I'm trying to draw the region of the surface area of the cylinder, $x^2+y^2 \le 2x$, limited by the cone $z=\sqrt{(x^2+y^2)}$ and the plane $z=0$ . I know that the cylinder's center is at $(1,0)$ $x^2+y^2\leq2x$ can be rewritten as $(x-1)^2+y^2\leq1$: a closed circle with center $(1,0)$ and radius $1$.
First parametrise by setting $x-1=\cos \theta$, $y=\sin \theta$
This gives $z=\sqrt{2x}=\sqrt{2(\cos \theta+1)}$.
We can now imagine the cylinder being unwrapped, so that the area is:
$$\int_0^{2\pi} \sqrt{2(\cos \theta+1)} d\theta$$
I have no idea how to visualize the part of the cylinder $x^2+y^2 \le 2x$ limited by the cone $z=\sqrt{(x^2+y^2)}$ and the plane $z=0$.
Trying to visualize, I'm guessing that the surface area given by the integral is the area
$$\int_0^{2\pi} \sqrt{2(\cos \theta+1)} d\theta$$
of a conic section?
I don’t think the cylinder is bounded above which is a problem.
The cylinder is tangent to the $y$ axes in $(0,0)$, which is the cone vertex. Of the cone you are considering the upper part $0 \le z$.
The portion of cylinder below that is a sort of "nail".
Working in cylindrical coordinates $(r, \phi, z)$ centered at $(1,0,0)$, the point $(r=1, -\pi < \phi \le \pi, 0)$ will be distant $\sqrt{(1+cos\phi)^2+sin^2\phi}$ from the origin, and the cone will be at the same distance above it.
Along the base circle $dl=d\phi$ and the elementary surface is $dA= \sqrt{(1+cos\phi)^2+sin^2\phi}\;d\phi$.
Can you continue from here ?