Find the area of the region bounded by the curves ๐‘ฅ 2 + ๐‘ฆ 2 = 4, y = √3๐‘ฅ ๐‘Ž๐‘›๐‘‘ ๐‘ฅ − ๐‘Ž๐‘ฅ๐‘–๐‘  ๐‘–๐‘› ๐‘กโ„Ž๐‘’ ๐‘“๐‘–๐‘Ÿ๐‘ ๐‘ก ๐‘ž๐‘ข๐‘Ž๐‘‘๐‘Ÿ๐‘Ž๐‘›๐‘ก


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  1. Class 12
  2. Solutions of Sample Papers and Past Year Papers - for Class 12 Boards


Question 34 (Choice 1) Find the area of the region bounded by the curves ๐‘ฅ^2+๐‘ฆ^2=4, ๐‘ฆ=โˆš3 ๐‘ฅ ๐‘Ž๐‘›๐‘‘ ๐‘ฅ โˆ’ ๐‘Ž๐‘ฅ๐‘–๐‘  ๐‘–๐‘› ๐‘กโ„Ž๐‘’ ๐‘“๐‘–๐‘Ÿ๐‘ ๐‘ก ๐‘ž๐‘ข๐‘Ž๐‘‘๐‘Ÿ๐‘Ž๐‘›๐‘ก Given Equation of Circle ๐‘ฅ2+๐‘ฆ2=4 ๐‘ฅ2+๐‘ฆ2=2^2 So, Radius = 2 โˆด Point A (2, 0) and B is (0, 2) Let point where line and circle intersect be point M Required Area = Area of shaded region = Area OMA Finding Point M Point M is intersection of line and circle Putting ๐’š=โˆš๐Ÿ‘ ๐’™ in Equation of Circle ๐‘ฅ2+๐‘ฆ2=4 ๐‘ฅ2+(โˆš3 ๐‘ฅ)2=4 4๐‘ฅ2=4 ๐‘ฅ2=1 โˆด ๐‘ฅ=ยฑ1 When ๐’™=๐Ÿ ๐‘ฆ=โˆš3 ๐‘ฅ=โˆš3 So, point is (1, โˆš3) When ๐’™=โˆ’๐Ÿ ๐‘ฆ=โˆš3 ๐‘ฅ=โˆ’โˆš3 So, point is (โ€“1, โ€“โˆš3) As Point M is in 1st Quadrant โˆด M = (1, โˆš3) โˆด Required Area = Area OMP + Area PMA = โˆซ1_0^1โ–’๐‘ฆ1๐‘‘๐‘ฅ+โˆซ1_1^2โ–’๐‘ฆ2๐‘‘๐‘ฅ Finding y1 and y2 For y1 โ€“ Equation of Line y = โˆš3x So, y1 = โˆš3x For y2 โ€“ Equation of Circle ๐‘ฅ^2+๐‘ฆ^2=4 ๐‘ฆ^2=4โˆ’๐‘ฅ^2 โˆด ๐‘ฆ=ยฑโˆš(2^2โˆ’๐‘ฅ^2 ) As Required Area is in first Quadrant โˆด ๐‘ฆ2=โˆš(2^2โˆ’๐‘ฅ^2 ) Thus, Required Area = โˆซ1_0^1โ–’ใ€–โˆš3 ๐‘ฅ ๐‘‘๐‘ฅใ€—+โˆซ1_1^2โ–’ใ€–โˆš(2^2โˆ’๐‘ฅ^2 ) ๐‘‘๐‘ฅใ€— =โˆš3 โˆซ1_0^1โ–’ใ€–๐‘ฅ ๐‘‘๐‘ฅใ€—+โˆซ1_1^2โ–’ใ€–โˆš(2^2โˆ’๐‘ฅ^2 ) ๐‘‘๐‘ฅใ€— =โˆš3 [๐‘ฅ^2/2]_0^1+โˆซ1_1^2โ–’ใ€–โˆš(2^2โˆ’๐‘ฅ^2 ) ๐‘‘๐‘ฅใ€— =โˆš3 [1^2/2โˆ’0^2/2]_0^1+โˆซ1_1^2โ–’ใ€–โˆš(2^2โˆ’๐‘ฅ^2 ) ๐‘‘๐‘ฅใ€— =โˆš3/2+โˆซ1_1^2โ–’ใ€–โˆš(2^2โˆ’๐‘ฅ^2 ) ๐‘‘๐‘ฅใ€— It is of form โˆซ1โ–’ใ€–โˆš(๐‘Ž^2โˆ’๐‘ฅ^2 ) ๐‘‘๐‘ฅ=1/2 ๐‘ฅโˆš(๐‘Ž^2โˆ’๐‘ฅ^2 )ใ€—+๐‘Ž^2/2 ใ€–๐‘ ๐‘–๐‘›ใ€—^(โˆ’1)โกใ€–๐‘ฅ/๐‘Ž+๐‘ใ€— Replacing a by 2 , we get ใ€–=โˆš3/2+[๐‘ฅ/2 โˆš((2)^2โˆ’๐‘ฅ^2 )+(2)^2/2 sin^(โˆ’1)โกใ€–๐‘ฅ/2ใ€— ]ใ€—_1^2 =โˆš3/2+2/2 โˆš((2)^2โˆ’(2)^2 )+2 sin^(โˆ’1)โกใ€–2/2ใ€—โˆ’1/2 โˆš((2)^2โˆ’("1" )^2 )โˆ’2 sin^(โˆ’1)โกใ€–1/2ใ€— =โˆš3/2+0+2 sin^(โˆ’1)โกใ€–(1)โˆ’1/2 โˆš(4โˆ’1)โˆ’ใ€— 2 sin^(โˆ’1)โก(1/2) =โˆš3/2+2 ร—๐œ‹/2โˆ’โˆš3/2โˆ’2ร—๐œ‹/6 =๐œ‹ โˆ’๐œ‹/3 =๐Ÿ๐…/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.