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Chapter 8 Class 12 Application of Integrals

Serial order wise

Last updated at March 22, 2023 by Teachoo

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 )