The area bounded by the curve y = x 2 - 9 and the line y = 3 is

We first find the points of intersection of the curves $y=\left|x^2-9\right|$ and $y=3$

which gives, $\left|x^2-9\right|=3$  $\Rightarrow x^2-9=\pm 3$

$\Rightarrow x^2=12$ (or) $x^2=6$ $\Rightarrow x=\pm \sqrt{12}$ (or) $x=\pm \sqrt{6}$

The graphs of the two curves are shown in figure

We have to find the area of the shaded region.

Required area $=2A_1+2A_2$  ($\because$ The areas $A_1$ & $A_2$ are symmetrical about $y-$ axis )

where $A_1=\underset{0}{\overset{\sqrt{6}}{\int }}\left[\left(9-x^2\right)-\left(3\right)\right]dx$ 

$A_1=\underset{0}{\overset{\sqrt{6}}{\int }}\left(6-x^2\right)dx$$={\left[6x-\frac{x^3}{3}\right]}_0^{\sqrt{6}}$ $=6\sqrt{6}-\frac{6\sqrt{6}}{3}=\frac{12\sqrt{6}}{3}=4\sqrt{6}$

$A_2=\underset{\sqrt{6}}{\overset{3}{\int }}\left[3-\left(9-x^2\right)\right]dx+\underset{3}{\overset{\sqrt{12}}{\int }}\left[3-\left(x^2-9\right)\right]dx$

$={\int }_{\sqrt{6}}^3\left(-6+x^2\right)dx+{\int }_3^{\sqrt{12}}\left(12-x^2\right)dx$

$={\left[-6x+\frac{x^3}{3}\right]}_{\sqrt{6}}^3+{\left[12x-\frac{x^3}{3}\right]}_3^{\sqrt{12}}$

$=\left(-18+9\right)-\left(-6\sqrt{6}+2\sqrt{6})\right)+\left(12\sqrt{12}-4\sqrt{12}\right)-\left(36-9\right)$

$=-36+4\sqrt{6}+8\sqrt{12}$

Hence, required area $=2A_1+2A_2$

$=2\left(4\sqrt{6}\right)+2\left(-36+4\sqrt{6}+8\sqrt{12}\right)$

$=8\sqrt{6}-72+8\sqrt{6}+16\sqrt{12}$

$=16\sqrt{6}+16\sqrt{12}-72$