The area bounded by the curve $y=\left|x^2-9\right|$ and the line y=3 is :

$\begin{array}{l}=2{\int }_0^3 (\sqrt{9+y}-\sqrt{9-y})dy+2{\int }_3^9 \left(\sqrt{9-y}\right)dy\\ =2\left[{\int }_0^3 (9+y{)}^{1/2}dy-{\int }_0^3 (9-y{)}^{1/2}dy+{\int }_3^9 (9-y{)}^{1/2}dy\right]\\ =2\left[\frac{2}{3}{\left[(9+y{)}^{3/2}\right]}_0^3+\frac{2}{3}{\left[(9-y{)}^{3/2}\right]}_0^3-\frac{2}{3}{\left[(9-y{)}^{3/2}\right]}_3^9\right]\\ =\frac{4}{3}[12\sqrt{12}-27+6\sqrt{6}-27-(0-6\sqrt{6}\left)\right]\\ \begin{array}{r}=\frac{4}{3}[24\sqrt{3}+12\sqrt{6}-54]\\ =8(4\sqrt{3}+2\sqrt{6}-9)\\ (NTA KEY\hspace{0.17em}BONUS)\end{array}\end{array}$