Centre of mass of bounded region conformation of numerical answer Hello I am looking for any help on solving the following;
I want to find the centre of mass of the uniform solid that is bounded by the regions $x^2+y^2=2x$, $z=\sqrt{x^2+y^{2}}$ and $z=0$, with the assumption of constant density $\delta$
My thoughts:
I know that the answer should be in form $(X,Y,Z)$ , where each is located by evaluating $\delta\iiint_{D} x dV$ (for the X coordinates, replaced with y and z respectability for theirs) , and dividing it by the total mass, $M=\delta\iiint_{D} dV$ ( noticing the $\delta$s should cancel)
But I am just having issues finalizing and continuing with the solution.
I think I would probably use cylindrical coordinates,
with $x=r\cos\theta$, $y=r\sin\theta$ and $z=z$
Then I would have,
$r^{2}cos^{2}\theta+r^{2}sin^{2}\theta=2rcos\theta$
$r^{2}=2rcos\theta$
$\frac{r}{2}=cos\theta$
we also have $z=0$ and $z=r$
I am now looking to see if anyone can help to conform if my numerical answers are correct or incorrect.
Update:
After solving , I obtained an answer of $$(\frac{4}{5},\frac{3}{10},\frac{9\pi}{16})$$ 
Can anyone please tell me if this is right or if I made a mistake and should look it over again?
This is now what I am wondering above else, if someone can tell me if it is correct or not.
Thank you
 A: Let's take a look on the integration domain:

The restriction $x^2 + y^2 = 2 x$ can be expressed in polar coordinates as $r(\theta) = 2 \cos \theta$. Then
$$
M = \delta \iiint_V dV = \color{red}{2} \delta \int_0^{\frac{\pi}2} \int_0^{2\cos \theta} \int_0^r r \,dz dr d\theta = \frac{\color{red}{32} \delta}{9}
$$
Edit
The volume $V$ can be written as
$$
V = \left\{(x,y,z) \in \mathbb{R}^3 \,\Bigg|\, 0 \le z \le \sqrt{x^2 + y^2},\,(x-1)^2 + y^2 \le 1 \right\}
$$
and so, the mass of the region is
$$
M = \iiint_V \delta d V = \delta \iiint_Vdxdydz. 
$$
There are several ways of doing this calculation. The first thing to note is that $z = z(x,y)$, while the pair $(x,y)$ should be given one in terms of the other. This tells us that the first thing to do is integrate with respect to $z$, and then with respect to $x$ or $y$. Of course, the $2x$ term makes more simple to express $y = y(x)$ than the other way around.
\begin{align}
M &= \delta \int_0^2 \left(\int_{-\sqrt{2x-x^2}}^{\sqrt{2x-x^2}} \left(\int_0^{\sqrt{x^2 + y^2}} d z \right)dy \right)dx \\ \\
&= \delta \int_0^2 \left(\int_{-\sqrt{2x-x^2}}^{\sqrt{2x-x^2}} \sqrt{x^2 + y^2} dy \right)dx
\end{align}
We could integrate this expression using trigonometric substitutions, but is a lot of hard work and in the end, it's the same as working with polar coordinates.
One way to do this in a clear manner is to go back to abstract notation, i.e,
$$
M = \delta \iint_\Omega r dxdy
$$
where
$$
\Omega = \left\{(x,y) \in \mathbb{R}^2 \,\Bigg|\, (x-1)^2 + y^2 \le 1 \right\}
$$
or, the circle of center $(1,0)$ and radius $1$. If we figure out how to define this circle properly in polar coordinates, we are done. Another way is to simply translate the circle to the origin and work from there. Let's do this method first.
Method 1
The integral at hand is
$$
M = \delta \iint_\Omega \sqrt{x^2 + y^2} dx dy = \delta \iint_{\Omega'} \sqrt{(\xi + 1)^2 + y^2}d\xi dy
$$
where $\Omega'$ is the unit circle at the origin, and $\xi = x-1$. Then, in polar coordinates
$$
M = \int_0^1 \int_0^{2\pi} \sqrt{r^2 + 2 r \cos \theta + 1}\,rdr d\theta.
$$
At first, this looked like a good idea, but with little effort, we can see it will take some serious work to calculate (at least in real variables).
Method 2
In polar coordinates, $\Omega$ is simply represented as
$$
\Omega = \left\{(r,\theta) \in \mathbb{R}^+\times[0,2 \pi) \,\Bigg|\, r \le 2 \cos \theta,\, 0 \le \theta  < \pi\right\}
$$
so,
$$
M = 2 \delta \int_0^{\pi \over 2} \int_0^{2 \cos \theta} r^2 dr d\theta = \frac{32 \delta}{9}.
$$
where we have correctly applied the domain symmetry. 
The coordinates
From here, the coordinates are relatively simple. Given that, in our formulation, $x$ and $y$ do not depend on $z$, we can go straight to
\begin{align}
\bar{x} &= \frac{9}{32} \iint_\Omega x r dx dy = \frac{9}{16} \int_0^{\pi \over 2} \int_0^{2\cos\theta}r^3 \cos \theta drd\theta = \frac{6}{5} \\
\bar{y} &= \frac{9}{32} \iint_\Omega y r dx dy = 0
\end{align}
and $\bar{z}$ has to be worked like the mass, i.e,
\begin{align}
\bar{z} &= \frac{9}{32} \int_0^2 \left(\int_{-\sqrt{2x-x^2}}^{\sqrt{2x-x^2}} \left(\int_0^{\sqrt{x^2 + y^2}} z d z \right)dy \right)dx \\ \\
&= \frac{9}{64}\iint_\Omega r^2 dx dy = \frac{9}{32} \int_0^{\pi \over 2} \int_0^{2 \cos \theta}r^3 dr d\theta = \frac{27 \pi}{128}.
\end{align}
In conclusion, the center of mass of $V$ is located at
$$
\left(\frac{6}{5}, 0,\frac{27 \pi}{128}\right)
$$
