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?
Best Answer
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 $\frac{5\pi}{24}$.