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

Transcript

Example 2 Find the area enclosed by the ellipse π‘₯^2/π‘Ž^2 +𝑦^2/𝑏^2 =1 We have to find Area Enclosed by ellipse Since Ellipse is symmetrical about both x-axis and y-axis ∴ Area of ellipse = 4 Γ— Area of OAB = 4 Γ— ∫_0^π‘Žβ–’γ€–π‘¦ 𝑑π‘₯γ€— We know that , π‘₯^2/π‘Ž^2 +𝑦^2/𝑏^2 =1 𝑦^2/𝑏^2 =1βˆ’π‘₯^2/π‘Ž^2 𝑦^2/𝑏^2 =(π‘Ž^2βˆ’π‘₯^2)/π‘Ž^2 𝑦^2=𝑏^2/π‘Ž^2 (π‘Ž^2βˆ’π‘₯^2 ) ∴ 𝑦=±√(𝑏^2/π‘Ž^2 (π‘Ž^2βˆ’π‘₯^2 ) ) 𝑦=±𝑏/π‘Ž √((π‘Ž^2βˆ’π‘₯^2 ) ) Since OAB is in 1st quadrant, value of y is positive ∴ 𝑦=𝑏/π‘Ž √(π‘Ž^2βˆ’π‘₯^2 ) Area of ellipse = 4 Γ— ∫_0^π‘Žβ–’γ€–π‘¦.𝑑π‘₯γ€— = 4∫_0^π‘Žβ–’π‘/π‘Ž √(π‘Ž^2βˆ’π‘₯^2 ) 𝑑π‘₯ = 4𝑏/π‘Ž ∫_0^π‘Žβ–’γ€–βˆš(π‘Ž^2βˆ’π‘₯^2 ) 𝑑π‘₯" " γ€— = 4𝑏/π‘Ž [π‘₯/2 √(π‘Ž^2βˆ’π‘₯^2 )+π‘Ž^2/2 sin^(βˆ’1)⁑〖π‘₯/π‘Žγ€— ]_0^π‘Ž = 4𝑏/π‘Ž [(π‘Ž/2 √(π‘Ž^2βˆ’π‘Ž^2 )+π‘Ž^2/2 sin^(βˆ’1)β‘γ€–π‘Ž/π‘Žγ€— )βˆ’(0/2 √(π‘Ž^2βˆ’0)βˆ’π‘Ž^2/2 sin^(βˆ’1)⁑(0) )] = 4𝑏/π‘Ž [0+π‘Ž^2/2 sin^(βˆ’1)⁑〖(1)γ€—βˆ’0βˆ’0] 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〗 = 4𝑏/π‘Ž Γ— π‘Ž^2/2 sin^(βˆ’1)⁑(1) = 2π‘Žπ‘ Γ—sin^(βˆ’1)⁑(1) = 2π‘Žπ‘ Γ— πœ‹/2 = πœ‹π‘Žπ‘ ∴ Required Area = 𝝅𝒂𝒃 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.