Find the area of the smaller region bounded by the ellipse 9x2+4y2=1 and the line 3x+2y=1

The area of the smaller region by the ellipse 9x2+4y2=1 and the line 3x+2y=1 represented by the shaded region BCAB ∴Area BCAB=Area (OBCAO)−Area (OBAO)=∫0321−9x2dx−∫032(1−3x)dx=32[∫039−x2dx]−32∫03(3−x)dx=32[2x9−x2+29sin−13x]03−32[3x−2x2]03=32[29(2π)]−32[9−29]=32[49π−29]=32×49(π−2)=23(π−2) sq. units

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Find the area of the smaller region bounded by the ellipse {x

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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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Charlotte Discussed

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

8 mins ago

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Text solutionVerified

The area of the smaller region by the ellipse

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

and the line

\frac{x}{3} + \frac{y}{2} = 1

represented by the shaded region

BC A B

∴ A re a BC A B = A re a (OBC A O) - A re a (OB A O)

= \int_{0}^{3} 2 1 - \frac{x ^{2}}{9} d x - \int_{0}^{3} 2 (1 - \frac{x}{3}) d x

= \frac{2}{3} [\int_{0}^{3} 9 - x^{2} d x] - \frac{2}{3} \int_{0}^{3} (3 - x) d x

= \frac{2}{3} [\frac{x}{2} 9 - x^{2} + \frac{9}{2} sin^{- 1} \frac{x}{3}]_{0}^{3} - \frac{2}{3} [3 x - \frac{x ^{2}}{2}]_{0}^{3}

= \frac{2}{3} [\frac{9}{2} (\frac{π}{2})] - \frac{2}{3} [9 - \frac{9}{2}]

= \frac{2}{3} [\frac{9 π}{4} - \frac{9}{2}]

= \frac{2}{3} \times \frac{9}{4} (π - 2)

= \frac{3}{2} (π - 2)

sq. units

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Question Text	Find the area of the smaller region bounded by the ellipse $\frac{x ^{2}}{9} + \frac{y ^{2}}{4} = 1$ and the line $\frac{x}{3} + \frac{y}{2} = 1$
Answer Type	Text solution:1
Upvotes	150

Find the area of the smaller region bounded by the ellipse 9x2​+4y2​=1 and the line 3x​+2y​=1

Text solutionVerified

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