Question 34 Using integration, find the area of the region {(𝑥, 𝑦)": x2 + y2 " ≤" 1, x + y " ≥" 1, x " ≥" 0, y " ≥" 0" }
Here, we are given
A circle and a line
And we need to find area enclosed
Circle - "x2 + y2 "≤" 1"
Circle is 𝑥2+𝑦2 =1
(𝑥−0)2+(𝑦−0)2 =1^2
So, Center = (0, 0) & Radius = 1
And "x2 + y2 "≤" 1"
means area enclosed inside the circle
Line -" x + y "≥" 1"
We draw line x + y = 1
And "x + y"≥" 1"
means area on right side of line
We see that circle and line intersect at two points
(1, 0) and (0, 1)
Now,
{(𝑥, 𝑦)": x2 + y2 " ≤" 1, x + y " ≥" 1, x " ≥" 0, y " ≥" 0" }
is the blue shaded region
Area required
Area required = Area OBCA − Area OAB
Area OBCA
Area OBCA = ∫1_0^1▒〖𝑦 𝑑𝑥〗
y → equation of circle
𝑥^2 + 𝑦^2 = 1
𝑦^2= 1 − 𝑥^2
y = √(1−𝑥^2 )
Therefore,
Area ACBO = ∫1_0^1▒〖√(1−𝑥^2 ) " " 𝑑𝑥〗
= ∫1_0^1▒〖√(12−𝑥^2 ) 𝑑𝑥〗
= [𝑥/2 √(12−𝑥^2 )+12/2 sin^(−1)〖𝑥/1〗 ]_0^1
= [𝑥/2 √(1−𝑥^2 )+1/2 sin^(−1)𝑥 ]_0^1
= [1/2 √(1−1^2 )+1/2 sin^(−1)1 ] – [0/2 √(1−0^2 )+1/2 sin^(−1)0 ]
= [0+1/2 sin^(−1)1 −0−0]
= 1/2 sin^(−1)1
= 1/2 × 𝜋/2
= 𝜋/4
Area OAB
Area OAB = ∫1_0^1▒〖𝑦 𝑑𝑥〗
y → equation of line
𝑥 + y = 1
y = 1 − x
Therefore,
Area OAB = ∫1_0^1▒(1−𝑥)𝑑𝑥
= [𝑥−𝑥^2/2]_0^1
= ["1 − " 1^2/2] − [0−0/2]
= 1 – 1/2 – 0 – 0
= 1/2
Thus,
Area required = Area OACB − AREA OAB
= (𝝅/𝟒−𝟏/𝟐) square units

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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