Question

Using integration, find the area of the region bounded by the curves x² + y² = 4, x = `sqrt(3)`y and x-axis lying in the first quadrant.

Answer 1

Given equation of circle x² + y² = 4

or x² + y² = (2)²

∴ Radius = 2

So, point A is (2, 0) and point B is (0, 2)

Let line x = `sqrt(3)`y intersect the circle at point C

On solving x² + y² = 4 and x = `sqrt(3)`y, we get

`(sqrt(3)y)^2 + y^2` = 4

⇒ 3y² + y² = 4

⇒ 4y²

⇒ y² = 1

⇒ y = ±1

For y = 1, x = `sqrt(3)` and y = –1, x = `-sqrt(3)`

Since point C is in 1^st quadrant

∴ C = `(sqrt(3), 1)`

∴ Required Area = `int_0^sqrt(3) y_"line" dx + int_sqrt(3)^2 y_"circle"dx`

= `int_0^sqrt(3) x/sqrt(3)dx + int_sqrt(3)^2 sqrt(4 - x^2)dx`

= `1/sqrt(3) int_0^sqrt(3) xdx + int_sqrt(3)^2 sqrt(4 - x^2)dx`

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

= `1/(2sqrt(3)){(sqrt(3))^2 - 0} + [{1/2(2)sqrt(4 - 2^2) + 2sin^-1(2/2)} - {1/2(sqrt(3))sqrt(4 - (sqrt(3))^2) - 2sin^-1  sqrt(3)/2}]`

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

= `2 π/2 - 2 π/3`

= `π - (2π)/3`

= `π/3` sq.units