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Home  >>  CBSE XII  >>  Math  >>  Application of Integrals
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Find the area of the smaller region bounded by the ellipse \( \frac{\large x^2}{\large 9} + \frac{\large y^2}{\large 4} =1\) and the line \( \frac{\large x}{\large 3}+ \frac{\large y}{\large 2}= 1\)

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Hence the required area is the area enclosed between the straight line and the ellipse.
To find the limits,the semi major axis extends from 0 to 3 and the x intercept of the straight line is also 3.so the limits are from 0 to 3
The required area is $A=\int_a^b[f(x)-g(x)]dx$
Here a=0,b=3,$f(x)=\frac{x^2}{9}+{y^2}{4}=1$
$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\frac{2}{3}[\sqrt{9-x^2}]$
$g(x)=\frac{x}{3}+\frac{y}{2}=1=\frac{2}{3}(3-x).$
$A=\frac{2}{3}\int_0^3\sqrt {9-x^2}dx-\frac{2}{3}\int_0^3(3-x)dx$
On integrating we get,
$A=\frac{2}{3}\begin{bmatrix}\frac{x}{2}\sqrt {9-x^2}+\frac{9}{2}\sin^{-1}(\frac{x}{3})\end{bmatrix}_0^3$
$\;=\frac{2}{3}\begin{bmatrix}3x-\frac{x^2}{2}\end{bmatrix}_0^3$
On applying the limits we get,
$A=\frac{2}{3}\begin{bmatrix}\frac{9}{2}(\frac{\pi}{2})\end{bmatrix}-\frac{2}{3}\begin{bmatrix}9-\frac{9}{2}\end{bmatrix}$
$\;\;\;=\frac{2}{3}\begin{bmatrix}\frac{9\pi}{4}-\frac{9}{2}\end{bmatrix}=\frac{2}{3}\times \frac{9}{4}(\pi-2)$
Hence the required area is $\frac{3}{2}(\pi-2)sq.units$
answered Dec 21, 2013 by yamini.v
 

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