How to find the center of mass of half a torus? Consider a halved solid torus (half a donut). The radius of the torus are $R_1$ and $R_2$.  I need to find its center of mass. The hint they give is that the center of mass of a homogeneous solid object $\Omega \subset \Bbb R^3$ is calculated as
$$\overline{x}=\int_{\Omega}\overline{r}d\overline{r}.$$

I really don't understand this formula, I don't know what $\overline{r}$ means and what is $\Omega$ in this case. I'd appreciate that someone explains what this formula means and how to apply it in this problem. Thanks in advance.
 A: I found it easiest to use cylindrical coordinates to set up the integrals needed for the center of mass.  Before we do so, however, I set my coordinate system up as follows.  I have positive $x$ coming out of the screen, positive $y$ going to the right, and positive $z$ up.  In cylindrical coordinates $(r,\phi,z)$:
$$x = r \cos{\phi}$$
$$y=r \sin{\phi}$$
where we have the limits defining the region $\Omega$:
$$\phi \in \left [ -\frac{\pi}{2},\frac{\pi}{2} \right]$$
$$z \in [-R_2,R_2]$$
$$r \in \left [ R_1 - \sqrt{R_2^2-z^2},  R_1 + \sqrt{R_2^2-z^2}\right]$$
Also, for an object of constant mass density, the expression for the $x$ component of the center of mass is
$$\bar{x} = \frac{\displaystyle \int_{\Omega} d^3 \vec{x} \, x}{\displaystyle \int_{\Omega} d^3 \vec{x}}$$
(Note that, by symmetry, we have $\bar{y}=0$ and $\bar{z}=0$.)  Let's first evaluate the denominator:
$$\begin{align} \int_{\Omega} d^3 \vec{x} &= \int_{-\pi/2}^{\pi/2} d\phi \, \int_{-R_2}^{R_2} dz \, \int_{R_1 - \sqrt{R_2^2-z^2}}^{R_1 + \sqrt{R_2^2-z^2}} dr \, r \\ &= \frac{\pi}{2}\int_{-R_2}^{R_2} dz \left [\left (R_1 + \sqrt{R_2^2-z^2} \right )^2 - \left (R_1 - \sqrt{R_2^2-z^2} \right )^2 \right ]\\ &= 4 \pi R_1 \int_0^{R_2} dz \, \sqrt{R_2^2-z^2}\\ &= \pi^2 R_1 R_2^2 \end{align}$$
Now we evaluate the center of mass:
$$\begin{align}\bar{x} &= \frac{1}{\pi^2 R_1 R_2^2} \int_{-\pi/2}^{\pi/2} d\phi \, \int_{-R_2}^{R_2} dz \, \int_{R_1 - \sqrt{R_2^2-z^2}}^{R_1 + \sqrt{R_2^2-z^2}} dr \, r^2 \cos{\phi} \\ &= \frac{4}{3 \pi^2 R_1 R_2^2} \int_0^{R_2} dz \, \left [\left (R_1 + \sqrt{R_2^2-z^2} \right )^3 - \left (R_1 - \sqrt{R_2^2-z^2} \right )^3 \right ]\\ &= \frac{8}{3 \pi^2 R_1 R_2^2} \int_0^{R_2} dz \,\left [3 R_1^2 \sqrt{R_2^2-z^2} + \left (R_2^2-z^2 \right )^{3/2} \right ] \\ &= \frac{8}{3 \pi^2 R_1 R_2} \left ( \frac{3 \pi}{4} R_1^2 R_2 + \frac{3 \pi}{16} R_2^3 \right )\end{align}$$
Simplifying, I get
$$\bar{x} = \frac{4 R_1^2+R_2^2}{2 \pi R_1}$$
ADDENDUM
As a quick note, in the limits as $R_2 \to 0$, we find that the center of mass becomes
$$\bar{x}=\frac{2}{\pi} R_1$$
which agrees with the center of mass of a uniform wire bent into a semicircle.
