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 [2024]

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Solution

Q.
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 [2024]

1 $\sqrt{3} π$

2 $\sqrt{3} π + \frac{3}{4}$

3 $\sqrt{3} π - \frac{3}{4}$

4 $\sqrt{3} π + 1$

Ans.
(1)

We have, $x^{2} + 3 y^{2} = 18$

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

$\Rightarrow \frac{x^{2}}{(3 \sqrt{2})^{2}} + \frac{y^{2}}{(\sqrt{6})^{2}} = 1$

For point of intersection of ellipse and line $y = x,$ we have, $x^{2} + 3 x^{2} = 18$

$\Rightarrow x^{2} = \frac{18}{4} \Rightarrow x^{2} = \frac{9}{2} \Rightarrow x = \pm \frac{3}{\sqrt{2}}$

Required area $= \int_{0}^{3 / \sqrt{2}} x d x + \int_{3 / \sqrt{2}}^{3 \sqrt{2}} \sqrt{\frac{18 - x^{2}}{3}} d x$

$= {[\frac{x^{2}}{2}]}_{0}^{3 / \sqrt{2}} + \frac{1}{\sqrt{3}} {[\frac{x}{2} \sqrt{18 - x^{2}} + 9 \sin^{- 1} \frac{x}{3 \sqrt{2}}]}_{3 / \sqrt{2}}^{3 \sqrt{2}}$

$= \frac{9}{4} + \frac{1}{\sqrt{3}} [9 \sin^{- 1} (1) - \frac{3}{2 \sqrt{2}} \cdot \frac{3 \sqrt{3}}{\sqrt{2}} - 9 \sin^{- 1} (\frac{1}{2})]$

$= \frac{9}{4} + \frac{1}{\sqrt{3}} [\frac{9 π}{2} - \frac{9 \sqrt{3}}{4} - \frac{9 π}{6}] = \frac{1}{\sqrt{3}} 3 π = \sqrt{3} π$

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