surface area using double integral

my problem is like this: find the surface area of the portion of the cone $\displaystyle z=\sqrt{x^2+y^2}$ that lies inside the cylinder $\displaystyle x^2+y^2=2x $.

My solution:

$\displaystyle x^2+y^2=2x $

=> $\displaystyle (x-1)^2+y^2=1 $

So my circle now has center at (1,0), radius = 1

=> $\displaystyle y=\sqrt{1-(x-1)^2} $

$\displaystyle A= \int_{0}^{2} \int_{-\sqrt{1-(x-1)^2}}^{\sqrt{1-(x-1)^2}}

\sqrt {(\frac{\partial z}{\partial x})^2 + (\frac{\partial z}{\partial y})^2 +1}

dydx$

$\displaystyle A= \int_{0}^{2} \int_{-\sqrt{1-(x-1)^2}}^{\sqrt{1-(x-1)^2}}

\sqrt 2}

dydx$

$\displaystyle A= \int_{0}^{2} 2\sqrt2(\sqrt{1-(x-1)^2}dx$

I stuck at this point, I dont know if I set up a right limit of integration, please help?