Using integration, find the area of the region
{(x, y) : x ^{ 2 } + y ^{ 2 } ≤ 1, x + y ≥ 1, x ≥ 0, y ≥ 0 }
CBSE Class 12 Sample Paper for 2020 Boards
Question 1
Question 2
Question 3
Question 4
Question 5
Question 6
Question 7
Question 8
Question 9
Question 10 Important
Question 11 Important
Question 12
Question 13
Question 14 (OR 1st Question)
Question 14 (OR 2nd Question) Important
Question 15 (OR 1st Question)
Question 15 (OR 2nd Question) Important
Question 16
Question 17 Important
Question 18 (OR 1st Question)
Question 18 (OR 2nd Question)
Question 19
Question 20
Question 21 (OR 1st Question) Important
Question 21 (OR 2nd Question) Important
Question 22
Question 23
Question 24 (OR 1st Question)
Question 24 (OR 2nd Question)
Question 25
Question 26 Important
Question 27
Question 28 (OR 1st Question) Important
Question 28 (OR 2nd Question)
Question 29
Question 30 Important
Question 31 (OR 1st Question) Important
Question 31 (OR 2nd Question)
Question 32 Important
Question 33 (OR 1st Question) Important
Question 33 (OR 2nd Question) Important
Question 34 You are here
Question 35 (OR 1st Question) Important
Question 35 (OR 2nd Question) Important
Question 36 Important
CBSE Class 12 Sample Paper for 2020 Boards
Last updated at April 16, 2024 by Teachoo
Question 34 Using integration, find the area of the region {(๐ฅ, ๐ฆ)": x2 + y2 " โค" 1, x + y " โฅ" 1, x " โฅ" 0, y " โฅ" 0" } Here, we are given A circle and a line And we need to find area enclosed Circle - "x2 + y2 "โค" 1" Circle is ๐ฅ2+๐ฆ2 =1 (๐ฅโ0)2+(๐ฆโ0)2 =1^2 So, Center = (0, 0) & Radius = 1 And "x2 + y2 "โค" 1" means area enclosed inside the circle Line -" x + y "โฅ" 1" We draw line x + y = 1 And "x + y"โฅ" 1" means area on right side of line We see that circle and line intersect at two points (1, 0) and (0, 1) Now, {(๐ฅ, ๐ฆ)": x2 + y2 " โค" 1, x + y " โฅ" 1, x " โฅ" 0, y " โฅ" 0" } is the blue shaded region Area required Area required = Area OBCA โ Area OAB Area OBCA Area OBCA = โซ1_0^1โใ๐ฆ ๐๐ฅใ y โ equation of circle ๐ฅ^2 + ๐ฆ^2 = 1 ๐ฆ^2= 1 โ ๐ฅ^2 y = โ(1โ๐ฅ^2 ) Therefore, Area ACBO = โซ1_0^1โใโ(1โ๐ฅ^2 ) " " ๐๐ฅใ = โซ1_0^1โใโ(12โ๐ฅ^2 ) ๐๐ฅใ = [๐ฅ/2 โ(12โ๐ฅ^2 )+12/2 sin^(โ1)โกใ๐ฅ/1ใ ]_0^1 = [๐ฅ/2 โ(1โ๐ฅ^2 )+1/2 sin^(โ1)โก๐ฅ ]_0^1 = [1/2 โ(1โ1^2 )+1/2 sin^(โ1)โก1 ] โ [0/2 โ(1โ0^2 )+1/2 sin^(โ1)โก0 ] = [0+1/2 sin^(โ1)โก1 โ0โ0] = 1/2 sin^(โ1)โก1 = 1/2 ร ๐/2 = ๐/4 Area OAB Area OAB = โซ1_0^1โใ๐ฆ ๐๐ฅใ y โ equation of line ๐ฅ + y = 1 y = 1 โ x Therefore, Area OAB = โซ1_0^1โ(1โ๐ฅ)๐๐ฅ = [๐ฅโ๐ฅ^2/2]_0^1 = ["1 โ " 1^2/2] โ [0โ0/2] = 1 โ 1/2 โ 0 โ 0 = 1/2 Thus, Area required = Area OACB โ AREA OAB = (๐ /๐โ๐/๐) square units