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  • Find the area of the smaller region bounded by the ellipse x^2/9 + y^2/4 = 1 and the line x/3 + y/2 = 1.

Q : 8        Find the area of the smaller region bounded by the ellipse   \small \frac{x^2}{9}+\frac{y^2}{4}=1  and the line    \small \frac{x}{3}+\frac{y}{2}=1.
 

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best_answer

We have to find the area of the shaded region

The given ellipse and the given line intersect at following points

\left ( 0,2 \right )and \left ( 3,0 \right )

\\\frac{x^{2}}{9}+\frac{y^{2}}{4}=1\\ y=\frac{2}{3}\sqrt{9-x^{2}}

Since the shaded region lies above x axis we take y to be positive

\\\frac{x}{3}+\frac{y}{2}=1\\ y=\frac{2}{3}(3-x)

The required area is

\\\frac{2}{3}\int_{0}^{3}\left ( \sqrt{9-x^{2}}-(3-x) \right )dx\\ =\frac{2}{3}[\frac{x}{2}(\sqrt{9-x^{2}})+\frac{9}{2}sin^{-1}\frac{x}{3}-3x+\frac{x^{2}}{2}]_{0}^{3}\\ =\frac{2}{3}\left ( \left [ \frac{9}{2}\times \frac{\pi }{2}-9+\frac{9}{2} \right ]-0 \right )\\ =\frac{2}{3}(\frac{9\pi }{4}-\frac{9}{2})\\ =\frac{3}{2}(\pi -2)units

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Sayak

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