Ex 8.1

Chapter 8 Class 12 Application of Integrals
Serial order wise

### Transcript

Ex 8.1, 2 Find the area of the region bounded by the ellipse ๐ฅ^2/4+๐ฆ^2/9=1Given Equation of Ellipse ๐ฅ^2/4+๐ฆ^2/9=1 ๐^๐/(๐)^๐ +๐^๐/(๐)^๐ =๐ Area of Ellipse = Area of ABCD = 2 ร [Area of BCD] = 2 ร โซ_(โ๐)^๐โใ๐ ๐๐ใ We know that ๐ฅ^2/4+๐ฆ^2/9=1 ๐ฆ^2/9=1โ๐ฅ^2/4 ๐ฆ^2/9=(4 โ ๐ฅ^2)/4 ๐ฆ^2=9/4 (4โ๐ฅ^2 ) Taking Square Root on Both Sides ๐ฆ="ยฑ" โ(9/4 (4โ๐ฅ^2 ) ) ๐ ="ยฑ" ๐/๐ โ(๐โ๐^๐ ) Since Area BCD is above the x-axis, The value of y will be positive โด ๐=๐/๐ โ(๐โ๐^๐ ) Now, Area of Ellipse = 2 ร โซ_(โ2)^2โใ๐ฆ ๐๐ฅใ = 2 ร โซ_(โ2)^2โใ3/2 โ(4โ๐ฅ^2 ) ๐๐ฅใ = 3 โซ_(โ๐)^๐โใโ((๐)^๐โ๐^๐ ) ๐๐ใ It is of form โ(๐^2โ๐ฅ^2 ) ๐๐ฅ=๐ฅ/2 โ(๐^2โ๐ฅ^2 )+๐^2/2 ใ๐ ๐๐ใ^(โ1)โกใ ๐ฅ/๐+๐ใ Replacing a by 2 we get = 3 [๐ฅ/2 โ((2)^2โ๐ฅ^2 )+(2)^2/2 sin^(โ1)โกใ๐ฅ/2ใ ]_(โ2)^2 = 3 [๐ฅ/2 โ(4โ๐ฅ^2 )+2 sin^(โ1)โกใ๐ฅ/2ใ ]_(โ2)^2 = 3[(2/2 โ(4โ2^2 )+2 sin^(โ1)โกใ2/2ใ )โ((โ2)/2 โ(4โใ(โ2)ใ^2 )+2 sin^(โ1)โกใ(โ2)/2ใ )] = 3[(0+2 sin^(โ1)โก1 )โ(0+2 sin^(โ1)โกใ(โ1)ใ )] = 3[2 sin^(โ1)โก1โ2 ใ๐๐๐ใ^(โ๐)โกใ(โ๐)ใ ] = 3[2 sin^(โ1)โก1โ2รโใ๐๐๐ใ^(โ๐)โก๐ ] = 3[2 ใ๐ ๐๐ใ^(โ1)โก1+2 ใ๐ ๐๐ใ^(โ1)โก1 ] = 3 [4 ใ๐๐๐ใ^(โ๐)โก(๐) ] = 3 ร 4 ร ๐/๐ = 6ฯ โด Area of Ellipse = 6ฯ square units