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In Einstein summation convention, the curl of a cross product has componentsThe second line uses the contraction of two Levi-Civita symbols; the last uses the product rule. These components prove the curl of a cross product identity, including both divergence and directional-derivative terms.

Now apply that identity to $m\times v$. On the orientable surface, $m=n$, $m\cdot v=0$, and $|m|=1$. Differentiating $m\cdot m=1$ gives $m_i{\partial }_j{m}_i=0$. Taking a dot product with $m$ in the curl of a cross product formula therefore gives, on $S$,Only tangency on the surface is needed; no claim is made that the extension $v$ is tangent away from it. The right side is the surface divergence of the tangent vector field.

With the boundary oriented according to Stokes theorem, cyclic invariance of the scalar triple product giveswhere $u$ is the outward conormal of a surface boundary. Thus Stokes theorem provesThis is the surface divergence theorem for a tangent field.

For the disc in the $z=0$ plane, take $n=(0,0,1)$ and $v=\mathbf{r}=(x,y,z)$. The field is tangent on the disc, and ${\partial }_j{v}_i={\delta }_{ij}$. The contraction is $3-|n{|}^2=2$, so the surface integral is $2\pi R^2$. On the positively oriented circular boundary, $u=(\cos \theta ,\sin \theta ,0)$ and $v=Ru$, so the boundary flux is $R(2\pi R)=2\pi R^2$. Both sides agree.