$0 \leq 9 x \leq y^2 ~\&~ y \geq 3 x-6$

$$\begin{aligned} & A=\text { Required Area }=\left[\int_0^1(-3 \sqrt{x}) d x-\int_0^1(3 x-6) d x\right] \\ & A=-\left.3\left(\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right)\right|_0 ^1-\left.\left(\frac{3 x^2}{2}-6 x\right)\right|_0 ^1 \\ & A=-2[1-0]\left[\frac{3}{2}-6\right] \\ & A=-2-\frac{3}{2}+6=\frac{5}{2} \text { Sq. unit } \\ & \therefore 6 A=6 \times \frac{5}{2}=15 \end{aligned}$$