∴ Area = \(\int\limits^{\sqrt{12}}_{-\sqrt6} y \, dx\)
\(= \int\limits^0_{-\sqrt6} -(x^2 - 9)dx + \int\limits_0^{\sqrt{12}}(x^2 - 9)dx\)
\(= - \left[\frac{x^3}3 - 9x\right]^0_{-\sqrt6} + \left(\frac{x^3}3 - 9x\right)^{\sqrt{12}}_0\)
\(= \frac{-6\sqrt6}3 + 9\sqrt6 + \left|\frac{12\sqrt{12}}{3}- 9\sqrt{12}\right|\)
\(= - 2\sqrt6 + 9\sqrt6 + | 4\sqrt{12} - 9\sqrt{12}|\)
\(= 7\sqrt6 + 5\sqrt{12}\)
\( = 7\sqrt6 + 10\sqrt3 \) square units.