Using integration, find the area of the following region: $\{(x, y) : \frac{x^{2}}{9} + \frac{y^{2}}{4} \leq 1 \leq \frac{x}{3} + \frac{y}{2}\}$ .

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Solution

$Let R = \{(x, y) : \frac{x^{2}}{9} + \frac{y^{2}}{4} \leq 1 \leq \frac{x}{3} + \frac{y}{2}\} R_{1} = \{(x, y) : \frac{x^{2}}{9} + \frac{y^{2}}{4} \leq 1\} and R_{2} = \{(x, y) : 1 \leq \frac{x}{3} + \frac{y}{2}\} ∴ R = R_{1 \cap} R_{2} \frac{x^{2}}{9} + \frac{y^{2}}{4} = 1 represents an ellipse, with centre at O (0, 0), cutting the coordinate axis at A (3, 0), A' (- 3, 0), B (0, 2) and B' (0, - 2) Hence, R_{1} is area interior to the ellipse \frac{x}{3} + \frac{y}{2} = 1 \Rightarrow 2 x + 3 y = 6 represents a straight line cutting the coordinate axis at A (3, 0) and B (0, 2) Hence, R_{2} will be area above the line \Rightarrow A (3, 0) and B (0, 2) are points of intersection o f ellipse and straight line . Area of shaded region, A = \int_{0}^{3} [\sqrt{4 (1 - \frac{x^{2}}{9})} - 2 (1 - \frac{x}{3})] d x = \int_{0}^{3} [\sqrt{\frac{36 - 4 x^{2}}{9}} - (\frac{6 - 2 x}{3})] d x = \frac{1}{3} \int_{0}^{3} [2 \sqrt{9 - x^{2}} - (6 - 2 x)] d x = \frac{1}{3} {[2 \times \{\frac{1}{2} x \sqrt{9 - x^{2}} + \frac{1}{2} \times 9 \sin^{- 1} (\frac{x}{3})\} - (6 x - 2 \frac{x^{2}}{2})]}_{0}^{3} = \frac{1}{3} {[x \sqrt{9 - x^{2}} + 9 \sin^{- 1} (\frac{x}{3}) - 6 x + x^{2}]}_{0}^{3} = \frac{1}{3} [0 + 9 \sin^{- 1} (1) - 18 + 9 - 0] = \frac{1}{3} [9 \frac{π}{2} - 9] = (\frac{3 π}{2} - 3) sq units$

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