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I am having a problem trying to check if the identity below should be positive or negative.

$$ \int\; \boldsymbol{A} \times (\boldsymbol{\nabla} a ) = \pm \int (\boldsymbol{\nabla} \times \boldsymbol{A} )a$$

where $a$ is a scalar function and the vectors are defined in 2-dimensional space.

For me the correct development should be:

First change the order of the vectorial product: $$ =- \int (\boldsymbol{\nabla} a )\times \boldsymbol{A} $$ than integrating by parts and eliminating the surface term:

$$= + \int a (\boldsymbol{\nabla} \times \boldsymbol{A} ) $$ therefore the identity is positive. But I am not sure if this is correct. I try using index notation to prove, but got confused: $$ \int A_i \epsilon_{ij}(\nabla_j a )= -\int (\nabla_jA_i \epsilon_{ij}) a = \int (\nabla_jA_i \epsilon_{ji}) a$$ again positive.

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  • $\begingroup$ Hi, welcome to Math SE. It looks like you're working with $2$-dimensional vectors, which you didn't state explicitly. $\endgroup$
    – J.G.
    Commented Nov 1, 2023 at 16:13
  • $\begingroup$ Adding a constant to $a$ changes the right side but not the left? $\endgroup$
    – aschepler
    Commented Nov 1, 2023 at 16:20
  • $\begingroup$ yes I am in two dimensions. I edited. Thanks $\endgroup$
    – Nathan
    Commented Nov 1, 2023 at 19:38
  • $\begingroup$ Leaving out the boundary term, for starters, just makes this incorrect, unless you say explicitly that either $\mathbf A$ or $a$ has compact support. You refer to "eliminating the surface term" — this is referring to the boundary term as if you were in 3 dimensions. Very confusing. At any rate, the usual Stokes's/Green's Theorem gives $\int_{\partial R} a\vec A\cdot d\vec r = \iint_R (\nabla\times(a\vec A) = \iint_R \big(\nabla a\times\vec A + a(\nabla\times\vec A)\big)$. $\endgroup$
    – Ted Shifrin
    Commented Nov 1, 2023 at 19:49
  • $\begingroup$ Also, please do not use irrelevant tags. This is just multivariable/vector calculus. If you want a tensor-analysis symbolic proof, that's still not differential geometry. :) $\endgroup$
    – Ted Shifrin
    Commented Nov 1, 2023 at 19:56

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