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Using integration, find the area of the region

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

Using integration, find area of region  {(x,  y) : x^2 + y^2 ≤ 1, x+y
Question 34 - CBSE Class 12 Sample Paper for 2020 Boards - Part 2
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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_0^1โ–’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 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.