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Question
Find the area of the smaller region bounded by the ellipse \[\frac{x^2}{9} + \frac{y^2}{4} = 1\] and the line \[\frac{x}{3} + \frac{y}{2} = 1 .\]
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Solution
For the given curves, the graph is as follows:
Area of the region bounded by the given curves:
\[\frac{2}{3} \int_0^3 \sqrt{9 - x^2} d x - \frac{1}{3} \int_0^3 (6 - 2x) d x = \frac{2}{3} \left[ \frac{x}{2}\sqrt{9 - x^2} + \frac{9}{2} \sin^{- 1} \frac{x}{3} \right]_0^3 - \frac{1}{3} \left[ 6x - x^2 \right]_0^3 \]
\[ = \frac{2}{3}\left[ \frac{9}{2} \times \frac{\pi}{2} \right] - \frac{1}{3}\left[ 18 - 9 \right]\]
\[ = \left( \frac{3\pi}{2} - 3 \right) \text { sq units }\]
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