Find the area of the smaller region bounded b | vedclass

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(A) The area of the smaller region bounded by the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$ and the line $\frac{x}{3}+\frac{y}{2}=1$ is the area of

Vedclass · Accepted Answer

$\frac{3}{2}(\pi-2)$ sq. units

Vedclass · Answer

(A) The area of the smaller region bounded by the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$ and the line $\frac{x}{3}+\frac{y}{2}=1$ is the area of the region $ABC$ shown in the figure.The points of intersection are $A(0, 2)$ and $B(3, 0)$.Area of the region $ABC = \int_{0}^{3} (y_{	ext{ellipse}} - y_{	ext{line}}) dx$From the ellipse equation,$y = 2\sqrt{1-\frac{x^{2}}{9}} = \frac{2}{3}\sqrt{9-x^{2}}$.From the line equation,$y = 2(1-\frac{x}{3}) = 2 - \frac{2x}{3}$.Area $= \int_{0}^{3} \left( \frac{2}{3}\sqrt{9-x^{2}} - (2 - \frac{2x}{3}) ight) dx$$= \frac{2}{3} \int_{0}^{3} \sqrt{3^{2}-x^{2}} dx - \int_{0}^{3} (2 - \frac{2x}{3}) dx$$= \frac{2}{3} \left[ \frac{x}{2}\sqrt{9-x^{2}} + \frac{9}{2}\sin^{-1}(\frac{x}{3}) ight]_{0}^{3} - \left[ 2x - \frac{x^{2}}{3} ight]_{0}^{3}$$= \frac{2}{3} \left[ (0 + \frac{9}{2} \cdot \frac{\pi}{2}) - 0 ight] - \left[ (6 - 3) - 0 ight]$$= \frac{2}{3} \cdot \frac{9\pi}{4} - 3 = \frac{3\pi}{2} - 3 = \frac{3}{2}(\pi-2)$ sq. units.

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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