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  1. Chapter 8 Class 12 Application of Integrals
  2. Serial order wise

Transcript

Ex 8.1, 6 Find the area of the region in the first quadrant enclosed by π‘₯βˆ’axis, line π‘₯ = √3 𝑦 and the circle π‘₯2 + 𝑦2 = 4. Given Equation of circle π‘₯^2+𝑦^2=4 π‘₯^2+𝑦^2=(2)^2 ∴ Radius π‘Ÿ = 2 So, point A is (2, 0) and point B is (0, 2) Let line π‘₯=√3 𝑦 intersect the circle at point C Therefore, We have to find Area of AOC Finding point C We know that π‘₯=√3 𝑦 Putting value of x in equation of circle π‘₯^2+𝑦^2=4 (√3 𝑦)^2+𝑦^2=4 3𝑦^2+𝑦^2=4 4𝑦^2=4 𝑦^2=1 ∴ 𝑦=Β±1 Now, finding value of x When π’š=𝟏 π‘₯=√3 𝑦 π‘₯=√3 Γ— 1 π‘₯=√3 When π’š=βˆ’πŸ π‘₯=√3 𝑦 π‘₯=√3 Γ— –1 π‘₯=βˆ’βˆš3 Since point C is in 1st quadrant ∴ C is (√3 , 1) Area OAC Area of OAC = Area OCX + Area XCA Area OCX Area OCX = ∫_0^(√3)▒〖𝑦 𝑑π‘₯γ€— 𝑦 β†’ Equation of line Now, π‘₯=√3 𝑦 𝑦=π‘₯/√3 Therefore, Area OCX = ∫_0^(√3)▒〖𝑦 𝑑π‘₯γ€— = ∫_0^(√3)β–’γ€–π‘₯/√3 𝑑π‘₯γ€— = 1/√3 ∫_0^(√3)β–’γ€–π‘₯ 𝑑π‘₯γ€— = 1/√3 [π‘₯^2/2]_0^√3 = 1/(2√3) [π‘₯^2 ]_0^√3 = (1 )/(2√3) [(√3)^2βˆ’(0)^2 ] = (1 )/(2√3) [ 3 ] = √3/2 Area XCA Area XCA = ∫_(√3)^2▒〖𝑦 𝑑π‘₯γ€— 𝑦 β†’ Equation of circle Now, π‘₯^2+𝑦^2=4 𝑦^2=4βˆ’π‘₯^2 𝑦 = ±√(4βˆ’π‘₯^2 ) Since XCA is in 1st Quadrant, = (1 )/(2√3) [ 3 ] = √3/2 Area XCA Area XCA = ∫_(√3)^2▒〖𝑦 𝑑π‘₯γ€— 𝑦 β†’ Equation of circle Now, π‘₯^2+𝑦^2=4 𝑦^2=4βˆ’π‘₯^2 𝑦 = ±√(4βˆ’π‘₯^2 ) Since XCA is in 1st Quadrant, Value of 𝑦 will be positive ∴ 𝑦 = √(4βˆ’π‘₯^2 ) Area XCA = ∫_(√3)^2β–’ √(4βˆ’π‘₯^2 ) 𝑑π‘₯ = ∫_(√3)^2β–’ √((2)^2βˆ’π‘₯^2 ) 𝑑π‘₯ = [1/2 π‘₯√((2)^2βˆ’π‘₯^2 )+(2)^2/2 sin^(βˆ’1)⁑〖π‘₯/2γ€— ]_(√3)^2 = [1/2 π‘₯√(4βˆ’π‘₯^2 )+2 sin^(βˆ’1)⁑〖π‘₯/2γ€— ]_(√3)^2 = 1/2 (2) √(4βˆ’2^2 )+2 sin^(βˆ’1)⁑〖2/2γ€—βˆ’1/2 (√3) √(4βˆ’(√3)^2 )βˆ’2 sin^(βˆ’1)β‘γ€–βˆš3/2γ€— It is of form √(π‘Ž^2βˆ’π‘₯^2 ) 𝑑π‘₯=1/2 π‘₯√(π‘Ž^2βˆ’π‘₯^2 )+π‘Ž^2/2 〖𝑠𝑖𝑛〗^(βˆ’1)⁑〖 π‘₯/π‘Ž+𝑐〗 Replacing a by 2 , we get = 0 + 2 sin^(βˆ’1)⁑(1) – √3/2 √(4βˆ’3)βˆ’2 sin^(βˆ’1)β‘γ€–βˆš3/2γ€— = 2 sin^(βˆ’1)⁑(1) – √3/2 βˆ’ 2 sin^(βˆ’1)β‘γ€–βˆš3/2γ€— = (βˆ’βˆš3)/2+2[sin^(βˆ’1)⁑〖(1)βˆ’π‘ π‘–π‘›^(βˆ’1) γ€— √3/2] = (βˆ’βˆš3)/2+2[πœ‹/2βˆ’πœ‹/3] = (βˆ’βˆš3)/2+2[πœ‹/6] = (βˆ’βˆš3)/2+πœ‹/3 Therefore, Area of OAC = Area OCX + Area XCA = √3/2βˆ’βˆš3/2+(πœ‹ )/3 = (πœ‹ )/3 square units ∴ Required Area = 𝝅/πŸ‘ square units

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.