How a vol. integral becomes a vol. integral plus surface integral

[tom8]

Hello,

Please see this pdf at some universities website:

http://physics.ucsc.edu/~peter/110A/helmholtz.pdf

In line 14 the author claims using integration by parts...I do not understand
who could the integration by parts be used here.

I understand the general case where we have der(a*b)=a der(b) + b der(a)

but I am not able to see how this happened here..

Thanks!

 

[Muphrid]

You use the fact that, for example, for scalar $\psi$ and vector $C$,

$$\int \nabla (\psi C) dV = \oint \psi C \cdot dA$$

This is a generalization of the fundamental theorem of calculus, and it's used even in the 1D case:

$$\int_a^b u v' \; dx + \int_a^b u' v \; dx = \int_a^b (uv)' \; dx = uv|_a^b$$

 

[tom8]

I understand the divergence theorem, but in my example I have two terms on the right, one of which is a surface integral and one is vol. integral. Thus this cannot be div. theorem.

 

[Muphrid]

It's the divergence theorem and integration by parts (or the product rule).

$$\int \nabla(\psi C) \; dV = \int C \cdot \nabla \psi \; dV + \int \psi \nabla \cdot C \; dV = \oint \psi C \cdot dA$$

 

[tom8]

It's the divergence theorem and integration by parts (or the product rule).

$$\int \nabla(\psi C) \; dV = \int C \cdot \nabla \psi \; dV + \int \psi \nabla \cdot C \; dV = \oint \psi C \cdot dA$$

Perfect, can you tell me what do psi corresponds to in my example?

Does correspond only to the rational (1/|r-r'|) or tothe rational and to d3r' ?

 

[Muphrid]

$\psi$ is only the scalar function $1/|r-r'|$. Their $d^3r'$ is $dV$.

 

[tom8]

$\psi$ is only the scalar function $1/|r-r'|$. Their $d^3r'$ is $dV$.

Now I got it. It is two steps at once, thank you so much !