Find the area of the ellipse π‘₯ 2 + 9𝑦 2 = 36 using integration

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Note : This is similar to Ex 8.1, 4 of NCERT – Chapter 8 Class 12 Application of Integration

Check the answer here https:// www.teachoo.com /3328/730/Ex-8.1--4---Find-area-bounded-by-ellipse-x2-16---y2-9--1/category/Ex-8.1/

  1. Class 12
  2. Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

Transcript

Question 34 (Choice 2) Find the area of the ellipse π‘₯2 + 9𝑦2 = 36 using integration Equation of Ellipse is :- π‘₯2 + 9𝑦2 = 36 π‘₯^2/36+(9𝑦^2)/36=1 π‘₯^2/36+𝑦^2/4=1 π‘₯^2/6^2 +𝑦^2/2^2 =1 Since Ellipse is symmetric along x and y-axis Area of ellipse = Area of ABCD = 4 Γ— [Area Of OBC] = 2 Γ— ∫_0^6▒〖𝑦.γ€— 𝑑π‘₯ Finding y π‘₯2 + 9𝑦2 = 36 9𝑦2 = 36 βˆ’ π‘₯2 𝑦^2=1/9 (36βˆ’π‘₯^2 ) Taking square root on both sides y = Β± √(1/9 (36βˆ’π‘₯^2 ) ) y = Β± 1/3 √(36βˆ’π‘₯^2 ) Since OBC is above x-axis y will be positive ∴ π’š=𝟏/πŸ‘ √(πŸ‘πŸ”βˆ’π’™^𝟐 ) Area of ellipse = 4 Γ— ∫_𝟎^πŸ”β–’γ€–π’š.γ€— 𝒅𝒙 = 4 Γ— ∫_0^6β–’γ€– 1/3 √(36βˆ’π‘₯^2 )γ€— 𝑑π‘₯ = 4/3 ∫_0^6β–’βˆš((6)^2βˆ’π‘₯^2 ) 𝑑π‘₯ 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 6 we get = 4/3 [π‘₯/2 √((6)^2βˆ’π‘₯^2 )+(6)^2/2 sin^(βˆ’1)⁑〖 π‘₯/6γ€— ]_0^6 = 4/3 [6/2 √((6)^2βˆ’(6)^2 )+18 γ€– sinγ€—^(βˆ’1)⁑(6/6)βˆ’0/2 √((6)^2βˆ’(0)^2 )βˆ’18sin^(βˆ’1) (0/6)] = 4/3 [18 sin^(βˆ’1) (1)] = 4/3 Γ— 18 Γ—πœ‹/2 = 12πœ‹ 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.