Area cylinder limited by cone

Question

I'm ask to find the surface area of the cylinder $x^2+y^2=2x$ limited by the cone $z=\sqrt{(x^2+y^2)}$ and the plane $z=0$ and . I know that the cilinder's center is at $(1,0)$, I understand how the shape look like but I'm having a lot of problems parameterizing that surface, how can I do it?

Answer 1

$x^2+y^2=2x$ can be rewritten as $(x-1)^2+y^2=1$: a circle with centre $(1,0)$ 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:

$$\int_0^{2\pi} \sqrt{2(cos \theta+1)} d\theta$$