$$\begin{aligned} & y^{\prime}=3-\frac{3}{2} \sqrt{x} \\ & y^{\prime}=3\left(1-\frac{\sqrt{x}}{2}\right) \end{aligned}$$

$$\begin{aligned} & \text { Area }=\int_\limits0^3\left(3 x-\left(3 x-x^{3 / 2}\right)\right) d x+\int_\limits3^9 \frac{27-3 x}{2}-\left(3 x-x^{3 / 2}\right) d x \\ & =\int_\limits0^3 x^{3 / 2} d x+\frac{1}{2} \int_\limits3^9\left(27-9 x+2 x^{3 / 2}\right) d x \\ & =\left[\frac{2}{5} x^{5 / 2}\right]_0^3+\frac{1}{2}\left[27 x-\frac{9 x^2}{2}+2 \times \frac{2}{5} x^{5 / 2}\right]_3^9 \\ & =\frac{2}{5} \times 9 \sqrt{3}+\frac{1}{2}\left[243-\frac{729}{2}+\frac{4}{5} \times 81 \times 3-81+\frac{81}{2}-\frac{4}{5} \times 9 \sqrt{3}\right] \\ & =\frac{1}{2}\left[\frac{486-729-81}{2}+\frac{972}{5}\right]=\frac{81}{5} \\ & A=\frac{81}{5} \\ & \therefore 10 A=10 \times \frac{81}{5}=162 \end{aligned}$$