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 cylinder's center is at $(1,0)$
$x^2+y^2=2x$ 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:
$$\int_0^{2\pi} \sqrt{2(\cos \theta+1)} d\theta$$
Basically from my understanding of the above:the surface area of the cylinder limited by the cone is
$$ \int_0^{2\pi}\int_0^\sqrt{2(\cos\theta+1)}drd\theta$$
I was told that when doing integration in polar coordinates,
$$dx dy=r dr d\theta$$ .
Is the "$r$ " in the integral: $$\int_0^{2\pi} \sqrt{2(\cos \theta+1)} d\theta$$ missing or equal to one?
I expected the surface area of the cylinder limited by the cone to be instead $$ \int_0^{2\pi}\int_0^\sqrt{2(\cos\theta+1)}rdrd\theta$$
and not :$$\int_0^{2\pi} \sqrt{2(\cos \theta+1)} d\theta$$
What's going on with "$ r$ " from the jacobian?
This is a followup question on :Area cylinder limited by cone.