Its a good idea to know what these circles look like.
The first is easy, its or r=12, which is a circle of radius 12.
if you add 36 to both sides of the second you get .
Now that's a circle that has diameter 12. It traces from (0,0) to (12,0).
NOW that we know what it looks like we can integrate it from zero to .
The outer radius is 12 and the inner one is that messy circle which we need to solve for in terms of r.
I would do the cal 3 integral of rdrd .
Actually, forget about it.
It's one circle inside another, just use geometry.
Unless I drew it incorrectly, it's one quarter of the area of the bigger circle of radius 12,
minus half the area of the smaller circle of radius 6...