Ex 8.1

Chapter 8 Class 12 Application of Integrals
Serial order wise

### Transcript

Ex 8.1, 3 Area lying in the first quadrant and bounded by the circle ๐ฅ2+๐ฆ2=4 and the lines ๐ฅ = 0 and ๐ฅ = 2 is (A) ฯ (B) ๐/2 (C) ๐/3 (D) ๐/4Given Equation of Circle :- ๐ฅ^2+๐ฆ^2=4 ๐^๐+๐^๐=(๐)^๐ โด Radius = ๐=๐ Now, Line ๐=๐ is y-axis & Line x = 2 passes through point A (๐ , ๐) So, Required area = Area of shaded region = Area OAB = โซ_๐^๐โใ๐.๐๐ใ We know that, ๐ฅ^2+๐ฆ^2=4 ๐ฆ^2=4โ๐ฅ^2 โด ๐=ยฑโ(๐โ๐^๐ ) As, OBA is in 1st Quadrant Value of y will be positive โด ๐=โ(๐โ๐^๐ ) Now, Required area = โซ_0^2โใ๐ฆ.๐๐ฅใ = โซ_0^2โใโ(4โ๐ฅ^2 ) ๐๐ฅใ = โซ_๐^๐โใโ((๐)^๐โ๐^๐ ) ๐๐ใ = [(๐ฅ )/2 โ((2)^2โ๐ฅ^2 )+2 sin^(โ1)โกใ๐ฅ/2 ใ ]_0^2 = [2/2 โ((2)^2โ2^2 )+2 sin^(โ1)โกใ2/2 ใโ0/2 โ((2)^2โ(0)^2 )โ2 sin^(โ1)โกใ0/2ใ ] = [0+2 sin^(โ1)โกใ(1)โ0โ4โ2 sin^(โ1)โก(0) ใ ] = 2 sin^(โ1)โกใ(1)โ2 sin^(โ1)โก(0) ใโ0 It is of form โ(๐^๐โ๐^๐ ) ๐๐=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 Replacing a by 2 , we get = 2[ใ๐๐๐ใ^(โ๐)โกใ(๐)โใ๐๐๐ใ^(โ๐)โก(๐) ใ ] = 2[๐/2โ0] = 2 . ๐/2 = ฯ Therefore, Area Required = ฯ square units So, the correct answer is (a)