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Question
Find the area enclosed between the circle x2 + y2 = 9, along X-axis and the line x = y, lying in the first quadrant
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Solution
Given equation of the circle is x2 + y2 = 9 .....(i)
and equation of the line is x = y ......(ii)
From (i), we get
y2 = 9 −x2
∴ y = `sqrt(9 - x^2)` .....(iii) .....[∵ In first quadrant, y > 0]
Substituting (ii) in (i), we get
x2 + x2 = 9
∴ 2x2 = 9
∴ x2 = `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
