find the volume below the cone $z=2-\sqrt{x^2+y^2}$ and inside the cylinder $(x-1)^2+y^2=1$ I know I need to transform it into cylindrical system to integrate it but I don't know how to deal with r, since the cylinder is not center at (0,0)
Assuming the volume is bounded below by the $xy$ plane, you need to ask yourself what is the projection of the volume in the $xy$ plane. And indeed, cylindrical coordinates are the way to go. 

So obviously, the projection is the disc $(x-1)^2+y^2\le 1$, which you need to convert to polar coordinates. Expand, and you get
$$
x^2-2x+1 +y^2 \le 1\quad \Rightarrow \quad r^2\le 2 r \cos \theta \quad \Rightarrow \quad 0 \le r\le 2  \cos \theta
$$
And careful with $\theta$, the cylinder is shifted along the $x$ axis, so 
$$
-\pi/2 \le \theta \le \pi/2
$$

Can you take it from there?
$$\int_{0}^{2\pi}\int_0^1 (2-\sqrt[n]{x^2+y^2}) dx dy$$ and put

$$x=1+r\cos{\theta}$$ and $$y=r\sin{\theta}$$ and $$dxdy=rdrd{\theta}$$

answer:$2\pi$