$$ \begin{aligned} & y^2=\frac{3 x}{2}, x+y=3, y=0 \\\\ & 2 y^2=3(3-y) \\\\ & 2 y^2+3 y-9=0 \\\\ & 2 y^2-3 y+6 y-9=0 \\\\ & (2 y-3)(y+2)=0 ; y=3 / 2 \\\\ & \text { Area }\left(\int_0^{\frac{3}{2}}\left(x_R-x_2\right) d y\right)-A_1 \\\\ & =\int_0^{\frac{3}{2}}\left((3-y)-\frac{2 y^2}{3}\right) d y-\frac{\pi}{8}(2) \\\\ & A=\left(3 y-\frac{y^2}{2}-\frac{2 y^3}{9}\right)_0^{\frac{3}{2}}-\frac{\pi}{4} \\\\ & 4 \mathrm{~A}+\pi=4\left[\frac{9}{2}-\frac{9}{8}-\frac{3}{4}\right]=\frac{21}{2}=10.50 \\\\ & \therefore 4(4 \mathrm{~A}+\pi)=42 \end{aligned} $$