Question

Using integration, find the area of the region bounded by the parabola y² = 4x and the circle 4x² + 4y² = 9.

Answer 1

The area bounded by the parabola  y² = 4x and circle 4x² + 4y² = 9, is represented as

The points of intersection of both the curves are `(1/2,sqrt2)` and `(1/2,sqrt2)`.

The required area is given by OABCO.

It can be observed that area OABCO is symmetrical about the x-axis.

∴ Area OABCO = 2 × Area OBC

Area OBCO = Area OMC + Area MBC

`= int_0^(1/2)2sqrt"x"  "dx"+int_(1/2)^(3/2) 1/2 sqrt(9-4"x"^2)  "dx"`

`= int_0^(1/2)2sqrt"x"  "dx"+int_(1/2)^(3/2) 1/2 sqrt((3)^2-(2"x")^2)  "dx"`

Put 2x = t ⇒ dx = `"dt"/2`

When x = `3/2`, t = 3 and when x = `1/2`, t = 1

`= int_0^(1/2)2sqrt"x"  "dx"+1/4int_3^1 sqrt((3)^2-("t")^2) "dt"`

`=2["x"^(3/2)/(3/2)]_0^(1/2) + 1/4["t"/2 sqrt(9-"t"^2) + 9/2sin^-1("t"/3)]_1^3`

`=2[2/3(1/2)^(3/2)]+1/4[{3/2sqrt(9-(3)^2) +9 /2 sin^-1 (3/3)}-{1/2sqrt(9-(1)^2)+9/2sin^-1(1/3)}]`

`=2/(3sqrt2)+1/4[{0+9/2sin^-1(1)}-{1/2sqrt8 +9/ 2sin^-1 (1/3)}]`

`=sqrt2/3+1/4[(9pi)/4-sqrt2-9/2sin^-1(1/3)]`

`=sqrt2/3+(9pi)/16-sqrt2/4-9/8sin^-1(1/3)`

`=(9pi)/16-9/8sin^-1(1/3)+sqrt2/12`

Therefore, the required area is `[2xx((9pi)/16-9/8sin^-1 (1/3)+sqrt2/12)]=(9pi)/8-  9/4s in^-1(1/3) +1/(3sqrt2)` sq.units.