^2\mathbf{S}_2) \cdot
                                  \left[-\mathbf{S}^2 \; \mathbf{S}^1\right]
                                 \tilde{S}'(t) S \\
      = & \int_{t_0}^{t_1} \begin{pmatrix} v^1 \\ v^2 \end{pmatrix}^\top
                           \left[\begin{array}{cc}
                               0 & S \\
                              -S & 0
                           \end{array}\right] \tilde{S}'(t)
                           \text{d}t \\
      = & \int_{t_0}^{t_1} \begin{pmatrix} v^1 \\ v^2 \end{pmatrix}^\top 
                           \begin{pmatrix} 
                               \tilde{S}^2{}'(t) \\ -\tilde{S}^1{}'(t) 
                           \end{pmatrix}
                           S \text{d}t \\
      = & \int_{t_0}^{t_1} \begin{pmatrix} v^1 S \\ v^2 S \end{pmatrix}^\top 
                           \frac{
                           \begin{pmatrix} 
                               \tilde{S}^2{}'(t) \\ -\tilde{S}^1{}'(t) 
                           \end{pmatrix}
                           }{\left\Vert \tilde{S}'(t) \right\Vert}
                           \left\Vert \tilde{S}'(t) \right\Vert \text{d}t \\
      = & \oint_\limits{C} \begin{pmatrix} v^1 S \\ v^2 S \end{pmatrix}^\top
                           \bar{\mathbf{n}} \text{d}C \\
      = & \int_\limits{A}\left(
                             \frac{\partial}{\partial S^1}\left(v^1 S\right) +
                             \frac{\partial}{\partial S^2}\left(v^2 S\right)
                         \right)\text{d}A 
                         \quad \text{by the divergence theorem in}\ \mathbb{R}^2 \\
      = & \int_\limits{A}\frac{1}{S}
                         \left(
                             \frac{\partial}{\partial S^1}\left(v^1 S\right) +
                             \frac{\partial}{\partial S^2}\left(v^2 S\right)
                         \right) S \text{d}A \\
      = & \int_\limits{\Omega}\frac{1}{S}
                         \left(
                             \frac{\partial}{\partial S^1}\left(v^1 S\right) +
                             \frac{\partial}{\partial S^2}\left(v^2 S\right)
                         \right) \text{d}\Omega
    \end{aligned}
\end{equation}
which is the desired result.
