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Given a Diffeomorphisem $\Phi$, consider the boundary integral

$ \int_{\partial\Phi(\Omega)}f\cdot\nu ~dS$,

where $\nu$ is the outer unit normal vector on $\Phi(\Omega)$. I now want to do a change of variables for this integarl to get

$ \int_{\partial\Phi(\Omega)}f\cdot\nu ~dS= \int_{\partial\Omega}[f\circ\Phi]\cdot[\nu\circ\Phi]\left|\operatorname{det}D\Phi\right|dS$.

My Problem now is to understand the right hand site of this equation. Since $\nu\circ\Phi$ is not necessary a outer unit normal vector on $\Omega$ I am not sure if this transformation can be done in this way.

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    $\begingroup$ The change of variables isn't right, anyhow, as you need to pull back $dS$ by $\Phi|_\Omega$, not by the diffeomorphism $\Phi$. $\endgroup$ – Ted Shifrin Jan 25 at 20:34
  • $\begingroup$ I am not really sure what you mean by that. Could you explaine it? $\endgroup$ – Bara Jan 26 at 16:23
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    $\begingroup$ You've changed notation. So now I need to say you need to pull back $dS$ by $\Phi|_{\partial\Omega}$. $|det D\Phi|$ gives the correct transformation law if you're integrating over the entire region $\Omega$. The best way to approach this problem entirely is with integration and pullback of differential forms. $\endgroup$ – Ted Shifrin Jan 26 at 19:27
  • $\begingroup$ Okay, I have actually no Idea how to do that. Can you recommend some literature on this topic? I dont even know where to start. $\endgroup$ – Bara Jan 26 at 22:12
  • $\begingroup$ I don't know your background, so this is very hard to address. I can point you to my lectures on YouTube linked in my profile (and to the textbook on which this course was based). There are all sorts of multivariable analysis and differential topology texts that are more advanced, as well. My approach to differential forms is concrete and does not use tensor products. $\endgroup$ – Ted Shifrin Jan 26 at 22:29

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