Answer
$12\pi$
Work Step by Step
On the $xy-$plane the circle $x^2+y^2=4$ is centered at $(0,0)$ and has the radius of $2$.
Let us verify the second circle.
$x^2+y^2-4y-12=0$
$x^2+y^2-4y=12$
$x^2+y^2-4y+4=16$
$(x-0)^2+(y-2)^2=4^2$
It is a circle centered at $(0,2)$ with the radius of $4$.
Area of the first circle: $\pi\cdot 2^2=4\pi$
Area of the second circle: $\pi\cdot 4^2=16\pi$
From the graph below, the first circle is inside the second circle.
Then, the area of the region that lies outside the first circle but inside the second circle is $16\pi-4\pi=12\pi$.
