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Find the area of the region bounded by the curves π‘₯ 2 + 𝑦 2 = 4, y = √3π‘₯ π‘Žπ‘›π‘‘ π‘₯ βˆ’ π‘Žπ‘₯𝑖𝑠 𝑖𝑛 π‘‘β„Žπ‘’ π‘“π‘–π‘Ÿπ‘ π‘‘ π‘žπ‘’π‘Žπ‘‘π‘Ÿπ‘Žπ‘›π‘‘

Find area of region bounded by curves x^2 + y^2 = 4, y = √3x and x-axi

Question 34  (Choice 1) - CBSE Class 12 Sample Paper for 2021 Boards - Part 2
Question 34  (Choice 1) - CBSE Class 12 Sample Paper for 2021 Boards - Part 3
Question 34  (Choice 1) - CBSE Class 12 Sample Paper for 2021 Boards - Part 4
Question 34  (Choice 1) - CBSE Class 12 Sample Paper for 2021 Boards - Part 5
Question 34  (Choice 1) - CBSE Class 12 Sample Paper for 2021 Boards - Part 6

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Question 34 (Choice 1) Find the area of the region bounded by the curves π‘₯^2+𝑦^2=4, 𝑦=√3 π‘₯ π‘Žπ‘›π‘‘ π‘₯ βˆ’ π‘Žπ‘₯𝑖𝑠 𝑖𝑛 π‘‘β„Žπ‘’ π‘“π‘–π‘Ÿπ‘ π‘‘ π‘žπ‘’π‘Žπ‘‘π‘Ÿπ‘Žπ‘›π‘‘ Given Equation of Circle π‘₯2+𝑦2=4 π‘₯2+𝑦2=2^2 So, Radius = 2 ∴ 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 π’š=βˆšπŸ‘ 𝒙 in Equation of Circle π‘₯2+𝑦2=4 π‘₯2+(√3 π‘₯)2=4 4π‘₯2=4 π‘₯2=1 ∴ π‘₯=Β±1 When 𝒙=𝟏 𝑦=√3 π‘₯=√3 So, point is (1, √3) When 𝒙=βˆ’πŸ 𝑦=√3 π‘₯=βˆ’βˆš3 So, point is (–1, β€“βˆš3) As Point M is in 1st Quadrant ∴ M = (1, √3) ∴ Required Area = Area OMP + Area PMA = ∫1_0^1▒𝑦1𝑑π‘₯+∫1_1^2▒𝑦2𝑑π‘₯ Finding y1 and y2 For y1 – Equation of Line y = √3x So, y1 = √3x For y2 – Equation of Circle π‘₯^2+𝑦^2=4 𝑦^2=4βˆ’π‘₯^2 ∴ 𝑦=±√(2^2βˆ’π‘₯^2 ) As Required Area is in first Quadrant ∴ 𝑦2=√(2^2βˆ’π‘₯^2 ) Thus, Required Area = ∫1_0^1β–’γ€–βˆš3 π‘₯ 𝑑π‘₯γ€—+∫1_1^2β–’γ€–βˆš(2^2βˆ’π‘₯^2 ) 𝑑π‘₯γ€— =√3 ∫1_0^1β–’γ€–π‘₯ 𝑑π‘₯γ€—+∫1_1^2β–’γ€–βˆš(2^2βˆ’π‘₯^2 ) 𝑑π‘₯γ€— =√3 [π‘₯^2/2]_0^1+∫1_1^2β–’γ€–βˆš(2^2βˆ’π‘₯^2 ) 𝑑π‘₯γ€— =√3 [1^2/2βˆ’0^2/2]_0^1+∫1_1^2β–’γ€–βˆš(2^2βˆ’π‘₯^2 ) 𝑑π‘₯γ€— =√3/2+∫1_1^2β–’γ€–βˆš(2^2βˆ’π‘₯^2 ) 𝑑π‘₯γ€— It is of form ∫1β–’γ€–βˆš(π‘Ž^2βˆ’π‘₯^2 ) 𝑑π‘₯=1/2 π‘₯√(π‘Ž^2βˆ’π‘₯^2 )γ€—+π‘Ž^2/2 〖𝑠𝑖𝑛〗^(βˆ’1)⁑〖π‘₯/π‘Ž+𝑐〗 Replacing a by 2 , we get γ€–=√3/2+[π‘₯/2 √((2)^2βˆ’π‘₯^2 )+(2)^2/2 sin^(βˆ’1)⁑〖π‘₯/2γ€— ]γ€—_1^2 =√3/2+2/2 √((2)^2βˆ’(2)^2 )+2 sin^(βˆ’1)⁑〖2/2γ€—βˆ’1/2 √((2)^2βˆ’("1" )^2 )βˆ’2 sin^(βˆ’1)⁑〖1/2γ€— =√3/2+0+2 sin^(βˆ’1)⁑〖(1)βˆ’1/2 √(4βˆ’1)βˆ’γ€— 2 sin^(βˆ’1)⁑(1/2) =√3/2+2 Γ—πœ‹/2βˆ’βˆš3/2βˆ’2Γ—πœ‹/6 =πœ‹ βˆ’πœ‹/3 =πŸπ…/πŸ‘ square units

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.