# Example 8 - Chapter 8 Class 12 Application of Integrals

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

Examples

Example 1

Example 2 Important

Example 3

Example 4

Example 5 Important

Example 6 Important Deleted for CBSE Board 2021 Exams only

Example 7 Important Deleted for CBSE Board 2021 Exams only

Example 8 Important Deleted for CBSE Board 2021 Exams only You are here

Example 9 Deleted for CBSE Board 2021 Exams only

Example 10 Important Deleted for CBSE Board 2021 Exams only

Example 11

Example 12

Example 13 Important Deleted for CBSE Board 2021 Exams only

Example 14 Important Deleted for CBSE Board 2021 Exams only

Example 15 Important

About the Author

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.