Exam

JEE Main

Year

2024

Date

9 April

Shift

Evening Shift

Subject

Mathematics

Chapter

Area Under Curves

JEE Main 2024 · 9 April · Evening Shift · Mathematics · Area Under Curves

The area (in square units) of the region enclosed by the ellipse...

The area (in square units) of the region enclosed by the ellipse $x^2+3 y^2=18$ in the first quadrant below the line $y=x$ is

Step-by-step solution

$\frac{x^2}{18}+\frac{y^2}{6}=1$

$$\begin{aligned} & \frac{x^2}{18}+\frac{3 x^2}{18}=1 \Rightarrow 4 x^2=18 \Rightarrow x^2=\frac{9}{2} \\ & \int_{\frac{3}{\sqrt{2}}}^{\sqrt[3]{2}} \frac{\sqrt{18-x^2}}{\sqrt{3}} d x \\ & =\frac{1}{\sqrt{3}}\left(\frac{x \sqrt{18-x^2}}{2}+\frac{18}{2} \sin ^{-1} \frac{x}{3 \sqrt{2}}\right)_{\frac{3}{\sqrt{2}}}^{3 \sqrt{2}} \\ & =\frac{1}{\sqrt{3}}\left(9 \times \frac{\pi}{2}-\frac{3}{2 \sqrt{2}} \times \frac{3 \sqrt{3}}{\sqrt{2}}-9 \times \frac{\pi}{6}\right) \end{aligned}$$
Required Area $$\begin{aligned} & =\frac{1}{2} \times \frac{9}{2}+\left(\frac{18 \pi}{6}-\frac{9 \sqrt{3}}{4}\right) \frac{1}{\sqrt{3}} \\ & =\sqrt{3} \pi \end{aligned}$$