is it correct that the following integral is, actually, a triple integral of the variable $r$? (it has to do with electromagnetic calculations) The integral is symbolised with a single S integration symbol (integration to "everywhere" -meaning- to all the area the charge extends- then comes the function $f(r)$ where $r\ge 0$ and the integrity element is not dr but $d^3r$. Note that the $f(r)$ function is actually a product of two functions which are both able to be integrated seperatedly, and are continuous functions.

$$\int_D f(r) \, d^3r,\qquad r>0$$

The above are referring to Green's theorem. Thank you.

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    $\begingroup$ I can't see the integral! $\endgroup$ – leo May 24 '12 at 2:19

The integral is over three dimensional space.

I assume that $r=\sqrt{x^2+y^2+z^2}$. In a common abuse of notation, $d^3r$ is an infinitesimal volume element. There are many ways to parametrize such an integral. In Cartesian coordinates the volume element is $dx dy dz$. In spherical coordinates, which seem to be relevant to your problem, it is $r^2 \sin\theta dr d\theta d\phi$. See here for more details. (Note that $\rho$ on that page is your $r$.) Here is a better link that follows our convention for spherical coordinates.


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