Using integration, find the area of the region

{(x,  y) : x 2 + y 2 ≤ 1,  x + y ≥ 1,  x ≥ 0, y ≥ 0 }

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  1. Class 12
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Transcript

Question 34 Using integration, find the area of the region {(๐‘ฅ, ๐‘ฆ)": x2 + y2 " โ‰ค" 1, x + y " โ‰ฅ" 1, x " โ‰ฅ" 0, y " โ‰ฅ" 0" } Here, we are given A circle and a line And we need to find area enclosed Circle - "x2 + y2 "โ‰ค" 1" Circle is ๐‘ฅ2+๐‘ฆ2 =1 (๐‘ฅโˆ’0)2+(๐‘ฆโˆ’0)2 =1^2 So, Center = (0, 0) & Radius = 1 And "x2 + y2 "โ‰ค" 1" means area enclosed inside the circle Line -" x + y "โ‰ฅ" 1" We draw line x + y = 1 And "x + y"โ‰ฅ" 1" means area on right side of line We see that circle and line intersect at two points (1, 0) and (0, 1) Now, {(๐‘ฅ, ๐‘ฆ)": x2 + y2 " โ‰ค" 1, x + y " โ‰ฅ" 1, x " โ‰ฅ" 0, y " โ‰ฅ" 0" } is the blue shaded region Area required Area required = Area OBCA โˆ’ Area OAB Area OBCA Area OBCA = โˆซ1_0^1โ–’ใ€–๐‘ฆ ๐‘‘๐‘ฅใ€— y โ†’ equation of circle ๐‘ฅ^2 + ๐‘ฆ^2 = 1 ๐‘ฆ^2= 1 โˆ’ ๐‘ฅ^2 y = โˆš(1โˆ’๐‘ฅ^2 ) Therefore, Area ACBO = โˆซ1_0^1โ–’ใ€–โˆš(1โˆ’๐‘ฅ^2 ) " " ๐‘‘๐‘ฅใ€— = โˆซ1_0^2โ–’ใ€–โˆš(12โˆ’๐‘ฅ^2 ) ๐‘‘๐‘ฅใ€— = [๐‘ฅ/2 โˆš(12โˆ’๐‘ฅ^2 )+12/2 sin^(โˆ’1)โกใ€–๐‘ฅ/1ใ€— ]_0^1 = [๐‘ฅ/2 โˆš(1โˆ’๐‘ฅ^2 )+1/2 sin^(โˆ’1)โก๐‘ฅ ]_0^1 = [1/2 โˆš(1โˆ’1^2 )+1/2 sin^(โˆ’1)โก1 ] โ€“ [0/2 โˆš(1โˆ’0^2 )+1/2 sin^(โˆ’1)โก0 ] = [0+1/2 sin^(โˆ’1)โก1 โˆ’0โˆ’0] = 1/2 sin^(โˆ’1)โก1 = 1/2 ร— ๐œ‹/2 = ๐œ‹/4 Area OAB Area OAB = โˆซ1_0^1โ–’ใ€–๐‘ฆ ๐‘‘๐‘ฅใ€— y โ†’ equation of line ๐‘ฅ + y = 1 y = 1 โˆ’ x Therefore, Area OAB = โˆซ1_0^1โ–’(1โˆ’๐‘ฅ)๐‘‘๐‘ฅ = [๐‘ฅโˆ’๐‘ฅ^2/2]_0^1 = ["1 โˆ’ " 1^2/2] โˆ’ [0โˆ’0/2] = 1 โ€“ 1/2 โ€“ 0 โ€“ 0 = 1/2 Thus, Area required = Area OACB โˆ’ AREA OAB = (๐…/๐Ÿ’โˆ’๐Ÿ/๐Ÿ) square units

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.