You can rewrite the formula of the cylinder as $ (x-1)^2+y^2=1 $ so, we have the surface $S:z=\sqrt {x^2+y^2}, (x,y)\in D$ wherein $D:=(x-1)^2+y^2≤1$. You know that $Area(S)=\iint_D d\sigma$ wherein $d\sigma=\frac{||\nabla f ||}{|\frac{\partial f}{\partial z}|}dx dy$. Here $f=x^2+y^2-z^2=0$ as you noted so, $d\sigma=\frac{||\nabla f ||}{|\frac{\partial f}{\partial z}|}dx dy=\frac{||(2x,2y,-2z)||}{|-2z|}dx dy$$=\frac{\sqrt{4x^2+4y^2+4(x^2+y^2)}}{2|z|}=\frac{\sqrt{8}}{2}dxdxy=\sqrt{2}dxdy$ Now, I think the rest is easy since your problem is really a homework. :)