My Math Forum Question in Stokes and Divergence theorem.

 Applied Math Applied Math Forum

 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

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post dmandman Real Analysis 0 August 18th, 2011 05:53 AM ZardoZ Applied Math 2 June 21st, 2011 05:31 AM george gill Calculus 5 May 14th, 2011 02:13 PM Curupira Calculus 0 October 14th, 2010 03:32 PM ggyyree Calculus 1 April 23rd, 2009 05:39 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top