its a bit tedious if you don't use the Levi-Civita tensor, but letting dxdy=-dydx (I know it's a bit weird but it's the only way this thing works) the LHS =
\int \int \frac{1}{2} \epsilon_{ijk} \epsilon_{klm} \partial_{j} T dx^l dx^m
= \int \int \frac{1}{2} (\delta_{il} \delta_{jm} - \delta_{im} \delta_{jl}) \partial_{j} T dx^l dx^m
= \int \int \frac{1}{2} (\partial_{m} T dx^i dx^m - \partial_{l} T dx^l dx^i)
= \int \int \frac{1}{2} (-\partial_{m} T dx^m dx^i - \partial_{l} T dx^l dx^i) = - \oint T dx^i
where we used the contracted epsilon identity and changed the last integral into a closed one because going from integrating with respect to an area to a line integral the boundaries change so that the integral becomes closed
Anyway hope this helped the way to do it using standard vector operations is to rewrite d\bold a = (dydz,dxdz,dxdy) as d\bold a = \frac{1}{2} (dydz -dzdy,dxdz-dzdx,dxdy-dydx) which makes sense because the area between two vectors is \left|\bold a \times \bold b \right|
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