Area bounded by the curve $x^2+y^2≤ 36$ and outside the parabola $y^2 = 9x$ is

$(24π-3\sqrt{3})$ sq units

The x-coordinates of the points of intersection of two curves $x^2 + y^2 = 36$ and $y^2 = 9x$ are roots of the equation

$x^2+y^2-36$   [On eliminating $y^2$]

$⇒ x^2+9x-36=0$

$⇒(x+12) (x-3)=0⇒ x-3=0⇒ x=3$    $[∵x>0]$

Putting $x = 3$ in $y^2 = 9x$, we obtain $y = ± 3\sqrt{3}$

Thus, two curves intersect at $A (3, 3\sqrt{3})$ and $B (3, -3\sqrt{3})$

∴ Require area = Area of the shaded region

= Area of semi-circle CDEOC + 2 Area of the region OAEO

$=\frac{1}{2}×π×6^2+2\int\limits_0^3(y_2-y_1)dx$

$=18π+2\int\limits_0^3(\sqrt{36-x^2}-3\sqrt{x})dx$

$=18π+\left[x\sqrt{36-x^2}+36\sin^{-1}\frac{x}{6}-6×\frac{2}{3}x^{3/2}\right]_0^3$

$=18π+9\sqrt{3}+36×\frac{π}{6-12\sqrt{3}}=(24π-3\sqrt{3})$