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  1. Chapter 8 Class 12 Application of Integrals
  2. Serial order wise

Transcript

Example 5 Find the area bounded by the ellipse ๐‘ฅ^2/๐‘Ž^2 +๐‘ฆ^2/๐‘^2 =1 and the ordinates ๐‘ฅ=0 and ๐‘ฅ=๐‘Ž๐‘’, where, ๐‘2=๐‘Ž2 (1 โ€“ ๐‘’2) and e < 1 Required Area = Area of shaded region = Area BORQSP = 2 ร— Area OBPS = 2 ร— โˆซ_0^๐‘Ž๐‘’โ–’ใ€–๐‘ฆ.๐‘‘๐‘ฅใ€— We know that , ๐‘ฅ^2/๐‘Ž^2 +๐‘ฆ^2/๐‘^2 =1 (As ellipse is symmetric about its axis ) ๐‘ฆ^2/๐‘^2 =(๐‘Ž^2โˆ’ใ€– ๐‘ฅใ€—^2)/๐‘Ž^2 ๐‘ฆ^2=๐‘^2/๐‘Ž^2 (๐‘Ž^2โˆ’๐‘ฅ^2 ) ๐‘ฆ=ยฑโˆš(๐‘^2/๐‘Ž^2 (๐‘Ž^2โˆ’๐‘ฅ^2 ) ) ๐‘ฆ=ยฑ๐‘/๐‘Ž โˆš((๐‘Ž^2โˆ’๐‘ฅ^2 ) ) Since OBPS is in 1st quadrant, value of y is positive โˆด ๐‘ฆ=๐‘/๐‘Ž โˆš(๐‘Ž^2โˆ’๐‘ฅ^2 ) Required Area = 2 ร— โˆซ_0^๐‘Ž๐‘’โ–’ใ€–๐‘ฆ.๐‘‘๐‘ฅใ€— = 2โˆซ_0^๐‘Ž๐‘’โ–’ใ€–๐‘/๐‘Ž โˆš(๐‘Ž^2โˆ’๐‘ฅ^2 )ใ€— ๐‘‘๐‘ฅ = 2๐‘/๐‘Ž โˆซ_0^๐‘Ž๐‘’โ–’โˆš(๐‘Ž^2โˆ’๐‘ฅ^2 ) ๐‘‘๐‘ฅ = 2๐‘/๐‘Ž [1/2 ๐‘ฅโˆš(๐‘Ž^2โˆ’๐‘ฅ^2 )+๐‘Ž^2/2 sin^(โˆ’1)โกใ€–๐‘ฅ/๐‘Žใ€— ]_0^๐‘Ž๐‘’ =2๐‘/๐‘Ž [(๐‘Ž๐‘’/2 โˆš(๐‘Ž^2โˆ’(๐‘Ž๐‘’)^2 )+๐‘Ž^2/2 sin^(โˆ’1)โกใ€–๐‘Ž๐‘’/๐‘Žใ€— )โˆ’(0/2 โˆš(๐‘Ž^2โˆ’0)+๐‘Ž^2/2 sin^(โˆ’1)โก(0/๐‘Ž) )] =2๐‘/๐‘Ž [๐‘Ž๐‘’/2 โˆš(๐‘Ž^2โˆ’๐‘Ž^2 ๐‘’^2 )+๐‘Ž^2/2 sin^(โˆ’1)โกใ€–(๐‘’)โˆ’0โˆ’๐‘Ž^2/2 sin^(โˆ’1)โก(0) ใ€— ] =2๐‘/๐‘Ž [๐‘Ž๐‘’/2.๐‘Žโˆš(1โˆ’๐‘’^2 )+๐‘Ž^2/2 sin^(โˆ’1)โกใ€–๐‘’โˆ’0ใ€— ] It is of form โˆš(๐‘Ž^2โˆ’๐‘ฅ^2 ) ๐‘‘๐‘ฅ=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€— =2๐‘/๐‘Ž [(๐‘Ž^2 ๐‘’)/2 โˆš(1โˆ’๐‘’^2 )+๐‘Ž^2/2 sin^(โˆ’1)โก๐‘’ ] =2๐‘/๐‘Ž (๐‘Ž^2/2)[๐‘’โˆš(1โˆ’๐‘’^2 )+sin^(โˆ’1)โก๐‘’ ] =๐‘Ž๐‘[๐‘’โˆš(1โˆ’๐‘’^2 )+sin^(โˆ’1)โก๐‘’ ] โˆด Required Area =๐’‚๐’ƒ[๐’†โˆš(๐Ÿโˆ’๐’†^๐Ÿ )+ใ€–๐’”๐’Š๐’ใ€—^(โˆ’๐Ÿ)โก๐’† ] 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.