Question

Using integration, 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.`

Answer 1

Given the equation of the ellipse is `"x"^2/9+"y"^2/4=1`

Let `"y"_1=2/3sqrt(9-"x"^2)` and

equation of the line is `"x"/3+"y"/2=1`

Let `"y"_2=2/3(3-"x")`

we have (3, 0) and (0, 2) as the points of intersection of ellipse and line.

Therefore, the area of a smaller region, A `=int_0^3("y"_1-"y"_2)"dx"`

`"A" =int_0^3[2/3sqrt(9-"x"^2)-2/3(3-"x")]"dx"`

`=int_0^3(2/3sqrt(9-"x"^2))"dx"-int_0^3[2/3(3-"x")]"dx"`

`=2/3["x"/2sqrt(9-"x"^2)+9/2sin^-1("x"/3)]_0^3-2/3(3"x"-"x"^2/2)_0^3`

`=2/3[(0+9/2xxpi/2)-0]-2/3(9-9/2-0)`

`=((3pi)/2-3)` sq. unit