Find the area of the region bounded by the curves ๐ฅ 2 + ๐ฆ 2 = 4, y = √3๐ฅ ๐๐๐ ๐ฅ − ๐๐ฅ๐๐ ๐๐ ๐กℎ๐ ๐๐๐๐ ๐ก ๐๐ข๐๐๐๐๐๐ก





CBSE Class 12 Sample Paper for 2021 Boards
CBSE Class 12 Sample Paper for 2021 Boards
Last updated at Dec. 16, 2024 by Teachoo
Transcript
Question 34 (Choice 1) Find the area of the region bounded by the curves ๐ฅ^2+๐ฆ^2=4, ๐ฆ=โ3 ๐ฅ ๐๐๐ ๐ฅ โ ๐๐ฅ๐๐ ๐๐ ๐กโ๐ ๐๐๐๐ ๐ก ๐๐ข๐๐๐๐๐๐ก Given Equation of Circle ๐ฅ2+๐ฆ2=4 ๐ฅ2+๐ฆ2=2^2 So, Radius = 2 โด Point A (2, 0) and B is (0, 2) Let point where line and circle intersect be point M Required Area = Area of shaded region = Area OMA Finding Point M Point M is intersection of line and circle Putting ๐=โ๐ ๐ in Equation of Circle ๐ฅ2+๐ฆ2=4 ๐ฅ2+(โ3 ๐ฅ)2=4 4๐ฅ2=4 ๐ฅ2=1 โด ๐ฅ=ยฑ1 When ๐=๐ ๐ฆ=โ3 ๐ฅ=โ3 So, point is (1, โ3) When ๐=โ๐ ๐ฆ=โ3 ๐ฅ=โโ3 So, point is (โ1, โโ3) As Point M is in 1st Quadrant โด M = (1, โ3) โด Required Area = Area OMP + Area PMA = โซ1_0^1โ๐ฆ1๐๐ฅ+โซ1_1^2โ๐ฆ2๐๐ฅ Finding y1 and y2 For y1 โ Equation of Line y = โ3x So, y1 = โ3x For y2 โ Equation of Circle ๐ฅ^2+๐ฆ^2=4 ๐ฆ^2=4โ๐ฅ^2 โด ๐ฆ=ยฑโ(2^2โ๐ฅ^2 ) As Required Area is in first Quadrant โด ๐ฆ2=โ(2^2โ๐ฅ^2 ) Thus, Required Area = โซ1_0^1โใโ3 ๐ฅ ๐๐ฅใ+โซ1_1^2โใโ(2^2โ๐ฅ^2 ) ๐๐ฅใ =โ3 โซ1_0^1โใ๐ฅ ๐๐ฅใ+โซ1_1^2โใโ(2^2โ๐ฅ^2 ) ๐๐ฅใ =โ3 [๐ฅ^2/2]_0^1+โซ1_1^2โใโ(2^2โ๐ฅ^2 ) ๐๐ฅใ =โ3 [1^2/2โ0^2/2]_0^1+โซ1_1^2โใโ(2^2โ๐ฅ^2 ) ๐๐ฅใ =โ3/2+โซ1_1^2โใโ(2^2โ๐ฅ^2 ) ๐๐ฅใ It is of form โซ1โใโ(๐^2โ๐ฅ^2 ) ๐๐ฅ=1/2 ๐ฅโ(๐^2โ๐ฅ^2 )ใ+๐^2/2 ใ๐ ๐๐ใ^(โ1)โกใ๐ฅ/๐+๐ใ Replacing a by 2 , we get ใ=โ3/2+[๐ฅ/2 โ((2)^2โ๐ฅ^2 )+(2)^2/2 sin^(โ1)โกใ๐ฅ/2ใ ]ใ_1^2 =โ3/2+2/2 โ((2)^2โ(2)^2 )+2 sin^(โ1)โกใ2/2ใโ1/2 โ((2)^2โ("1" )^2 )โ2 sin^(โ1)โกใ1/2ใ =โ3/2+0+2 sin^(โ1)โกใ(1)โ1/2 โ(4โ1)โใ 2 sin^(โ1)โก(1/2) =โ3/2+2 ร๐/2โโ3/2โ2ร๐/6 =๐ โ๐/3 =๐๐ /๐ square units