# Triple integral to find the mass of the intersection between two spheres

I've got two unit spheres, one is centered at $(0,0,0)$ and the other at $(0,0,1)$, the intersection of these two spheres is my region $R$. I would like to find the integral:

$$\iiint\limits_R z\;dV$$ I think that it corresponds to the mass of the intersection of my two spheres with density z. I get two equations for the spheres: $$x^2+y^2+z^2=1 \\ x^2+y^2+(z-1)^2=1$$ I change them to spherical coordinates to obtain: $$\iiint\limits_R \rho^3\left(\cos\phi\right)\left(\sin\phi\right)d\rho\,d\phi\,d\theta$$ Now, I would like to find the limits of integration and this is where I would need help. I think that the limits for $\rho$ are: $\rho=1$ and $\rho=2\cos\phi$

I don't know the limits of $\phi$, I would say that one of them is $\,{\pi}/{3}\,$ as $\,2\cos\phi=1$.

And finally, the limits of $\theta$ are just from $0$ to $2\pi$.

So, how can I find the limits of integration and also do I have to separate the integral for each sphere?

• It is possible, and simpler, to work with only one sphere. Do you see why ? – Jean Marie Mar 6 '16 at 15:19
• I am afraid I don't see why – Akou Mar 6 '16 at 15:24
• You have a symmetry with respect to plane $z=1/2$, so all you have to do is to work only on sphere centered in $(0,0,0)$ with a single spherical cap mathworld.wolfram.com/SphericalCap.html and double the result at the end. – Jean Marie Mar 6 '16 at 15:44
• Thank you, but wouldn't this work if I was looking for the volume? Whereas here I am looking for the mass which is dependent on z so I can't just multiply by 2. – Akou Mar 6 '16 at 15:56
• All right. Sorry. I overlooked this. For another time, place in evidence, in the title or first sentence that you are working with inhomogeneous medium, it is rather non physical... – Jean Marie Mar 6 '16 at 16:09

The problem is $R$ cannot be described as only one region: in spherical coordinates it is the union of two regions $R_1$ and $R_2$: $$R_1=\{(\rho,\theta,\phi)\;|\; 0 \le \theta \le 2\pi ,0 \le \rho \le 1,0\le \phi \le \pi/3\}\\ R_2=\{(\rho,\theta,\phi)\;|\; 0 \le \theta \le 2\pi ,0 \le \rho \le 2\cos\phi,\pi/3\le \phi \le \pi/2\}$$ Therefore, the integral equals $$\iiint_Rz\; dV = \iiint_{R_1\cup R_2}z\; dV = \iiint_{R_1}z\; dV+\iiint_{R_2}z\; dV = \frac{5\pi}{24}$$ Alternatively, you can do this in one shot in cylindrical coordinates: $$R= \{(r,\theta,z)\;|\; 0 \le \theta \le 2\pi ,0 \le r \le \frac{\sqrt{3}}{2},1-\sqrt{1-r^2}\le z \le \sqrt{1-r^2}\}$$ You can check that the answer is again $\frac{5\pi}{24}$.
• Thank you very much, could you please explain to me how you get $\phi$ for R2? – Akou Mar 6 '16 at 17:47
• The spheres intersect at $\phi=\pi/3$, and then the sphere centered at $(0,0,1)$ never crosses the $xy$ plane, so $R$ is always above $z=0$; i.e. $\phi\le\pi/2$. – Kuifje Mar 6 '16 at 18:19