# Example 8 - Chapter 8 Class 12 Application of Integrals (Term 2)

Last updated at Dec. 12, 2019 by Teachoo

Last updated at Dec. 12, 2019 by Teachoo

Transcript

Example 8 In Fig AOBA is the part of the ellipse 9๐ฅ2+๐ฆ2=36 in the first quadrant such that OA = 2 and OB = 6. Find the area between the arc AB and the chord AB Here, Area Area Required = Area ABC Area ABC Area ABC = Area OACB โ Area โ OAB Area OACB Area OACB = โซ_0^2โ๐ฆ๐๐ฅ Here, ๐ฆ โ Equation of Ellpise 9๐ฅ2+๐ฆ2=36 ๐ฆ2=36โ9๐ฅ2 ๐ฆ="ยฑ" โ(36โ9๐ฅ^2 ) As OACB is in 1st quadrant, Value of ๐ฆ will be positive โด ๐ฆ=โ(36โ9๐ฅ^2 ) Area OACB = โซ_0^2โ๐ฆ๐๐ฅ = โซ_0^2โโ(36โ9๐ฅ^2 ) = โซ_0^2โโ(9(36/9โ๐ฅ^2)) = โซ_0^2โโ(9(4โ๐ฅ^2)) = โซ_0^2โโ(3^2 (2^2โ๐ฅ^2)) = 3โซ_0^2โใโ(2^2โ๐ฅ^2 ) ๐๐ฅ ใ = 3[๐ฅ/2 โ(2^2โ๐ฅ^2 )+2^2/2 sin^(โ1)โกใ๐ฅ/2ใ ]_0^2 = 3 [2/2 โ(2^2โ2^2 )+2 sin^(โ1)โกใ2/2โ0/2ใ โ(2^2โ0^2 )โ2 sin^(โ1)โก0 ] It is of form โซ1โใโ(๐^2โ๐ฅ^2 ) ๐๐ฅ=1/2 ๐ฅโ(๐^2โ๐ฅ^2 )ใ+๐^2/2 ใ๐ ๐๐ใ^(โ1)โกใ๐ฅ/๐+๐ใ Replacing a by 2 , we get = 3 [0+2 sin^(โ1)โกใ1โ0โ0ใ ] = 3 [2 sin^(โ1)โก1 ] = 3 ร 2 ร ๐/2 = 3๐ Area OAB Area OAB =โซ_0^2โ๐ฆ๐๐ฅ Here, ๐ฆ โ equation of line AB Equation of line between A(2, 0) & B(0, 6) is (๐ฆ โ 0)/(๐ฅ โ 2)=(6 โ 0)/(0 โ 2) ๐ฆ/(๐ฅโ2)=6/(โ2) Eq. of line b/w (x1, y1) & (x2, y2) is (๐ฆ โ ๐ฆ1)/(๐ฅ โ ๐ฅ1)=(๐ฆ2 โ ๐ฆ1)/(๐ฅ2 โ ๐ฅ1) ๐ฆ/(๐ฅ โ 2)=โ3 ๐ฆ=โ3(๐ฅโ2) Area OBC =โซ_0^2โใ๐ฆ ๐๐ฅใ =โซ_0^2โใโ3(๐ฅโ2) ๐๐ฅใ = โ3โซ_0^2โใ(๐ฅโ2) ๐๐ฅใ = โ3 [๐ฅ^2/2 โ2๐ฅ ]_0^2 = โ 3 [2^2/2โ2 ร2โ0^2/2โ2 ร0] = โ3 [โ2] = 6 Area Required = Area OACB โ Area โ OAB =๐๐ โ๐ square units

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Chapter 8 Class 12 Application of Integrals (Term 2)

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.