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Given a integral $$\int_vd^3{r} \;\vec{r}\;\rho(r)$$ and How do you convert it to spherical coordinate system, noting that $\rho(r)$ is indeed as it is without vector, i.e. it is spherically symmetric $\rho(\vec{r})=\rho(r)$.

$$\int_0^{\infty}\int_0^{2\pi}\int_o^\pi \dots \rho(r)r^2\sin(\phi)d\phi d\theta dr$$

I guess all I have to do is convert $\vec{r}$ from radial vector coordinate system to spherical coordinate system. But, I am stuck.

ADDED There should not be $v$. $$\int d^3{r} \;\vec{r}\;\rho(r)$$

Actually, I was asked to calculate dipole moment in this case.

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The integral of a vector, as you have written it, is just shorthand notation for a vector of integrals. Concretely, if we write $\vec r= (x,y,z)$ in cartesian coordinates, then \begin{align} \int d^3r\, \vec r\,\rho(r) &= \left(\int d^3 r\, x\,\rho(r),\int d^3 r\, y\,\rho(r),\int d^3 r\, z\,\rho(r) \right) \end{align} Now, we simply note the transformation between cartesian coordinates and spherical coordinates, and use this to evaluate each of these component integrals. In the convention for $\phi$ and $\theta$ as the polar and azimuthal coordinates respectively, we have \begin{align} x = r\sin\phi\cos\theta, \qquad y = r\sin\phi\sin\theta, \qquad z = r\cos\phi \end{align} so we get, for example, \begin{align} \int d^3 r\, x\,\rho(r) = \int dr\,d\phi\,d\theta \,(r^2\sin\phi)(r\sin\phi\cos\theta)\rho(r) \end{align} and similarly for the other components.

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    $\begingroup$ This is all well and good, so +1, but do you think the OP was really supposed to just examine the parity of $\vec{r}$ and $\rho$ over a spherically symmetric domain? Unless I'm mistaken, the answer needs no integration at all, right? $\endgroup$ – user10851 Oct 29 '13 at 2:20
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    $\begingroup$ @ChrisWhite The thought crossed my mind. What threw me off was the subscript $v$ under the first integral sign in the question which made me think that integration over an arbitrary domain is desired. Also, there is a lot of talk in the question about converting to spherical coordinates. Perhaps the OP will clarify re. the domain of integration. $\endgroup$ – joshphysics Oct 29 '13 at 2:26
  • $\begingroup$ @ChrisWhite, I have edited. By no integration is needed, do you mean, that it is trivially zero? But how would that be shown mathematically? $\endgroup$ – user59756 Oct 29 '13 at 6:08
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    $\begingroup$ @user59756 Think of what happens to the various objects under the integral sign when you send $\vec r\to -\vec r$. Alternatively, look at the angular integrations for each component of the integral in spherical coordinates. $\endgroup$ – joshphysics Oct 29 '13 at 6:13
  • $\begingroup$ @user59756 Sure thing. $\endgroup$ – joshphysics Oct 29 '13 at 7:15

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