Find the area of the ellipse 𝑥 2 + 9𝑦 2 = 36 using integration

Find the area of the ellipse x^2 + 9y^2 = 36 using integration [Video]

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

 

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/


Transcript

Question 34 (Choice 2) Find the area of the ellipse 𝑥2 + 9𝑦2 = 36 using integration Equation of Ellipse is :- 𝑥2 + 9𝑦2 = 36 𝑥^2/36+(9𝑦^2)/36=1 𝑥^2/36+𝑦^2/4=1 𝑥^2/6^2 +𝑦^2/2^2 =1 Since Ellipse is symmetric along x and y-axis Area of ellipse = Area of ABCD = 4 × [Area Of OBC] = 2 × ∫_0^6▒〖𝑦.〗 𝑑𝑥 Finding y 𝑥2 + 9𝑦2 = 36 9𝑦2 = 36 − 𝑥2 𝑦^2=1/9 (36−𝑥^2 ) Taking square root on both sides y = ± √(1/9 (36−𝑥^2 ) ) y = ± 1/3 √(36−𝑥^2 ) Since OBC is above x-axis y will be positive ∴ 𝒚=𝟏/𝟑 √(𝟑𝟔−𝒙^𝟐 ) Area of ellipse = 4 × ∫_𝟎^𝟔▒〖𝒚.〗 𝒅𝒙 = 4 × ∫_0^6▒〖 1/3 √(36−𝑥^2 )〗 𝑑𝑥 = 4/3 ∫_0^6▒√((6)^2−𝑥^2 ) 𝑑𝑥 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 6 we get = 4/3 [𝑥/2 √((6)^2−𝑥^2 )+(6)^2/2 sin^(−1)⁡〖 𝑥/6〗 ]_0^6 = 4/3 [6/2 √((6)^2−(6)^2 )+18 〖 sin〗^(−1)⁡(6/6)−0/2 √((6)^2−(0)^2 )−18sin^(−1) (0/6)] = 4/3 [18 sin^(−1) (1)] = 4/3 × 18 ×𝜋/2 = 12𝜋 square units

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.