Ex 8.1, 3 (MCQ) - Chapter 8 Class 12 Application of Integrals
Last updated at April 16, 2024 by Teachoo
Area bounded by curve and horizontal or vertical line
Area bounded by curve and horizontal or vertical line
Last updated at April 16, 2024 by Teachoo
Ex 8.1, 3 Area lying in the first quadrant and bounded by the circle ๐ฅ2+๐ฆ2=4 and the lines ๐ฅ = 0 and ๐ฅ = 2 is (A) ฯ (B) ๐/2 (C) ๐/3 (D) ๐/4Given Equation of Circle :- ๐ฅ^2+๐ฆ^2=4 ๐^๐+๐^๐=(๐)^๐ โด 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 = โซ_๐^๐โใ๐.๐ ๐ใ We know that, ๐ฅ^2+๐ฆ^2=4 ๐ฆ^2=4โ๐ฅ^2 โด ๐=ยฑโ(๐โ๐^๐ ) As, OBA is in 1st Quadrant Value of y will be positive โด ๐=โ(๐โ๐^๐ ) Now, Required area = โซ_0^2โใ๐ฆ.๐๐ฅใ = โซ_0^2โใโ(4โ๐ฅ^2 ) ๐๐ฅใ = โซ_๐^๐โใโ((๐)^๐โ๐^๐ ) ๐ ๐ใ = [(๐ฅ )/2 โ((2)^2โ๐ฅ^2 )+2 sin^(โ1)โกใ๐ฅ/2 ใ ]_0^2 = [2/2 โ((2)^2โ2^2 )+2 sin^(โ1)โกใ2/2 ใโ0/2 โ((2)^2โ(0)^2 )โ2 sin^(โ1)โกใ0/2ใ ] = [0+2 sin^(โ1)โกใ(1)โ0โ4โ2 sin^(โ1)โก(0) ใ ] = 2 sin^(โ1)โกใ(1)โ2 sin^(โ1)โก(0) ใโ0 It is of form โ(๐^๐โ๐^๐ ) ๐ ๐=1/2 ๐ฅโ(๐^2โ๐ฅ^2 )+๐^2/2 ใ๐ ๐๐ใ^(โ1)โกใ ๐ฅ/๐+๐ใ Replacing a by 2 , we get = 2[ใ๐๐๐ใ^(โ๐)โกใ(๐)โใ๐๐๐ใ^(โ๐)โก(๐) ใ ] = 2[๐/2โ0] = 2 . ๐/2 = ฯ Therefore, Area Required = ฯ square units So, the correct answer is (a)