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

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$4 (4 \sqrt{3} + \sqrt{6} - 4)$

$8 (4 \sqrt{3} + \sqrt{6} - 9)$

$4 (2 \sqrt{3} + \sqrt{6} - 4)$

$8 (4 \sqrt{3} + 3 \sqrt{6} - 9)$

answer is D.

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Detailed Solution

$\begin{array}{l} = 2 \int_{0}^{3} (\sqrt{9 + y} - \sqrt{9 - y}) d y + 2 \int_{3}^{9} (\sqrt{9 - y}) d y \\ = 2 [\int_{0}^{3} (9 + y)^{1 / 2} d y - \int_{0}^{3} (9 - y)^{1 / 2} d y + \int_{3}^{9} (9 - y)^{1 / 2} d y] \\ = 2 [\frac{2}{3} {[(9 + y)^{3 / 2}]}_{0}^{3} + \frac{2}{3} {[(9 - y)^{3 / 2}]}_{0}^{3} - \frac{2}{3} {[(9 - y)^{3 / 2}]}_{3}^{9}] \\ = \frac{4}{3} [12 \sqrt{12} - 27 + 6 \sqrt{6} - 27 - (0 - 6 \sqrt{6})] \\ \begin{aligned} = \frac{4}{3} [24 \sqrt{3} + 12 \sqrt{6} - 54] \\ = 8 (4 \sqrt{3} + 2 \sqrt{6} - 9) \\ (N T A K E Y B O N U S) \end{aligned} \end{array}$