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