Fdtxjr

$S=(x,y,z):x^2+y^2+z^2=4, (x-1)^2+y^2 leq 1 $.

$x^2+y^2+z^2=4 iff fracx^24+fracy^24 +fracz^24=1$

I'm not exactly sure what to parametrize the set $S$ by I thought of using spherical coordinates, in particular $G(theta, phi)=(frac12costheta sinphi, frac12sintheta sinphi, frac12cosphi)$

$(x-1)^2+y^2 leq 1$ I'm not sure how this is contributing to the work..

The set here is a pair of spherical caps - the result of intersecting a sphere with the inside of an off-center cylinder. The projection of the sphere in the $xy$-plane is a circle of radius $2$, while the projection of the cylinder is a circle of radius $1$, off-center by $1$. They touch on the boundary, but the smaller circle is entirely contained in the larger one - which means we get a cap with $z>0$ and another with $z<0$.

So, then, parametrizing. First, rectangular coordinates:

We have $(x-1)^2+y^2le 1$, and $z=pmsqrt4-x^2-y^2$. That gives us two parametrized regions:
$$0le xle 2, -sqrt1-(x-1)^2le ylesqrt1-(x-1)^2, z(x,y)=sqrt4-x^2-y^2$$
$$0le xle 2, -sqrt1-(x-1)^2le ylesqrt1-(x-1)^2, z(x,y)=-sqrt4-x^2-y^2$$

Well, that's a lot of square roots. Definitely not the easiest to work with.

How about cylindrical coordinates? We've got two circles here, so there are two ways to go. First, cylindrical with respect to the small circle:
$$0le rle 1, 0lethetale 2pi, x = 1+rcostheta, y = rsintheta, z=sqrt4-(1+rcostheta)^2-r^2sin^2theta$$
$$0le rle 1, 0lethetale 2pi, x = 1+rcostheta, y = rsintheta, z=-sqrt4-(1+rcostheta)^2-r^2sin^2theta$$
Those $z$ formulas simplify a bit, but there will still be a $costheta$ term in there.

Next, cylindrical with respect to the large circle:
$$(rcostheta-1)^2+(rsintheta)^2le 1,x=rcostheta,y=rsintheta,z=sqrt4-r^2$$
$$(rcostheta-1)^2+(rsintheta)^2le 1,x=rcostheta,y=rsintheta,z=-sqrt4-r^2$$
In this option, we've shunted the more complicated parts off into the bounds for $r$ and $theta$. Expanding that inequality out, it becomes
beginalign*r^2cos^2theta-2rcostheta+1+r^2sin^2theta &le 1\
r^2 &le 2rcosthetaendalign*
So then, $-dfracpi2lethetaledfracpi2$ and $0le rle 2costheta$. That looks like the way to go.

So, now that you have a parametrization, can you set up and solve the area integral?

Using spherical coordinates could be a better option. Use
$$x = rcostheta cosphi$$
$$y = rsintheta cosphi $$
$$z = rsinphi $$
Where in this case $r = 2$

Then the second equation becomes
$$r^2cos^2phi - 2rcostheta cosphi = 0$$
$$implies rcosphi = 2costheta $$
$$implies cosphi = costheta $$
$$implies phi = theta $$

The integral for the surface is

$$2int_0^fracpi2int_0^theta r^2cosphi dphi dtheta $$
$$= 8int_0^fracpi2sintheta dtheta = 8$$