Consider a point $P = (x,y,z) = (x,\sqrt{4-x^2},z)$ on the surface of the second cylinder. Suppose that $y\ge|z|$, so that $P$ is between the planes $y=z$ and $y=-z$. Then $z^2\le y^2$, so $x^2+z^2\le x^2+y^2=4$, and hence $P$ is automatically within the first cylinder. In other words, every point $(x,y,z)$ that
lies on the surface of the second cylinder;
is between the planes $y=z$ and $y=-z$; and
has $y\ge 0$ is on the surface of the intersection of the two cylinders.
This is a cylindrical face of the intersection; it contains vertical straight line segments that lie on the surface of the cylinder $x^2+y^2=4$, and it has curved top and bottom edges that are symmetric in the $xy$-plane. The mirror image of this face in the $xz$-plane is another face of the intersection, and there are two more such faces; they are also mirror images of each other, but in the $xy$-plane, and they contain straight line segments that lie on the surface of the cylinder $x^2+z^2=4$.
These four faces are pairwise congruent, so you need only get the area of one of them and multiply it by $4$.