Prove that $$\int_S\left(d\vec\sigma\times \vec\nabla\right)\times \vec P=\int_{\partial S}d\vec r\times \vec P~.$$ Can someone help me with this problem? I'm having troubles with using the correct identities to get the answer.


1 Answer 1


A good way to deal with these things is to learn how to use the Levi-Civita symbols, but since you are asking this question I will assume you are not familiar with it.

Another good idea is to simplify by making your expressions scalar. Let there be some unit vector $\mathbf{\hat{a}}$ that is pointing in the same direction everywhere. I will use bold-face to denote vectors. You can have $\mathbf{\hat{a}}=\mathbf{\hat{x}}$ or $\mathbf{\hat{y}}$ or $\mathbf{\hat{z}}$ or any combination of them.


$$ \mathbf{\hat{a}}.\int_S d^2 r \: \left(\mathbf{\hat{n}}\times\boldsymbol{\nabla}\right)\times \boldsymbol{P} $$

Where $\mathbf{\hat{n}}$ is the unit vector normal to the surface $S$, whilst $d^2 r$ is the surface element (in your notation $d\boldsymbol{\sigma}=d^2 r\: \mathbf{\hat{n}}$).

The neat thing about $\mathbf{\hat{a}}$ is that it is constant, and so can be taken in and out of the integral as well as the derivative. You can then show (try it component by component if you wish):

$$ \mathbf{\hat{a}}.\left(\mathbf{\hat{n}}\times\boldsymbol{\nabla}\right)\times\boldsymbol{P}=-\mathbf{\hat{n}}.\boldsymbol{\nabla}\times\left(\mathbf{\hat{a}}\times\boldsymbol{P}\right) $$

It follows that:

$$ \mathbf{\hat{a}}.\int_S d^2 r \: \left(\mathbf{\hat{n}}\times\boldsymbol{\nabla}\right)\times \boldsymbol{P} = - \int_S d^2r \: \mathbf{\hat{n}}.\boldsymbol{\nabla}\times\left(\mathbf{\hat{a}}\times\boldsymbol{P}\right)=-\int_{\partial S} dl \:\mathbf{\hat{l}}.\left(\mathbf{\hat{a}}\times\boldsymbol{P}\right) $$

Where $dl$ is the arc-length element and $\mathbf{\hat{l}}$ is the tangent vector. All you need to do now is to re-arange $\mathbf{\hat{l}}.\left(\mathbf{\hat{a}}\times\boldsymbol{P}\right)\to\mathbf{\hat{a}}.\left(\mathbf{\hat{l}}\times\boldsymbol{P}\right)$, and take $\mathbf{\hat{a}}$ out of the integral. Since it is valid for any $\mathbf{\hat{a}}$, it is valid as a vector identity.


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