For any constant vector c,
[tex]\mathbf{c} \cdot \int_C \mathbf{t} \times \mathbf{n} ds = \int_C (\mathbf{n} \times \mathbf{c}) \cdot \mathbf{t} ds =\int_S \nabla \times (\mathbf{n} \times \mathbf{c}) \cdot \mathbf{n}da[/tex]
Using the vector identity
[tex]\nabla \times (\mathbf{a} \times \mathbf{b}) = \mathbf{b} \cdot \nabla \mathbf{a} - \mathbf{a} \cdot \nabla \mathbf{b} - (\nabla \cdot \mathbf{a} ) \mathbf{b} + (\nabla \cdot \mathbf{b} )\mathbf{a}[/tex]
and the fact that c is constant, the above integral becomes
[tex]\int_S [\mathbf{c}\cdot \nabla \mathbf{n} - (\nabla \cdot \mathbf{n})\mathbf{c} ] \cdot \mathbf{n} da <br />
=<br />
\mathbf{c} \cdot \int_S (\nabla \mathbf{n} ) \cdot \mathbf{n} - (\nabla \cdot \mathbf{n }) \mathbf{n} da[/tex]
The first term in the integrand is zero, since it is just the gradient of the length of the unit normal. Also, since the vector c was arbitrary, its dot product with the integral can be dropped:
[tex]\int_C \mathbf{t} \times \mathbf{n} ds = -\int_S (\nabla \cdot \mathbf{n} ) \mathbf{n} da[/tex]
Depending on how you define the normal (outward or inward), div n = -2H up to a sign*. So that gives the result
[tex]\int_C \mathbf{t} \times \mathbf{n} ds = \int_S 2H \mathbf{n} da[/tex]
I had an extra minus sign in the first post.
* The mean curvature H of a surface is equal to half the trace of the second fundamental form B: 2H = gjkbjk. The components of the second fundamental form can be defined as
[tex]b_{jk} = -\mathbf{x}_j \cdot \mathbf{n}_{,k} = -\mathbf{x}_j \cdot \mathbf{n}_{,k} = - n^i_{{ };k}\mathbf{x}_j \cdot \mathbf{x}_i = -n^i_{{ };k}g_{ij}[/tex]
Multiplying both sides by the inverse of g_ij and then summing the diagonal elements gives
[tex]g^{jk}b_{jk} = - n^k_{{ };k} = -\nabla \cdot \mathbf{n}[/tex]