Using integration, find the area of the region
{(x, y) : x ^{ 2 } + y ^{ 2 } ≤ 1, x + y ≥ 1, x ≥ 0, y ≥ 0 }
Last updated at Oct. 24, 2019 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
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CBSE Class 12 Sample Paper for 2020 Boards
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