JEE Main · 2024 · 09 Apr 2024 · Shift 2 · Area Under Curves

Ellipse: $\frac{x^2}{18}+\frac{y^2}{6}=1$, i.e. $y=\sqrt{\frac{18-x^2}{3}}$ in the first quadrant. Line $y=x$ meets it where $x^2+3x^2=18⟹x_0=\frac{3}{\sqrt{2}}$.

For $x<x_0$ the ellipse lies above the line ($y_{\text{ellipse}}(0)=\sqrt{6}>0$); for $x>x_0$ the ellipse lies below the line. So the region enclosed by the ellipse, in the first quadrant, below $y=x$ is

First term: $\frac{x_0^2}{2}=\frac{9/2}{2}=\frac{9}{4}$.

Second term: using $\int \sqrt{18-x^2}dx=\frac{x}{2}\sqrt{18-x^2}+9{\sin }^{-1}\frac{x}{\sqrt{18}}$,
at $x=\sqrt{18}$: value $=0+9\cdot \frac{\pi }{2}=\frac{9\pi }{2}$.
at $x=x_0=\frac{3}{\sqrt{2}}$: $\sqrt{18-x_0^2}=\sqrt{18-4.5}=\sqrt{13.5}=\frac{3\sqrt{6}}{2}$, so first part $=\frac{x_0}{2}\cdot \frac{3\sqrt{6}}{2}=\frac{9\sqrt{3}}{4}$; and ${\sin }^{-1}\left(\frac{x_0}{\sqrt{18}}\right)={\sin }^{-1}\frac{1}{2}=\frac{\pi }{6}$, giving second part $=\frac{3\pi }{2}$.
So value at $x_0=\frac{9\sqrt{3}}{4}+\frac{3\pi }{2}$.


Dividing by $\sqrt{3}$: second term of the area $=\sqrt{3}\pi -\frac{9}{4}$.