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Using integration, find the area of the region{(x,y):0โ‰คyโ‰คโˆš3 x, x^2+y^2ย  โ‰ค4}

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

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Transcript

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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.