I need help solving this problem.
Using polar coordinates, evaluate the integral which gives the area which lies in the first quadrant between the circles $\displaystyle x^2+y^2=144$ and $\displaystyle x^2-12x+y^2=0 $
Its a good idea to know what these circles look like.
The first is easy, its $\displaystyle x^2+y^2=144$ or r=12, which is a circle of radius 12.
if you add 36 to both sides of the second you get $\displaystyle (x-6)^2+y^2=36$.
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 $\displaystyle \pi/2$.
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$\displaystyle \theta$.
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...
$\displaystyle \pi\biggl({144\over 4}-{36\over 2}\biggr)$.