The area (in square units) of the region enclosed by the ellipse x2 + 3y2 = 18 in the first quadrant below the line y = x is

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Mathematics

Introduction to Determinants

Question

The area (in square units) of the region enclosed by the ellipse x² + 3y² = 18 in the first quadrant below the line y = x is

$\sqrt{3} π$

$\sqrt{3} π - \frac{3}{4}$

$\sqrt{3} π + 1$

$\sqrt{3} π + \frac{3}{4}$

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

$\frac{x^{2}}{18} + \frac{x^{2}}{6} = 1$ ⇒ 4x² = 18

$x = \pm \frac{3}{\sqrt{2}} = y$

Area $= \int_{0}^{\frac{3}{\sqrt{2}}} \sqrt{18 (1 - \frac{y^{2}}{6})} dy - \int_{0}^{3 / \sqrt{2}} ydy = \int_{0}^{\frac{3}{\sqrt{2}}} \sqrt{18 - 3 y^{2}} dy - \frac{1}{2} \times \frac{9}{2} = \sqrt{3} π$

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