# Thread: Center of mass of two spheres

1. ## Center of mass of two spheres

Here is the problem:

Find the center of mass of a solid with constant density (D) located in the upper semispace ($z\geq0$) and between the two spheres $x^{2}+y^{2}+z^{2}=1$ and $x^{2}+y^{2}+z^{2}=9$

My first instinct was to use spherical coordinates and set up the triple integral

$\int_{\phi=0}^{\pi/2} \int_{\theta=0}^{2\pi} \int_{\rho=1}^{3} \rho^{2}sin({\phi}) D {d\rho} {d\theta} {d\phi}$

(Sorry, that's the best I can do with my limited Latex knowledge).

I'm wondering if I have this setup right. Or do I need to create a double integral instead? I always get confused with triple integrals because the boundaries are not the actual volume of the shape, but rather the domain of some function f(x,y,z) where the triple integral is asking us to find sum.

2. ## Re: Center of mass of two spheres

and what about superposition?

3. ## Re: Center of mass of two spheres

Yes, that integral is the mass, M, of the object. It should be obvious, by symmetry, that the x and y coordinates of the center of mass are 0. To find the z coordinate, since $\displaystyle z= \rho cos(\phi)$,
$\displaystyle Mz= D\int_0^{\pi/2}\int_0^{2\pi}\int_1^3 \rho^3 sin(\phi)cos(\phi)d\rho d\theta d\phi$
$\displaystyle = 2\pi D\int_0^{\pi/2} sin(\phi)cos(\phi)d\phi\int_1^3 \rho^3 d\rho$

4. ## Re: Center of mass of two spheres

Originally Posted by ameva
and what about superposition?
Well the density is constant so there is no function to integrate......is that what you mean?

Originally Posted by HallsofIvy
Yes, that integral is the mass, M, of the object. It should be obvious, by symmetry, that the x and y coordinates of the center of mass are 0. To find the z coordinate, since $\displaystyle z= \rho cos(\phi)$,
$\displaystyle Mz= D\int_0^{\pi/2}\int_0^{2\pi}\int_1^3 \rho^3 sin(\phi)cos(\phi)d\rho d\theta d\phi$

$\displaystyle = 2\pi D\int_0^{\pi/2} sin(\phi)cos(\phi)d\phi\int_1^3 \rho^3 d\rho$
Thanks! I did find that the x and y coordinates were 0.

For z I had $15/13$......does that sound right?