JEE Main · 2023 · 15 Apr 2023 · Shift 1 · Area Under Curves

The parabola $2y^2=3x$ meets $x+y=3$ where $2y^2=3(3-y)$, i.e. $2y^2+3y-9=0$, giving $y=\frac{3}{2}$ and $x=\frac{3}{2}$. The area between the line and the parabola above the $x$-axis is ${\int }_0^{3/2}\left[(3-y)-\frac{2y^2}{3}\right]dy={\left[3y-\frac{y^2}{2}-\frac{2y^3}{9}\right]}_0^{3/2}=\frac{9}{2}-\frac{9}{8}-\frac{3}{4}=\frac{21}{8}$. The circle $(x-3{)}^2+y^2=2$ has centre $(3,0)$ and radius $\sqrt{2}$; at that corner the region lies between the $x$-axis and the line of slope $-1$, so the part removed is a sector of angle $\frac{\pi }{4}$ with area $\frac{1}{2}(\sqrt{2}{)}^2\cdot \frac{\pi }{4}=\frac{\pi }{4}$. Hence $A=\frac{21}{8}-\frac{\pi }{4}$ and $4(\pi +4A)=4\pi +16\left(\frac{21}{8}-\frac{\pi }{4}\right)=42$.