Given the density in a given spherical co-ordinate

$\rho(r,\theta,\phi) = \rho_0 e^{-r/R} (1-cos \theta)$

find the center of mass of the sphere.

I managed to get using infinitesimal sized the equation of the volume in a given location:

$dV = r^2sin(\phi)dr\ d\theta\ d\phi$

However I still do not know how to find the center of mass using spherical co-ordinates. I thought of using multiple integrals, and it works for the total Mass, but I'm clueless on the subject of finding the location using the co-ordinates of the cent


1 Answer 1


Center of mass calculation is essentially the same as other averages such as expectation value in probability theory. Essentially it is the weighted average of the space coordinate.

The definition is best wrote down in Cartesian coordinates:

$$\mathbf{r}_\text{center} = \int_V \mathbf{r} f(\mathbf{r}) \mathrm{d} V$$

with $\mathbf{r} = (x,y,z)$ and $\mathrm{d} V = \mathrm{d} x \mathrm{d} y \mathrm{d} z$. And $$f(\mathbf{r}) = \frac{1}{\int_V\rho(\mathbf{r}) \mathrm{d} V} \rho(\mathbf{r})$$ the volume normalized distribution function.

This hint should allow you to solve the stated problem. It is now just a matter of correct calculation of the integrals with spherical coordinates. You will get the center of mass in Cartesian coordinates, though. But I dont if there exists any simple integral expression to obtain the center of mass directly in spherical coordinates.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .