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  • Using integration, find the area of the region :

Using integration, find the area of the region :
\left \{ \left ( x,y \right ):9x^{2}+4y^{2} \leq 36,3x+2y\geq 6\right \}

 

 

 

 
 
 
 
 

Answers (1)

given region is \left \{ \left ( x,y \right ):3x^{2}+4y^{2}\leq 36\: and\: 3x+2y\geq 6 \right \}
we draw the curves corresponding to equations
9x^{2}+4y^{2}= 36\; \; or\; \; \frac{x^{2}}{4}+\frac{y^{2}}{9}= 1\: and \: 3x+2y= 6
The diagram is  The curve internal at (2,0) and (0,3)

The shaded area is the area enhanced by the curves and is
Required area = \int_{0}^{2}\sqrt{9\left ( 1-\frac{x^{2}}{4} \right )dx}-\int_{0}^{2}\left ( \frac{6}{2}-\frac{3x}{2} \right )dx
                    = 3\left [ \frac{x}{4} \sqrt{4-x^{2}}+\frac{4}{4}\sin^{-1}\frac{x}{2}-x+\frac{x^{2}}{4}\right ]^{2}_{0}
                   = 3\left [ \left ( \frac{2}{4}\sqrt{4-4} +\frac{4}{4}\sin^{-1}\left ( \frac{2}{2} \right )-2+\frac{4}{4}-0\right ) \right ]
                   = 3\left [ \frac{\pi }{2}-1 \right ]   

Posted by

Ravindra Pindel

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