Find the area enclosed between the circle x^{2} + y^{2} = 9, along X-axis and the line x = y, lying in the first quadrant

#### Solution

Given equation of the circle is x^{2} + y^{2} = 9 .....(i)

and equation of the line is x = y ......(ii)

From (i), we get

y^{2} = 9 −x^{2}

∴ y = `sqrt(9 - x^2)` .....(iii) .....[∵ In first quadrant, y > 0]

Substituting (ii) in (i), we get

x^{2} + x^{2} = 9

∴ 2x^{2 }= 9

∴ x^{2 }= `9/2`

∴ x = `3/sqrt(2)` .....[∵ In first quadrant, x > 0]

When x = `3/sqrt(2)`, y = `3/sqrt(2)`

∴ The point of intersection is `"B"(3/sqrt(2), 3/sqrt(2))`.

Required area = area of the region OCABO

= area of the region OCBO + area of the region ABCA

= `int_0^(3/sqrt(2)) x "d"x + int_(3/(sqrt(2)))^3 sqrt(9 - x^2) "d"x` .....[From (iii) and (ii)]

= `1/2[x^2]_0^(3/sqrt(2)) + [x/2 sqrt(9 - x^2) + 9/2 sin^-1 (x/3)]_(3/sqrt(2))^3`

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

= `9/4 + 9/2(pi/2) - 3/(2sqrt(2)) (3/sqrt(2)) - 9/2(pi/4)`

= `9/4 + (9pi)/4 - 9/4 - (9pi)/8`

= `(9pi)/8` sq.units