Question

Using integration, find the area of the region bounded by line y = `sqrt(3)x`, the curve y = `sqrt(4 - x^2)` and Y-axis in first quadrant.

Answer 1

Given, y = `sqrt(4 - x^2)`

`\implies` x² + y² = 4

 For finding point of intersection put y = `sqrt(3)x` in y = `sqrt(4 - x^2)`, we get

`sqrt(3)x = sqrt(4 - x^2)`

`\implies` 3x² = 4 – x²

`\implies` 4x² = 4

`\implies` x² = 1

`\implies` x = ± 1

∴ y = `sqrt(3)`

∴ Coordinates of A is `(1, sqrt(3))`

∴ Required Area = `int_0^sqrt(3) y/sqrt(3) dy + int_sqrt(3)^2 sqrt(4 - y^2) dy`

= `1/sqrt(3) [y^2/2]_0^sqrt(3) + [y/2 sqrt(4 - y^2) + 4/2 sin^-1 (y/2)]_sqrt(3)^2`

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

= `sqrt(3)/2 + 2 xx π/2 - sqrt(3)/2 - 2 xx π/3`

= `π - (2π)/3`

= `π/3` sq. units.