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Ex 8.1, 3 Area lying in the first quadrant and bounded by the circle ๐‘ฅ2+๐‘ฆ2=4 and the lines ๐‘ฅ = 0 and ๐‘ฅ = 2 is (A) ฯ€ (B) ๐œ‹/2 (C) ๐œ‹/3 (D) ๐œ‹/4Given Equation of Circle :- ๐‘ฅ^2+๐‘ฆ^2=4 ๐’™^๐Ÿ+๐’š^๐Ÿ=(๐Ÿ)^๐Ÿ โˆด Radius = ๐’“=๐Ÿ Now, Line ๐’™=๐ŸŽ is y-axis & Line x = 2 passes through point A (๐Ÿ , ๐ŸŽ) So, Required area = Area of shaded region = Area OAB = โˆซ_๐ŸŽ^๐Ÿโ–’ใ€–๐’š.๐’…๐’™ใ€— We know that, ๐‘ฅ^2+๐‘ฆ^2=4 ๐‘ฆ^2=4โˆ’๐‘ฅ^2 โˆด ๐’š=ยฑโˆš(๐Ÿ’โˆ’๐’™^๐Ÿ ) As, OBA is in 1st Quadrant Value of y will be positive โˆด ๐’š=โˆš(๐Ÿ’โˆ’๐’™^๐Ÿ ) Now, Required area = โˆซ_0^2โ–’ใ€–๐‘ฆ.๐‘‘๐‘ฅใ€— = โˆซ_0^2โ–’ใ€–โˆš(4โˆ’๐‘ฅ^2 ) ๐‘‘๐‘ฅใ€— = โˆซ_๐ŸŽ^๐Ÿโ–’ใ€–โˆš((๐Ÿ)^๐Ÿโˆ’๐’™^๐Ÿ ) ๐’…๐’™ใ€— = [(๐‘ฅ )/2 โˆš((2)^2โˆ’๐‘ฅ^2 )+2 sin^(โˆ’1)โกใ€–๐‘ฅ/2 ใ€— ]_0^2 = [2/2 โˆš((2)^2โˆ’2^2 )+2 sin^(โˆ’1)โกใ€–2/2 ใ€—โˆ’0/2 โˆš((2)^2โˆ’(0)^2 )โˆ’2 sin^(โˆ’1)โกใ€–0/2ใ€— ] = [0+2 sin^(โˆ’1)โกใ€–(1)โˆ’0โˆš4โˆ’2 sin^(โˆ’1)โก(0) ใ€— ] = 2 sin^(โˆ’1)โกใ€–(1)โˆ’2 sin^(โˆ’1)โก(0) ใ€—โˆ’0 It is of form โˆš(๐’‚^๐Ÿโˆ’๐’™^๐Ÿ ) ๐’…๐’™=1/2 ๐‘ฅโˆš(๐‘Ž^2โˆ’๐‘ฅ^2 )+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 = 2[ใ€–๐’”๐’Š๐’ใ€—^(โˆ’๐Ÿ)โกใ€–(๐Ÿ)โˆ’ใ€–๐’”๐’Š๐’ใ€—^(โˆ’๐Ÿ)โก(๐ŸŽ) ใ€— ] = 2[๐œ‹/2โˆ’0] = 2 . ๐œ‹/2 = ฯ€ Therefore, Area Required = ฯ€ square units So, the correct answer is (a)

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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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.