Integral/Vector calculus $\oint_{\partial S} u \vec \nabla v \cdot d \vec \lambda=\int_S (\vec \nabla u)\times (\vec \nabla v)\cdot d\vec S.$ I am trying to show that
$$
\oint_{\partial S} u \vec \nabla v \cdot d \vec \lambda=\int_S (\vec \nabla u)\times (\vec \nabla v)\cdot d\vec S
$$
using Levi Cevita notation methods only.  The Levi Cevita tensor is given by $\epsilon_{ijk}$ and is a totally anti symmetric tensor.
The functions u and v are dependent on the radius vector in 3 dimensions ($\vec r$).  I am stuck on this I know  the general idea of using this notation is to write things like
$$
(\vec \nabla \times \vec A)_i= \epsilon_{ijk} \partial_j A_k
$$
but I still am stuck.  Thank you
 A: We can prove  the identity by using Stoke's theorem and tensor notation
\begin{equation}
\oint_{\partial S}u \vec{\nabla} v \cdot d\vec{\lambda}=\int_S \big(\vec{\nabla} u\big)\times \big( \vec{\nabla}v \big)\cdot d\vec{S}.
\end{equation}
We now use Stoke's theorem to relate line integral to surface integral by
$$
\oint_{\partial S}u \vec{\nabla} v \cdot d\vec{\lambda}=\int_{S}\vec{\nabla}\times (u \vec{\nabla} v) \cdot \hat{n} dS.
$$
Now I will use your desired $\epsilon_{ijk}$ notation to show that 
$$
\int_{S}\vec{\nabla}\times (u \vec{\nabla} v) \cdot \hat{n} dS=\int_S \big(\vec{\nabla} u\big)\times \big( \vec{\nabla}v \big)\cdot d\vec{S}.
$$
We obtain
$$
\big( \vec{\nabla}\times (u \vec{\nabla}v) \big)_i=\epsilon_{ijk}\partial_j (u\partial_k v)=\epsilon_{ijk}( \partial_j u \partial_k v+u\partial_j \partial_k v)
$$
where I used the product rule.  However the term $\partial_j \partial_k v$ is symmetric, so we know this will vanish!  We can see this by using
$$
\partial_j \partial_k v=\frac{1}{2} (\partial_j \partial_k +\partial_k \partial_j)v
$$
since it is symmetric w.r.t j and k.  Thus we can see by swapping the indices j and k we obtain
$$
\epsilon_{ijk}\partial_j\partial_k v=-\epsilon_{ijk}\partial_k \partial_jv
$$
which is only true if this term is zero.  Also you know that a total antisymmetric tensor times a symmetric quantity is zero.  We are left with
$$
\big( \vec{\nabla}\times (u \vec{\nabla}v) \big)_i=\epsilon_{ijk}\partial_j u \partial_k v=(\vec{\nabla}u \times \vec{\nabla}v)_i
$$
which concludes the desired proof.  
