JEE Main 2023 (Online) 15th April Morning ShiftQuestion 1
If the area bounded by the curve $2y^{2}=3x$, lines $x+y=3$, $y=0$ and outside the circle $(x-3)^{2}+y^{2}=2$ is $A$, then $4(\pi+4A)$ is equal to $\_\_\_\_\_\_$ .
Application of Integrals
Answer$42$
Solution
Parabola and line meet where $2y^2+3y-9=0\Rightarrow y=\frac32$, point $(\frac32,\frac32)$. Region (before removing circle): $\int_0^{3/2}\sqrt{\frac{3x}{2}}dx+\frac12\cdot\frac32\cdot\frac32=\frac32+\frac98=\frac{21}{8}$. Circle centre $(3,0)$, $r=\sqrt2$ (leftmost $x=3-\sqrt2>\frac32$): the part inside the region is the sector between $y=0$ and the line $x+y=3$ (angle $\frac\pi4$): area $\frac12\cdot2\cdot\frac\pi4=\frac\pi4$. $A=\frac{21}8-\frac\pi4$, so $4(\pi+4A)=4\pi+42-4\pi=42$.