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Vector / Integral question - probably simple, but can't see answer. Watch

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    http://www.maths.ox.ac.uk/system/fil.../Sheet3_13.pdf

    It's question 6 of the above, just the bit at the start of step c) where you prove \nabla\phi(r)=\int \frac{k\nabla'\rho(r')}{|r-r'|}dV'.

    Basically, you have \phi(r), and you can assume the identities proved in a) and b). Prime on a vector operator means treat r' as variable and r as constant.

    What I did was;

    \nabla\phi(r)=\int\nabla(\frac{k  \rho(r')}{|r-r'|})dV'
    =\int\nabla(\frac{1}{|r-r'|})k\rho(r')dV'
    =-k\int\nabla'(\frac{1}{|r-r'|})\rho(r')dV'

    Then using integration by parts or the product rule;

    =-k\int\nabla'(\frac{\rho(r')}{|r-r'|})\rho(r')dV'+k\int(\frac{ \nabla'(\rho(r'))}{|r-r'|})dV'

    However, this is the wrong answer, as the first term won't always be zero and the second term is the required integral. The class notes I copied off the board (without understanding at the same time) seem to do the same thing I have, but the answer is definitely the one asked for, as that's what it says in the textbook.

    Can anyone tell me what I'm missing. I don't know where to start looking. Thanks


    EDIT: I've solved the problem - there is a typo in the question. It should say rho vanishes outside and on the boundary, which causes the extra term to disappear, by the divergence theorem. I am now really happy. Feel free to delete the thread.
 
 
 
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