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$\displaystyle\int \left[ \nabla \times \dfrac {M(r')}{r} \right] d\tau =\oint \dfrac{1}{r} \left[ M(r')×da' \right]$

I came across this integral from the David Griffiths introduction to electrodynamics, page no 278 ,fourth edition . But I really don't understand the procedure they came up with. It would be a great help if someone explains it step by step.

Joe
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F.sharmin
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  • I would like a little more context here. For instance, what is $M$, and what are $r$ and $r'$? Are these integrals over curves, surfaces or volumes? Is the ambient space 2 or 3 dimensional? Without this kind of information, that equation could mean anything. – Arthur Apr 01 '19 at 10:05
  • well here M means magnetic dipole moment, r' is unit vector and r is distance – F.sharmin Apr 01 '19 at 10:07
  • at first it was volume integral then they converted to surface integral. I just dont understand the conversion procedure. – F.sharmin Apr 01 '19 at 10:10
  • There are many theorems (with names like Gauss, Green and Stokes connected to them) that link some integral over a volume to a related integral over the surface of that volume, and they are all in some shape or form a version of the fundamental theorem of calculus. I think for you, this one is a pretty good fit. – Arthur Apr 01 '19 at 10:18
  • thank you! Really appreciate it. – F.sharmin Apr 03 '19 at 13:56

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