1
$\begingroup$

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.

$\endgroup$
  • 3
    $\begingroup$ I can't see the integral! $\endgroup$ – leo May 24 '12 at 2:19
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.