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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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Ex 8.1, 3 Area lying in the first quadrant and bounded by the circle 𝑥2+𝑦2=4 and the lines 𝑥 = 0 and 𝑥 = 2 is (A) π (B) 𝜋/2 (C) 𝜋/3 (D) 𝜋/4Given Equation of Circle :- 𝑥^2+𝑦^2=4 𝒙^𝟐+𝒚^𝟐=(𝟐)^𝟐 ∴ Radius = 𝒓=𝟐 Now, Line 𝒙=𝟎 is y-axis & Line x = 2 passes through point A (𝟐 , 𝟎) So, Required area = Area of shaded region = Area OAB = ∫_𝟎^𝟐▒〖𝒚.𝒅𝒙〗 We know that, 𝑥^2+𝑦^2=4 𝑦^2=4−𝑥^2 ∴ 𝒚=±√(𝟒−𝒙^𝟐 ) As, OBA is in 1st Quadrant Value of y will be positive ∴ 𝒚=√(𝟒−𝒙^𝟐 ) Now, Required area = ∫_0^2▒〖𝑦.𝑑𝑥〗 = ∫_0^2▒〖√(4−𝑥^2 ) 𝑑𝑥〗 = ∫_𝟎^𝟐▒〖√((𝟐)^𝟐−𝒙^𝟐 ) 𝒅𝒙〗 = [(𝑥 )/2 √((2)^2−𝑥^2 )+2 sin^(−1)⁡〖𝑥/2 〗 ]_0^2 = [2/2 √((2)^2−2^2 )+2 sin^(−1)⁡〖2/2 〗−0/2 √((2)^2−(0)^2 )−2 sin^(−1)⁡〖0/2〗 ] = [0+2 sin^(−1)⁡〖(1)−0√4−2 sin^(−1)⁡(0) 〗 ] = 2 sin^(−1)⁡〖(1)−2 sin^(−1)⁡(0) 〗−0 It is of form √(𝒂^𝟐−𝒙^𝟐 ) 𝒅𝒙=1/2 𝑥√(𝑎^2−𝑥^2 )+𝑎^2/2 〖𝑠𝑖𝑛〗^(−1)⁡〖 𝑥/𝑎+𝑐〗 Replacing a by 2 , we get = 2[〖𝒔𝒊𝒏〗^(−𝟏)⁡〖(𝟏)−〖𝒔𝒊𝒏〗^(−𝟏)⁡(𝟎) 〗 ] = 2[𝜋/2−0] = 2 . 𝜋/2 = π Therefore, Area Required = π square units So, the correct answer is (a)

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.