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Using integration, find the area of the region{(x,y):0≤y≤√3 x, x^2+y^2  ≤4}

This question is similar to Question 34 CBSE Class 12 - Sample Paper 2020 Boards

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Question 12 (Choice 2) Using integration, find the area of the region {(๐‘ฅ,๐‘ฆ):0โ‰ค๐‘ฆโ‰คโˆš3 ๐‘ฅ, ๐‘ฅ^2+๐‘ฆ^2 โ‰ค4} Here, we are given A circle and a line And we need to find area enclosed Circle - "x2 + y2 "โ‰ค" 4" Circle is ๐‘ฅ2+๐‘ฆ2 =4 (๐‘ฅโˆ’0)^2+(๐‘ฆโˆ’0)^2 =2^2 So, Center = (0, 0) & Radius = 2 And "x2 + y2 "โ‰ค" 2" means area enclosed inside the circle Line - ๐ŸŽโ‰ค๐’šโ‰คโˆš๐Ÿ‘ ๐’™ We draw line ๐‘ฆ=โˆš3 ๐‘ฅ And ๐’šโ‰ฅ๐ŸŽ means value of y is always positive Now, {(๐‘ฅ,๐‘ฆ):0โ‰ค๐‘ฆโ‰คโˆš3 ๐‘ฅ, ๐‘ฅ^2+๐‘ฆ^2 โ‰ค4} is the blue shaded region โˆด Required Area = Area OAB Now, Point A is (2, 0) Finding point B We know that ๐‘ฆ=โˆš3 ๐‘ฅ Putting value of y in equation of circle ๐‘ฅ^2+๐‘ฆ^2=4 ๐’™^๐Ÿ+(โˆš๐Ÿ‘ ๐’™)^๐Ÿ=๐Ÿ’ x^2+3๐‘ฅ^2=4 4๐‘ฅ^2=4 ๐‘ฅ^2=1 โˆด ๐’™=ยฑ๐Ÿ Since x is in 1st quadrant Thus, x = 1 Since ๐‘ฆ=โˆš3 ๐‘ฅ Putting x = 1 ๐’š=โˆš๐Ÿ‘ Thus, point B = (1, โˆš๐Ÿ‘) Now, Area required = Area OBX + Area XBA Area OBX Area OBX = โˆซ_0^1โ–’ใ€–๐‘ฆ ๐‘‘๐‘ฅใ€— ๐‘ฆ โ†’ Equation of line ๐’š=โˆš๐Ÿ‘ ๐’™ Therefore, Area OBX = โˆซ_0^1โ–’ใ€–โˆš3 ๐‘ฅ ๐‘‘๐‘ฅใ€— = โˆš3 โˆซ_0^1โ–’ใ€–๐‘ฅ ๐‘‘๐‘ฅใ€— = โˆš3 [๐‘ฅ^2/2]_0^1 = โˆš3/2 [๐‘ฅ^2 ]_0^1 = โˆš3/2 [(1)^2โˆ’(0)^2 ] = โˆš๐Ÿ‘/๐Ÿ Area XBA Area XBA = โˆซ_1^2โ–’ใ€–๐‘ฆ ๐‘‘๐‘ฅใ€— ๐’š โ†’ Equation of circle Now, ๐‘ฅ^2+๐‘ฆ^2=4 ๐‘ฆ^2=4โˆ’๐‘ฅ^2 ๐‘ฆ = ยฑโˆš(4โˆ’๐‘ฅ^2 ) Since XBA is in 1st Quadrant, Value of ๐‘ฆ will be positive โˆด ๐’š = โˆš(๐Ÿ’โˆ’๐’™^๐Ÿ ) Area XBA = โˆซ_1^2โ–’ โˆš(4โˆ’๐‘ฅ^2 ) ๐‘‘๐‘ฅ = โˆซ_๐Ÿ^๐Ÿโ–’ โˆš((๐Ÿ)^๐Ÿโˆ’๐’™^๐Ÿ ) ๐’…๐’™ = [1/2 ๐‘ฅโˆš((2)^2โˆ’๐‘ฅ^2 )+(2)^2/2 sin^(โˆ’1)โกใ€–๐‘ฅ/2ใ€— ]_1^2 = [1/2 ๐‘ฅโˆš(4โˆ’๐‘ฅ^2 )+2 sin^(โˆ’1)โกใ€–๐‘ฅ/2ใ€— ]_1^2 = 1/2 (2) โˆš(4โˆ’2^2 )+2 sin^(โˆ’1)โกใ€–2/2ใ€—โˆ’1/2 (1) โˆš(4โˆ’(1)^2 )โˆ’2 sin^(โˆ’1)โกใ€–1/2ใ€— = 0 + 2 sin^(โˆ’1)โก(1) โ€“ 1/2 โˆš(4โˆ’1)โˆ’2 sin^(โˆ’1)โกใ€–1/2ใ€— = 2 sin^(โˆ’1)โก(1) โ€“ โˆš3/2 โˆ’ 2 sin^(โˆ’1)โกใ€–1/2ใ€— = (โˆ’โˆš3)/2+2[sin^(โˆ’1)โกใ€–(1)โˆ’๐‘ ๐‘–๐‘›^(โˆ’1) ใ€— 1/2]= (โˆ’โˆš3)/2+2[sin^(โˆ’1)โกใ€–(1)โˆ’๐‘ ๐‘–๐‘›^(โˆ’1) ใ€— 1/2] = (โˆ’โˆš3)/2+2[๐œ‹/2โˆ’๐œ‹/6] = (โˆ’โˆš3)/2+2[2๐œ‹/6] = (โˆ’โˆš๐Ÿ‘)/๐Ÿ+๐Ÿ๐…/๐Ÿ‘ Therefore, Required Area = Area OBX + Area XBA = โˆš3/2โˆ’โˆš3/2+2๐œ‹/3 = ๐Ÿ๐…/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.