# Example 5 - 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 5 Find the area bounded by the ellipse ๐ฅ^2/๐^2 +๐ฆ^2/๐^2 =1 and the ordinates ๐ฅ=0 and ๐ฅ=๐๐, where, ๐2=๐2 (1 โ ๐2) and e < 1 Required Area = Area of shaded region = Area BORQSP = 2 ร Area OBPS = 2 ร โซ_0^๐๐โใ๐ฆ.๐๐ฅใ We know that , ๐ฅ^2/๐^2 +๐ฆ^2/๐^2 =1 (As ellipse is symmetric about its axis ) ๐ฆ^2/๐^2 =(๐^2โใ ๐ฅใ^2)/๐^2 ๐ฆ^2=๐^2/๐^2 (๐^2โ๐ฅ^2 ) ๐ฆ=ยฑโ(๐^2/๐^2 (๐^2โ๐ฅ^2 ) ) ๐ฆ=ยฑ๐/๐ โ((๐^2โ๐ฅ^2 ) ) Since OBPS is in 1st quadrant, value of y is positive โด ๐ฆ=๐/๐ โ(๐^2โ๐ฅ^2 ) Required Area = 2 ร โซ_0^๐๐โใ๐ฆ.๐๐ฅใ = 2โซ_0^๐๐โใ๐/๐ โ(๐^2โ๐ฅ^2 )ใ ๐๐ฅ = 2๐/๐ โซ_0^๐๐โโ(๐^2โ๐ฅ^2 ) ๐๐ฅ = 2๐/๐ [1/2 ๐ฅโ(๐^2โ๐ฅ^2 )+๐^2/2 sin^(โ1)โกใ๐ฅ/๐ใ ]_0^๐๐ =2๐/๐ [(๐๐/2 โ(๐^2โ(๐๐)^2 )+๐^2/2 sin^(โ1)โกใ๐๐/๐ใ )โ(0/2 โ(๐^2โ0)+๐^2/2 sin^(โ1)โก(0/๐) )] =2๐/๐ [๐๐/2 โ(๐^2โ๐^2 ๐^2 )+๐^2/2 sin^(โ1)โกใ(๐)โ0โ๐^2/2 sin^(โ1)โก(0) ใ ] =2๐/๐ [๐๐/2.๐โ(1โ๐^2 )+๐^2/2 sin^(โ1)โกใ๐โ0ใ ] It is of form โ(๐^2โ๐ฅ^2 ) ๐๐ฅ=1/2 ๐ฅโ(๐^2โ๐ฅ^2 )+๐^2/2 ใ๐ ๐๐ใ^(โ1)โกใ ๐ฅ/๐+๐ใ =2๐/๐ [(๐^2 ๐)/2 โ(1โ๐^2 )+๐^2/2 sin^(โ1)โก๐ ] =2๐/๐ (๐^2/2)[๐โ(1โ๐^2 )+sin^(โ1)โก๐ ] =๐๐[๐โ(1โ๐^2 )+sin^(โ1)โก๐ ] โด Required Area =๐๐[๐โ(๐โ๐^๐ )+ใ๐๐๐ใ^(โ๐)โก๐ ] square units

Examples

Example 1

Example 2 Important

Example 3

Example 4

Example 5 Important You are here

Example 6 Important Deleted for CBSE Board 2022 Exams

Example 7 Important Deleted for CBSE Board 2022 Exams

Example 8 Important Deleted for CBSE Board 2022 Exams

Example 9 Deleted for CBSE Board 2022 Exams

Example 10 Important Deleted for CBSE Board 2022 Exams

Example 11

Example 12

Example 13 Important

Example 14 Important Deleted for CBSE Board 2022 Exams

Example 15 Important

Chapter 8 Class 12 Application of Integrals (Term 2)

Serial order wise

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.