Integration by parts problem involving vector functions

[vabamyyr]

Hi,

I am trying to chew through the proof of reciprocity in MRI. At some point I come across to the following expression:

\Phi_{M}=\oint\vec{dl}\cdot\left[\frac{\mu_{0}}{4\pi}\int{d^{3}r'}\frac{\vec{\nabla'}\times\vec{M}(\vec{r'})}{\left|\vec{r}-\vec{r'}\right|}\right]

Now it says that by using integration by parts (where surface term can be ignored for finite current sources), and using vector identity, \vec{A}\cdot\left(\vec{B}\times\vec{C}\right)=-\left(\vec{A}\times\vec{C}\right)\cdot\vec{B}, we get

\Phi_{M}=\frac{\mu_{0}}{4\pi}\int{d^{3}r'\vec{M}(\vec{r'})\cdot\left[\vec{\nabla'}\times\left(\oint\frac{\vec{dl}}{\left|\vec{r}-\vec{r'}\right|}\right)\right]}

Can someone explain to me how to use integration by parts in this case and what does ignoring surface term mean in this context?

 

[Mandelbroth]

Hi,

I am trying to chew through the proof of reciprocity in MRI. At some point I come across to the following expression:

\Phi_{M}=\oint\vec{dl}\cdot\left[\frac{\mu_{0}}{4\pi}\int{d^{3}r'}\frac{\vec{\nabla'}\times\vec{M}(\vec{r'})}{\left|\vec{r}-\vec{r'}\right|}\right]

Now it says that by using integration by parts (where surface term can be ignored for finite current sources), and using vector identity, \vec{A}\cdot\left(\vec{B}\times\vec{C}\right)=-\left(\vec{A}\times\vec{C}\right)\cdot\vec{B}, we get

\Phi_{M}=\frac{\mu_{0}}{4\pi}\int{d^{3}r'\vec{M}(\vec{r'})\cdot\left[\vec{\nabla'}\times\left(\oint\frac{\vec{dl}}{\left|\vec{r}-\vec{r'}\right|}\right)\right]}

Can someone explain to me how to use integration by parts in this case and what does ignoring surface term mean in this context?

This link is helpful.