In page 3 of this articlehttp://faculty.uml.edu/cbaird/95.657%282012%29/Helmholtz_Decomposition.pdf(adsbygoogle = window.adsbygoogle || []).push({});

I have two question:

I use ##\vec r## for ##\vec x## and ##\vec r'## for ##\vec x'## in the article.

[tex]\vec F(\vec r)=\frac {1}{4\pi}\nabla\left[\int_{v'}\nabla'\cdot\left(\frac{\vec F(\vec r')}{|\vec r-\vec r'|}\right)d\tau'-\int_{v'}\frac{\nabla'\cdot \vec F(\vec r')}{|\vec r-\vec r'|}d\tau'\right]+\frac {1}{4\pi}\nabla\times\left[-\int_{v'}\nabla'\times\left(\frac{\vec F(\vec r')}{|\vec r-\vec r'|}\right)d\tau'+\int_{v'}\frac{\nabla'\times \vec F(\vec r')}{|\vec r-\vec r'|}d\tau'\right][/tex]

(1) it said

[tex]\frac {1}{4\pi}\nabla\int_{v'}\nabla'\cdot\left(\frac{\vec F(\vec r')}{|\vec r-\vec r'|}\right)d\tau'=\int_{s'}\left(\frac{\vec F(\vec r')}{|\vec r-\vec r'|}\right)\cdot d\vec s'[/tex]

where this term goes to zero as ##|\vec r-\vec r'|\rightarrow\;\infty##. But not the ##\int_{v'}\frac{\nabla'\cdot \vec F(\vec r')}{|\vec r-\vec r'|}d\tau'##. Why is this true?

(2)Also it said using a version of Stokes theorem to change the third term ##\int_{v'}\nabla'\times\left(\frac{\vec F(\vec r')}{|\vec r-\vec r'|}\right)d\tau' ## to a surface integral and it will goes to zero as ##|\vec r-\vec r'|\rightarrow\;\infty##. I don't know a version of Stokes theorem that change from a volume integral to surface integral.

Please help.

Thanks

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# Question regarding surface and volume integral

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