Question

Find area of the ellipse 4x² + 9y² = 36

Answer 1

By the symmetry of the ellipse, required area of the ellipse is 4 times the area of the region OPQO.

For the region OPQO, the limits of integration are x = 0 and x = 3.

Given equation of the ellipse is 4x² + 9y² = 36

∴ `x^2/9 + y^2/4` = 1

∴ `y^2/4 = 1 - x^2/9`

∴ y² = `4(1 - x^2/9) = 4/9(9 - x^2)`

∴ y = `+- 2/3 sqrt(9 - x^2)`

∴ y = `2/3 sqrt(9 - x^2)`   .....[∵ In first quadrant, y > 0]

∴ Required area = 4(area of the region OPQO)

= `4int_0^3 y  "d"x = 4 int_0^3 2/3 sqrt(9 - x^2)  "d"x`

= `(4 xx 2)/3 int_0^3 sqrt((3)^2 - x^2)  "d"x`

= `8/3 [x/2 sqrt((3)^2 - x^2) + (3)^2/2 sin^-1 (x/3)]_0^3`

= `8/3 {[3/2 sqrt((3)^2 - (3)^2) + (3)^2/2 sin^-1 (3/3)] - [0/2 sqrt((3)^2 - (0)^2) + (3)^2/2 sin^-1 (0/3)]}`

= `8/3 {[0 + 9/2 sin^-1 (1)] - [0 + 0]}`

= `8/3(9/2 xx pi/2)`

= 6π sq. units