Hello, kittycat!
The problem has be solved "head on".
Find the surface area of the portion of the sphere of radius 4
that lies inside the cylinder $\displaystyle x^2 +y^2 \:=\:12$ and above the xyplane.
Turn the diagram 90° "clockwise". Code:

* * *
*  *
***A
* B * :*
 * :
*  * : *
  *     +    + *  
* O C *

*  *
*+*
*  *
* * *

In right triangle $\displaystyle ABO\!:\;\;OA = 4,\:OB = \sqrt{12}$
. . Hence: .$\displaystyle BA\,=\,OC\,=\,2$ .**
We want the area of the "polar cap" formed by the segment at the far right.
The formula for the area of a surface of revolution is: .$\displaystyle S \;=\;2\pi\int^b_a y\,\sqrt{1 + \left(\frac{dy}{dx}\right)^2}\,dx$
The equation of the circle is: .$\displaystyle x^2+y^2\:=\:16\quad\Rightarrow\quad y \:=\:\sqrt{16x^2}$
Then: .$\displaystyle \frac{dy}{dx}\:=\:\frac{x}{\sqrt{16x^2}}$
. . . . $\displaystyle 1 + \left(\frac{dy}{dx}\right)^2\;=\;1 + \frac{x^2}{16x^2} \;=\;\frac{16}{16x^2}$
. . . .$\displaystyle \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \;=\;\frac{4}{\sqrt{16x^2}} $
Hence, we have: .$\displaystyle S \;=\;2\pi\int^4_2\sqrt{16x^2}\cdot\frac{4}{\sqrt{16x^2}}\,dx \;=\;8\pi\int^4_2\!\!dx \;=\;8\pi x\bigg]^4_2$
. . . . $\displaystyle = \;8\pi(4  2) \;=\;\boxed{16\pi} $
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
**
Note that: .$\displaystyle \angle AOC \,=\,60^o$
Fascinating bit of trivia:
A central cone with vertex angle 120° subtends ¼ the area of the sphere.
With such a cone at the north pole and the south pole,
. . the "equatorial band" has half the area of the sphere.