$S= \{(x,y,z): x^2+y^2+z^2=4, (x-1)^2+y^2 \leq 1 \}$
Parametrize the set $S$ by $G(\theta, \phi)=(2\cos\theta \sin\phi, 2\sin\theta \sin\phi, 2\cos\phi)$ where we know that $\theta \in [0,2\pi]$ and $\phi \in [0,\pi]$.
Then simplifying the second equation in the set $S$ we have:
$4\sin^2 \phi-4\cos\theta \sin\phi \leq 0 \implies \sin\phi \leq \cos\theta$ and because we know that for any angle $\sin,\cos$ is bounded below and above by $-1,1$ respectively, i.e. $-1 \leq \sin\phi \leq \cos\theta \leq 1$.
So for $\phi \in [-\pi/2, \pi/2]$ (from $\sin^{-1}-1 \leq \phi \leq \sin^{-1}1)$ and combining the intervals we have $\phi \in [0,\pi/2]$. So the limits of integration for $\phi$ is $\phi \in [0, \pi/2]$.
Similarly from calculating $-1 \leq \cos\theta \leq 1$ which is equivalent to $\pi \leq \theta \leq 2\pi$ and combining it with the interval $\theta \in [0, 2\pi]$ we have that the limits of integration for $\theta$ is $\theta \in [\pi, 2\pi]$.
If someone can verify my work for limits of integration portion that'll be greatly appreciated! Thank you.
The second equation is a cylinder on the positive x axis. It's equation in spherical form is $$sin\phi = cos \theta \implies \phi = \frac{\pi}{2} - \theta$$
If we consider the area above the $xy$ plane, the limits are
$$\phi = 0 to (\frac{\pi}{2} - \theta)$$
$$\theta = -\frac{\pi}{2} to \frac{\pi}{2}$$