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Question
Find the area of the region common to the circle x2 + y2 =9 and the parabola y2 =8x
Solution
`x^2+y^2=9 and y^2=8x`
`x^2+8x=9`
`x^2+8x-9=0`
`therefore (x+9)(x-1)=0`
`therefore x=1 or x=-9 `
`therefore y=+-2sqrt2`
∴ The points of intersections are
`P(1,2sqrt2) and Q(1,-2sqrt2)`
`y^2=8x`
`therefore y=sqrt8sqrtx=2sqrt2 x^(1/2)-> f_1(x)`
`and x^2+y^2=9 therefore y^2=9-x^2`
`therefore y=sqrt(9-x^2)-> f_2(x)`
Required area,
= Area OPAQO = 2 Area OPAMO
= 2(Area OPMO + Area APMA)
`=2[int_0^1f_1(x)dx+int_1^3f_2(x)dx]`
`=2[int_0^12sqrt2 x^(1/2)dx+int_1^3sqrt(9-x^2)dx]`
`=2[2sqrt2(x^(3/2)/(3/2))_0^1+(x/2sqrt(9-x^2)+9/2sin^-1(x/3)_1^3)]`
`=2[(4sqrt2)/3+(3/2(0)+9/2sin^-1 (1)-1/2sqrt8-9/2sin^-1 (1/3))]`
`=2[(4sqrt2)/3+9/2.x/2-sqrt2-9/2sin^-1 (1/3)]`
`=2[(sqrt2/3+(9pi)/4-9/2 sin^-1 (1/3)) sq. units`
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