The area (in sq. units) of the part of the circle x2+y2=36,which is outside the parabola y2=9x,is:

The circle and curve intersect at x=3∴Area =πr2−2∫039xdx+∫3636−x2dx             =36π−123−2x236−x2+18sin−1x636             =36π−123−29π−932+3π             =24π−33