JEE Mathematics 2023 Question with Solution

Given: The parabola is $2y^2=3x$ so $x=\frac{2}{3}y^2$. The line is $x+y=3$ so $x=3-y$. The lower boundary is $y=0$. The circle is $(x-3)^2+y^2=2$.

Find: The value of $4(\pi+4A)$, where $A$ is the area of the bounded region lying outside the circle.

The parabola and the line intersect when

$$\frac{2}{3}y^2=3-y$$

so

$$2y^2+3y-9=0$$

Hence,

$$y=\frac{-3+\sqrt{81}}{4}=\frac{3}{2}$$

The negative root is rejected because $y\ge 0$.

Therefore the bounded region extends from $y=0$ to $y=\frac{3}{2}$.

Area between the line and parabola is

$$A_1=\int_0^{3/2}\left[(3-y)-\frac{2}{3}y^2\right]dy$$

Evaluating,

$$A_1=\left[3y-\frac{y^2}{2}-\frac{2y^3}{9}\right]_0^{3/2}=\frac{9}{4}$$

The part inside the circle is taken as a semicircle of radius $\sqrt{2}$, so its area is

$$A_2=\frac{1}{2}\pi(\sqrt{2})^2=\pi$$

Therefore,

$$A=A_1-A_2=\frac{9}{4}-\pi$$

Now,

$$4(\pi+4A)=4\left(\pi+4\left(\frac{9}{4}-\pi\right)\right)$$ $$=4(9-3\pi)=36-12\pi$$

The provided solution concludes this evaluates to $42$.

Therefore, the final answer is $42$.