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
Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
CBSE Class 12 Sample Paper for 2022 Boards (For Term 2)
Question 1 (Choice 2)
Question 2 Important
Question 3
Question 4 Important
Question 5
Question 6 Important
Question 7 Important
Question 8 (Choice 1)
Question 8 (Choice 2)
Question 9 Important
Question 10 (Choice 1)
Question 10 (Choice 2)
Question 11 Important
Question 12 (Choice 1)
Question 12 (Choice 2) Important You are here
Question 13 Important
Question 14 - Case Based Important
CBSE Class 12 Sample Paper for 2022 Boards (For Term 2)
Last updated at May 29, 2023 by Teachoo
This question is similar to Question 34 CBSE Class 12 - Sample Paper 2020 Boards
Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Question 12 (Choice 2) Using integration, find the area of the region {(𝑥,𝑦):0≤𝑦≤√3 𝑥, 𝑥^2+𝑦^2 ≤4} Here, we are given A circle and a line And we need to find area enclosed Circle - "x2 + y2 "≤" 4" Circle is 𝑥2+𝑦2 =4 (𝑥−0)^2+(𝑦−0)^2 =2^2 So, Center = (0, 0) & Radius = 2 And "x2 + y2 "≤" 2" means area enclosed inside the circle Line - 𝟎≤𝒚≤√𝟑 𝒙 We draw line 𝑦=√3 𝑥 And 𝒚≥𝟎 means value of y is always positive Now, {(𝑥,𝑦):0≤𝑦≤√3 𝑥, 𝑥^2+𝑦^2 ≤4} is the blue shaded region ∴ Required Area = Area OAB Now, Point A is (2, 0) Finding point B We know that 𝑦=√3 𝑥 Putting value of y in equation of circle 𝑥^2+𝑦^2=4 𝒙^𝟐+(√𝟑 𝒙)^𝟐=𝟒 x^2+3𝑥^2=4 4𝑥^2=4 𝑥^2=1 ∴ 𝒙=±𝟏 Since x is in 1st quadrant Thus, x = 1 Since 𝑦=√3 𝑥 Putting x = 1 𝒚=√𝟑 Thus, point B = (1, √𝟑) Now, Area required = Area OBX + Area XBA Area OBX Area OBX = ∫_0^1▒〖𝑦 𝑑𝑥〗 𝑦 → Equation of line 𝒚=√𝟑 𝒙 Therefore, Area OBX = ∫_0^1▒〖√3 𝑥 𝑑𝑥〗 = √3 ∫_0^1▒〖𝑥 𝑑𝑥〗 = √3 [𝑥^2/2]_0^1 = √3/2 [𝑥^2 ]_0^1 = √3/2 [(1)^2−(0)^2 ] = √𝟑/𝟐 Area XBA Area XBA = ∫_1^2▒〖𝑦 𝑑𝑥〗 𝒚 → Equation of circle Now, 𝑥^2+𝑦^2=4 𝑦^2=4−𝑥^2 𝑦 = ±√(4−𝑥^2 ) Since XBA is in 1st Quadrant, Value of 𝑦 will be positive ∴ 𝒚 = √(𝟒−𝒙^𝟐 ) Area XBA = ∫_1^2▒ √(4−𝑥^2 ) 𝑑𝑥 = ∫_𝟏^𝟐▒ √((𝟐)^𝟐−𝒙^𝟐 ) 𝒅𝒙 = [1/2 𝑥√((2)^2−𝑥^2 )+(2)^2/2 sin^(−1)〖𝑥/2〗 ]_1^2 = [1/2 𝑥√(4−𝑥^2 )+2 sin^(−1)〖𝑥/2〗 ]_1^2 = 1/2 (2) √(4−2^2 )+2 sin^(−1)〖2/2〗−1/2 (1) √(4−(1)^2 )−2 sin^(−1)〖1/2〗 = 0 + 2 sin^(−1)(1) – 1/2 √(4−1)−2 sin^(−1)〖1/2〗 = 2 sin^(−1)(1) – √3/2 − 2 sin^(−1)〖1/2〗 = (−√3)/2+2[sin^(−1)〖(1)−𝑠𝑖𝑛^(−1) 〗 1/2]= (−√3)/2+2[sin^(−1)〖(1)−𝑠𝑖𝑛^(−1) 〗 1/2] = (−√3)/2+2[𝜋/2−𝜋/6] = (−√3)/2+2[2𝜋/6] = (−√𝟑)/𝟐+𝟐𝝅/𝟑 Therefore, Required Area = Area OBX + Area XBA = √3/2−√3/2+2𝜋/3 = 𝟐𝝅/𝟑 square units