CBSE Class 12 Sample Paper for 2022 Boards (For Term 2)

Class 12
Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

## Using integration, find the area of the region{(x,y):0≤y≤√3 x, x^2+y^2  ≤4}

This question is similar to Question 34 CBSE Class 12 - Sample Paper 2020 Boards

### Transcript

Question 12 (Choice 2) Using integration, find the area of the region {(๐ฅ,๐ฆ):0โค๐ฆโคโ3 ๐ฅ, ๐ฅ^2+๐ฆ^2 โค4} Here, we are given A circle and a line And we need to find area enclosed Circle - "x2 + y2 "โค" 4" Circle is ๐ฅ2+๐ฆ2 =4 (๐ฅโ0)^2+(๐ฆโ0)^2 =2^2 So, Center = (0, 0) & Radius = 2 And "x2 + y2 "โค" 2" means area enclosed inside the circle Line - ๐โค๐โคโ๐ ๐ We draw line ๐ฆ=โ3 ๐ฅ And ๐โฅ๐ means value of y is always positive Now, {(๐ฅ,๐ฆ):0โค๐ฆโคโ3 ๐ฅ, ๐ฅ^2+๐ฆ^2 โค4} is the blue shaded region โด Required Area = Area OAB Now, Point A is (2, 0) Finding point B We know that ๐ฆ=โ3 ๐ฅ Putting value of y in equation of circle ๐ฅ^2+๐ฆ^2=4 ๐^๐+(โ๐ ๐)^๐=๐ x^2+3๐ฅ^2=4 4๐ฅ^2=4 ๐ฅ^2=1 โด ๐=ยฑ๐ Since x is in 1st quadrant Thus, x = 1 Since ๐ฆ=โ3 ๐ฅ Putting x = 1 ๐=โ๐ Thus, point B = (1, โ๐) Now, Area required = Area OBX + Area XBA Area OBX Area OBX = โซ_0^1โใ๐ฆ ๐๐ฅใ ๐ฆ โ Equation of line ๐=โ๐ ๐ Therefore, Area OBX = โซ_0^1โใโ3 ๐ฅ ๐๐ฅใ = โ3 โซ_0^1โใ๐ฅ ๐๐ฅใ = โ3 [๐ฅ^2/2]_0^1 = โ3/2 [๐ฅ^2 ]_0^1 = โ3/2 [(1)^2โ(0)^2 ] = โ๐/๐ Area XBA Area XBA = โซ_1^2โใ๐ฆ ๐๐ฅใ ๐ โ Equation of circle Now, ๐ฅ^2+๐ฆ^2=4 ๐ฆ^2=4โ๐ฅ^2 ๐ฆ = ยฑโ(4โ๐ฅ^2 ) Since XBA is in 1st Quadrant, Value of ๐ฆ will be positive โด ๐ = โ(๐โ๐^๐ ) Area XBA = โซ_1^2โ โ(4โ๐ฅ^2 ) ๐๐ฅ = โซ_๐^๐โ โ((๐)^๐โ๐^๐ ) ๐๐ = [1/2 ๐ฅโ((2)^2โ๐ฅ^2 )+(2)^2/2 sin^(โ1)โกใ๐ฅ/2ใ ]_1^2 = [1/2 ๐ฅโ(4โ๐ฅ^2 )+2 sin^(โ1)โกใ๐ฅ/2ใ ]_1^2 = 1/2 (2) โ(4โ2^2 )+2 sin^(โ1)โกใ2/2ใโ1/2 (1) โ(4โ(1)^2 )โ2 sin^(โ1)โกใ1/2ใ = 0 + 2 sin^(โ1)โก(1) โ 1/2 โ(4โ1)โ2 sin^(โ1)โกใ1/2ใ = 2 sin^(โ1)โก(1) โ โ3/2 โ 2 sin^(โ1)โกใ1/2ใ = (โโ3)/2+2[sin^(โ1)โกใ(1)โ๐ ๐๐^(โ1) ใ 1/2]= (โโ3)/2+2[sin^(โ1)โกใ(1)โ๐ ๐๐^(โ1) ใ 1/2] = (โโ3)/2+2[๐/2โ๐/6] = (โโ3)/2+2[2๐/6] = (โโ๐)/๐+๐๐/๐ Therefore, Required Area = Area OBX + Area XBA = โ3/2โโ3/2+2๐/3 = ๐๐/๐ square units