JEE Main 2024 — Area Under Curves Question with Solution
JEE Main 2024 (09 Apr Shift 2)
Question
The area (in square units) of the region enclosed by the ellipse in the first quadrant below the line is
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Show full solutionCorrect option: C
Correct answer
C
Step-by-step explanation
$\begin{aligned} & \frac{x^2}{18}+\frac{3 x^2}{18}=1 \Rightarrow 4 x^2=18 \Rightarrow x^2=\frac{9}{2} \\ & \int_{\frac{3}{\sqrt{2}}}^{\sqrt[3]{2}} \frac{\sqrt{18-x^2}}{\sqrt{3}} d x \\ & =\frac{1}{\sqrt{3}}\left(\frac{x \sqrt{18-x^2}}{2}+\frac{18}{2} \sin ^{-1} \frac{x}{3 \sqrt{2}}\right)_{\frac{3}{\sqrt{2}}}^{3 \sqrt{2}} \\ & =\frac{1}{\sqrt{3}}\left(9 \times \frac{\pi}{2}-\frac{3}{2 \sqrt{2}} \times \frac{3 \sqrt{3}}{\sqrt{2}}-9 \times \frac{\pi}{6}\right) \end{aligned}$
Required Area $\begin{aligned} & =\frac{1}{2} \times \frac{9}{2}+\left(\frac{18 \pi}{6}-\frac{9 \sqrt{3}}{4}\right) \frac{1}{\sqrt{3}} \\ & =\sqrt{3} \pi \end{aligned}$
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This is a previous-year question from JEE Main 2024, covering the Area Under Curves chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.