Answer is: 15
\(0\le 9x \le y^2 \ \&\ y\ge 3x - 6\)
A = Required Area \(=\left[\int\limits_{0}^{1}(-3 \sqrt{x}) d x-\int\limits_{0}^{1}(3 x-6) d x\right]\)
\(A=-\left.3\left(\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right)\right|_{0} ^{1}-\left.\left(\frac{3 x^{2}}{2}-6 x\right)\right|_{0} ^{1}\)
\(A=-2[1-0]\left[\frac{3}{2}-6\right]\)
\(\mathrm{A}=-2-\frac{3}{2}+6=\frac{5}{2}\) Sq. unit
\(\therefore 6 \mathrm{A}=6 \times \frac{5}{2}=15\)
