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  • Using integration, find the area of the smaller region bounded by the ellipse <img alt="\frac{x^{2}}{9}+\frac{y^{2}}{4}= 1" src="https://entrancecorner.codecogs.com/gif.latex?%5Cfrac%7Bx%5E%7B2%7D%7D%7B9%7D&plus;%5Cfrac%7By%5E%7B2%7D%7D%7B4%7D

Using integration, 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\cdot

 

 

 

 
 
 
 
 

Answers (1)


given curves are  \frac{x^{2}}{9}+\frac{y^{2}}{4}= 1---(i)
\frac{x}{3}+\frac{y}{2}= 1---(ii)
curve (i) represents an ellipse of the form \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}= 1
whose major axis is along x-axis 
Also (ii) is a line with x and y intercepts as 3 & 2 respectively.
so required area = \int_{0}^{3}y\, dx-\int_{0}^{3}ydx
= \frac{2}{3}\int_{0}^{3}\sqrt{3^{2}-x^{2}dx}-\frac{2}{3}\int_{0}^{3}\left ( 3-x \right )dx
= \frac{2}{3}\left [ \frac{x}{2}\sqrt{9-x^{2}}+\frac{9}{2}\sin^{-1}\frac{x}{3} \right ]^{3}_{0}\frac{-2}{3}\left [ 3x\,- \frac{1}{2}x^{2} \right ]^{3}_{0}
= \frac{2}{3}\left [ 0+\frac{9}{2}\sin^{-1}1 \right ]-\left [ 0+0 \right ]-\frac{2}{3}\left [ \left ( 9-\frac{9}{2} \right )-0 \right ]
= \frac{2}{3}\times \frac{9\pi }{4}-\frac{2}{3}\times \frac{9}{2}= 3\left ( \frac{\pi }{2}-1 \right )\: sq.units.
 

Posted by

Ravindra Pindel

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