calculate the surface area of the part of a cylinder $ x^2 + (y-1)^2 = 1 $ that is inside the sphere $ x^2 + y^2 + z^2 = 4 $. calculate the surface area of the part of a cylinder $ x^2 + (y-1)^2 = 1 $ that is inside the sphere
$  x^2 + y^2 + z^2 = 4 $.
my trial :
The Domain of integration on the YZ plane is : 
solving : 
(*) $  x^2 + (y-1)^2 = 1 $
$ x^2 + (y)^2 + z^2 \leq 4 $
We get : $ 0 \leq z \leq \sqrt{4-2y}$
and   $ -2 \leq y \leq 2 $
$||\nabla{Cylinder(x,y,z)}|| = ||(2x,2(y-1),0)|| = 2 ~$ see(*)
S = $2~\int_{-2}^{2}\int_{0}^{\sqrt{4-2y}}~~2~dydz$ ( by symmetry *2)
I also tried solving by parmetrization and i didn't get the same answer i really need HELP is this way alright ? or is there something wrong

Let $\gamma : [0, 2\pi] \to \mathbb{R}^2, \gamma(t) = (\cos t, 1+\sin t)$ be the parameterization of the circle $x^2+(y-1)^2 = 1$ in the $xy$-plane.
The height of the cylinder over a point $(x,y)$ on this circle is given by $z = \sqrt{4-x^2-y^2}$ so the area is
$$A = 2\int_\gamma z\,d\gamma = 2\int_\gamma \sqrt{4-x^2-y^2}\,d\gamma = 2\int_0^{2\pi}\sqrt{4-\cos^2t -(1+\sin t)^2}\,dt = 2\int_0^{2\pi}\sqrt{2(1-\sin t)}\,dt$$
We can solve this integral by noting that $1-\sin t = \left(\sin\frac{t}2 - \cos\frac{t}2\right)^2$ so we get
$$A = 16$$
Is this result what you are getting?
