Question

Solve the following :

Find the area of the region in first quadrant bounded by the circle x² + y² = 4 and the X-axis and the line x = `ysqrt(3)`.

Answer 1

For finding the point of intersection of the circle and the  line, we solve
x² + y² = 4                   ...(1)
and x = `ysqrt(3)`                ...(2)
From (2), x² = 3y
From (1), x² = 4 - y²
∴  3y² = 4 - y²
∴ 4y² = 4
∴  y² = 1
∴ y = 1 in the first quadrant.
When y = , x = 1 x `sqrt(3) = sqrt(3)`
∴ the circle and the line intersect at `"A"(sqrt(3), 1)` in the first quadrant
Required area = area of the region OCAEDO
= area of the region OCADO + area of the region DAED
Now, area of the region OCADO

= area under the line x `ysqrt(3)`

i.e. y = `x/sqrt(y)` between x = 0 and x = `sqrt(3)`

= `int_0^(sqrt(3)) x/sqrt(3)*dx`

= `[x^2/(2sqrt(3))]_0^(sqrt(3))`

= `(3)/(2sqrt(3)) - 0`

= `sqrt(3)/(2)`

Area of the region DAED
= area under the circle x² + y² = 4 i.e. y = `+ sqrt(4 - x^2)` (in the first quadrant) between x = `sqrt(3)` and x = 2

= `int_sqrt(3)^2 sqrt(4 - x^2)*dx`

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

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

= `0 + 2(pi/2) - sqrt(3)/(2) - 2 (pi/3)`

= `pi - sqrt(3)/(2) - (2pi)/(3)`

= `pi/(3) - sqrt(3)/(2)`

∴ required area = `(sqrt3)/(2) + (pi/3 - sqrt(3)/(2))`

= `pi/(3)"sq units"`.