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On pg 74 of Dalarsson's Tensors, Relativity and Cosmology (The Integral theorems for tensor field chapter), the covariant surface vector was defined as: $$dS_k=\frac{1}{2}\epsilon_{kmn}dx^mdx^n=\frac{1}{2}\sqrt{g}e_{kmn}dx^mdx^n. \tag{10.41}$$

In the Descartes coordinates, where $\sqrt{g}=1$, the components of this vector are given by $$dS_1=dx^2dx^3, dS_2=dx^1dx^3 , dS_3=dx^1dx^2 . \tag{10.42}$$

However, I attempted calculating the components in (10.42), for instance $$dS_1=\frac{1}{2}(e_{123}dx^2dx^3+e_{132}dx^3dx^2)$$ which yielded $$dS_1=\frac{1}{2}(dx^2dx^3-dx^3dx^2)$$ which was clearly incorrect. So it would be great if anyone could explain the errors that I have made.

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By the way, this looks like this is a discussion in the context of 3-dimensional Euclidean space, but you might want to check that and edit the question to give that information explicitly.

Equation 10.41 can't be describing the area covector in terms of infinitesimal coordinate changes. If it was, then due to the antisymmetry of the Levi-Civita tensor $\epsilon_{kmn}$ on the indices $m$ and $n$, this expression would vanish identically.

So I think their notation $dx^m dx^n$ actually means the wedge product $dx^m\wedge dx^n$, i.e., these are differential forms. So your final result looks fine, because $dx^3\wedge dx^2=-dx^2\wedge dx^3$, so it reduces to $dS_1=dx^2\wedge dx^3$.

If you want to see a different presentation of this material, without differential forms, you could check out section 7.6.3 of my special relativity book, which is free online.

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The symbols $dx^1$ and $dx^2$ are presumably meant to be differential forms, and these anticommute: $$ dx^1\wedge dx^2 =- dx^2\wedge dx^1. $$ Many authors leave out the $\wedge$ symbol when writing products of forms, and this (again "presumably" because I have not seen the book) is what the author is doing here. With this understood, your computation is essentially correct. When you insert infinitesimal vectors $\delta {\bf x}_1$ and $\delta {\bf x}_2$ into the two-form $dS_3$ you have $$ dS_3 (\delta {\bf x}_1, \delta {\bf x}_2)=\frac 12 (dx^1\wedge dx^2-dx^2\wedge dx^1)(\delta {\bf x}_1, \delta {\bf x}_2)= \frac 12 (\delta x_1^1 \delta x_2^2- \delta x_1^2 \delta x_2^1) = \frac 12 (\delta {\bf x}_1 \times \delta {\bf x}_2)_3 $$ which is the third component of the vector products of the displacements $\delta {\bf x}_1$ and $\delta {\bf x}_2$. This is the third component of the vector element of area defined by the parallelogram whose sides are $\delta {\bf x}_1$ and $\delta {\bf x}_2$.

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