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
CBSE Class 12 Sample Paper for 2020 Boards
Last updated at Dec. 16, 2024 by Teachoo
Transcript
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