My Math Forum Question in Stokes and Divergence theorem.
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 August 10th, 2013, 09:14 AM #1 Newbie   Joined: Jun 2013 Posts: 22 Thanks: 0 Question in Stokes and Divergence theorem. In page 3 of this articlehttp://faculty.uml.edu/cbaird/95 ... sition.pdf I have two question: I use $\vec r$ for $\vec x$ and $\vec {r'}$ for $\vec {x'}$ in the article. $\vec F(\vec {r})=\frac {1}{4\pi}\nabla\left[\int_{v'}\nabla'\cdot\left(\frac{\vec {F}(\vec {r'})}{|\vec {r}-\vec {r'}|}\right)d\tau'-\int_{v'}\frac{\nabla'\cdot \vec {F}(\vec {r'})}{|\vec {r}-\vec {r'}|}d\tau'\right]+\frac {1}{4\pi}\nabla\times\left[-\int_{v'}\nabla'\times\left(\frac{\vec {F}(\vec {r'})}{|\vec {r}-\vec {r'}|}\right)d\tau'+\int_{v'}\frac{\na bla'\times \vec {F}(\vec {r'})}{|\vec {r}-\vec {r'}|}d\tau#39;\right]$ (1) it said $\frac {1}{4\pi}\nabla\int_{v'}\nabla'\cdot\left( \frac{\vec {F}(\vec {r'})}{|\vec {r}-\vec {r'}|}\right)d\tau'=\int_{s'}\left(\fr ac{\vec {F}(\vec {r'})}{|\vec {r}-\vec {r'}|}\right)\cdot d\vec {s#39;}$ 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)$\int_{v'}\nabla'\times\left(\frac{\vec {F}(\vec {r'})}{|\vec {r}-\vec {r'}|}\right)d\tau=\int_{s'}\left(\frac{\v ec {F}(\vec {r'})}{|\vec {r}-\vec {r'}|}\right)\times d\vec {s#39;}$. Again, how do I proof this surface integral goes to zero at infinity? Please help, Thanks.

 Tags divergence, question, stokes, theorem

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