Find the area of the smaller region bounded by the ellipse `(x^2)/9+(y^2)/4=1` and the line `x/3+y/2=1.`

Explanation

Given, curve `(x^(2))/(9)+(y^(2))/(4)=1 " " ` ...(1)
Represents an ellipse whose centre is (0, 0)
and equation of line `(x)/(3)+(y)/(2)=1 " " ` …(2)
For the points fo intersection of the ellipse and the line, put the value of x from equation (2) to equation (1),
`((y)/(2)-1)^(2)+(y^(2))/(4)=1`
`implies (y^(2))/(4)+1-y+(y^(2))/(4)=1`
`impliesy^(2)-2y=0`
`impliesy=0,2`
Now `y=0` then `x=3`
and `y=2` then `x=0`
i.e., A(3, 0) and B(0,2) are the points of intersection.

` :. ` Required area
`=int_(0)^(3)2sqrt(1-(x^(2))/(9))dx-int_(0)^(3)2(1-(x)/(3))dx`
`=(2)/(3)int_(0)^(3)sqrt(3^(2)-x^(2))dx-(2)/(3)int_(0)^(3)(3-x)dx`
`=(2)/(3)[(x)/(2)sqrt(3^(2)-x^(2))+(9)/(2)"sin"^(-1)(x)/(3)]_(0)^(3)-(2)/(3)[3x-(x^(2))/(2)]_(0)^(3)`
`=(2)/(3)[0+(9)/(2)"sin"^(-1)(1)-0]-(2)/(3)[9-(9)/(2)-0]`
`=3((pi)/(2))-(3)=3((pi)/(2)-1)`
`=(3)/(2)(pi-2)` sq. units.