Question

Using integration, find the area of the region `{(x, y): 0 ≤ y ≤ sqrt(3)x, x^2 + y^2 ≤ 4}`

Answer 1

Solving `y = sqrt(3)x` and `x^2 + y^2` = 4, we get the points of intersection as `(1, sqrt(3))` and `(-1, - sqrt(3))`

The required area = the shaded area = `int_0^1 sqrt(3)x  dx + int_1^2 sqrt(4 - x^2) dx`

= `sqrt(3)/2 [x^2]_0^1 + 1/2 [xsqrt(4 - x^2) + 4 sin^-1  x/2]_1^2`

= `sqrt(3)/2 + 1/2 [2pi - sqrt(3) - 2 pi/3]`

= `(2pi)/3` square units