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‘กℎ๐‘’ ๐‘“๐‘–๐‘Ÿ๐‘ ๐‘ก ๐‘ž๐‘ข๐‘Ž๐‘‘๐‘Ÿ๐‘Ž๐‘›๐‘ก

Find area of region bounded by curves x^2 + y^2 = 4, y = √3x and x-axi

Question 34  (Choice 1) - CBSE Class 12 Sample Paper for 2021 Boards - Part 2
Question 34  (Choice 1) - CBSE Class 12 Sample Paper for 2021 Boards - Part 3
Question 34  (Choice 1) - CBSE Class 12 Sample Paper for 2021 Boards - Part 4
Question 34  (Choice 1) - CBSE Class 12 Sample Paper for 2021 Boards - Part 5 Question 34  (Choice 1) - CBSE Class 12 Sample Paper for 2021 Boards - Part 6

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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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.