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 and above the xyplane.
Turn the diagram 90° "clockwise". Code:

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

*  *
*+*
*  *
* * *

In right triangle
. . Hence: . .**
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: .
The equation of the circle is: .
Then: .
. . . .
. . . .
Hence, we have: .
. . . .
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
**
Note that: .
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.