Examples

Chapter 8 Class 12 Application of Integrals
Serial order wise

### Transcript

Question 3 Find the area bounded by the ellipse ๐ฅ^2/๐^2 +๐ฆ^2/๐^2 =1 and the ordinates ๐ฅ=0 and ๐ฅ=๐๐, where, ๐2=๐2 (1 โ ๐2) and e < 1 Required Area = Area of shaded region = Area BORQSP = 2 ร Area OBPS = 2 ร โซ_0^๐๐โใ๐ฆ.๐๐ฅใ We know that , ๐ฅ^2/๐^2 +๐ฆ^2/๐^2 =1 (As ellipse is symmetric about its axis ) ๐ฆ^2/๐^2 =(๐^2โใ ๐ฅใ^2)/๐^2 ๐ฆ^2=๐^2/๐^2 (๐^2โ๐ฅ^2 ) ๐ฆ=ยฑโ(๐^2/๐^2 (๐^2โ๐ฅ^2 ) ) ๐ฆ=ยฑ๐/๐ โ((๐^2โ๐ฅ^2 ) ) Since OBPS is in 1st quadrant, value of y is positive โด ๐ฆ=๐/๐ โ(๐^2โ๐ฅ^2 ) Required Area = 2 ร โซ_0^๐๐โใ๐ฆ.๐๐ฅใ = 2โซ_0^๐๐โใ๐/๐ โ(๐^2โ๐ฅ^2 )ใ ๐๐ฅ = 2๐/๐ โซ_0^๐๐โโ(๐^2โ๐ฅ^2 ) ๐๐ฅ = 2๐/๐ [1/2 ๐ฅโ(๐^2โ๐ฅ^2 )+๐^2/2 sin^(โ1)โกใ๐ฅ/๐ใ ]_0^๐๐ =2๐/๐ [(๐๐/2 โ(๐^2โ(๐๐)^2 )+๐^2/2 sin^(โ1)โกใ๐๐/๐ใ )โ(0/2 โ(๐^2โ0)+๐^2/2 sin^(โ1)โก(0/๐) )] =2๐/๐ [๐๐/2 โ(๐^2โ๐^2 ๐^2 )+๐^2/2 sin^(โ1)โกใ(๐)โ0โ๐^2/2 sin^(โ1)โก(0) ใ ] =2๐/๐ [๐๐/2.๐โ(1โ๐^2 )+๐^2/2 sin^(โ1)โกใ๐โ0ใ ] It is of form โ(๐^2โ๐ฅ^2 ) ๐๐ฅ=1/2 ๐ฅโ(๐^2โ๐ฅ^2 )+๐^2/2 ใ๐ ๐๐ใ^(โ1)โกใ ๐ฅ/๐+๐ใ =2๐/๐ [(๐^2 ๐)/2 โ(1โ๐^2 )+๐^2/2 sin^(โ1)โก๐ ] =2๐/๐ (๐^2/2)[๐โ(1โ๐^2 )+sin^(โ1)โก๐ ] =๐๐[๐โ(1โ๐^2 )+sin^(โ1)โก๐ ] โด Required Area =๐๐[๐โ(๐โ๐^๐ )+ใ๐๐๐ใ^(โ๐)โก๐ ] square units