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Ex 8.1, 5 - Find area by ellipse x2/4 + y2/9 =1 - Class 12

Ex 8.1, 5 - Chapter 8 Class 12 Application of Integrals - Part 2
Ex 8.1, 5 - Chapter 8 Class 12 Application of Integrals - Part 3 Ex 8.1, 5 - Chapter 8 Class 12 Application of Integrals - Part 4

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Ex 8.1, 5 Find the area of the region bounded by the ellipse 𝑥^2/4+𝑦^2/9=1 Given Equation of Ellipse 𝑥^2/4+𝑦^2/9=1 𝑥^2/(2)^2 +𝑦^2/(3)^2 =1 Area of Ellipse = Area of ABCD = 2 × [Area of BCD] = 2 × ∫_(−2)^2▒〖𝑦 𝑑𝑥〗 We know that 𝑥^2/4+𝑦^2/9=1 𝑦^2/9=1−𝑥^2/4 𝑦^2/9=(4 − 𝑥^2)/4 𝑦^2=9/4 (4−𝑥^2 ) Taking Square Root on Both Sides 𝑦="±" √(9/4 (4−𝑥^2 ) ) 𝑦 ="±" 3/2 √(4−𝑥^2 ) Since Area BCD is above the x-axis, The value of y will be positive ∴ 𝑦=3/2 √(4−𝑥^2 ) Now, Area of Ellipse = 2 × ∫_(−2)^2▒〖𝑦 𝑑𝑥〗 = 3 ∫_(−2)^2▒〖 √((2)^2−𝑥^2 ) 𝑑𝑥〗 = 3 [𝑥/2 √((2)^2−𝑥^2 )+(2)^2/2 sin^(−1)⁡〖𝑥/2〗 ]_(−2)^2 It is of form √(𝑎^2−𝑥^2 ) 𝑑𝑥=𝑥/2 √(𝑎^2−𝑥^2 )+𝑎^2/2 〖𝑠𝑖𝑛〗^(−1)⁡〖 𝑥/𝑎+𝑐〗 Replacing a by 2 we get = 3 [𝑥/2 √(4−𝑥^2 )+2 sin^(−1)⁡〖𝑥/2〗 ]_(−2)^2 = 3[(2/2 √(4−2^2 )+2 sin^(−1)⁡〖2/2〗 )−((−2)/2 √(4−〖(−2)〗^2 )+2 sin^(−1)⁡〖(−2)/2〗 )] = 3[(0+2 sin^(−1)⁡1 )−(0+2 sin^(−1)⁡〖(−1)〗 )] = 3[2 sin^(−1)⁡1−2 〖𝒔𝒊𝒏〗^(−𝟏)⁡〖(−𝟏)〗 ] = 3[2 sin^(−1)⁡1−2×−〖𝒔𝒊𝒏〗^(−𝟏)⁡𝟏 ] = 3[2 〖𝑠𝑖𝑛〗^(−1)⁡1+2 〖𝑠𝑖𝑛〗^(−1)⁡1 ] = 3 [4 〖𝒔𝒊𝒏〗^(−𝟏)⁡(𝟏) ] = 3 × 4 × 𝝅/𝟐 = 6π ∴ Area of Ellipse = 6π square units (As sin-1 (–θ) = –sin–1 θ )

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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.