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 ⇒ {x}^{2} - 9 = \pm 3$</p><p>$⇒ {x}^{2} = 12$(or)${x}^{2} = 6 ⇒ x = \pm \sqrt{12}$(or)$x = \pm \sqrt{6}$</p><p>The graphs of the two curves are shown in figure</p><p><img src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/31f98fea-dd53-4345-bed9-cc24aff51813-image.png" style="width: 442px; height: 230px;" /></p><p>We have to find the area of the shaded region.</p><p>Required area$= 2{A}_{1} + 2{A}_{2}$($∵$The areas${A}_{1}&{A}_{2}$are symmetrical about$y -$axis )</p><p>where${A}_{1} = \mathop{∫}_{0}^{\sqrt{6}}\left[\left(9 - {x}^{2}\right) - \left(3\right)\right]dx$</p><p>${A}_{1} = \mathop{∫}_{0}^{\sqrt{6}}\left(6 - {x}^{2}\right)dx$$= {\left[6x - \dfrac{{x}^{3}}{3}\right]}_{0}^{\sqrt{6}} = 6\sqrt{6} - \dfrac{6\sqrt{6}}{3} = \dfrac{12\sqrt{6}}{3} = 4\sqrt{6}$

${A}_{2} = \mathop{∫}_{\sqrt{6}}^{3}\left[3 - \left(9 - {x}^{2}\right)\right]dx + \mathop{∫}_{3}^{\sqrt{12}}\left[3 - \left({x}^{2} - 9\right)\right]dx$

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

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

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

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

Hence, required area $= 2{A}_{1} + 2{A}_{2}$

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

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

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