Example 2 - Chapter 8 Class 12 Application of Integrals

Transcript

Example 2 Find the area enclosed by the ellipse 𝑥^2/𝑎^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 × Area of OAB = 4 × ∫_𝟎^𝒂▒〖𝒚 𝒅𝒙〗 We know that , 𝑥^2/𝑎^2 +𝑦^2/𝑏^2 =1 𝑦^2/𝑏^2 =1−𝑥^2/𝑎^2 𝑦^2/𝑏^2 =(𝑎^2−〖 𝑥〗^2)/𝑎^2 𝑦^2=𝑏^2/𝑎^2 (𝑎^2−𝑥^2 ) 𝑦=±√(𝑏^2/𝑎^2 (𝑎^2−𝑥^2 ) ) 𝒚=±𝒃/𝒂 √((𝒂^𝟐−𝒙^𝟐 ) ) Since OAB is in 1st quadrant, value of y is positive ∴ 𝒚=𝒃/𝒂 √(𝒂^𝟐−𝒙^𝟐 ) Area of ellipse = 4 × ∫_0^𝑎▒〖𝑦.𝑑𝑥〗 = 4∫_𝟎^𝒂▒𝒃/𝒂 √(𝒂^𝟐−𝒙^𝟐 ) 𝒅𝒙 = 4𝑏/𝑎 ∫_0^𝑎▒〖√(𝑎^2−𝑥^2 ) 𝑑𝑥" " 〗 = 𝟒𝒃/𝒂 [𝒙/𝟐 √(𝒂^𝟐−𝒙^𝟐 )+𝒂^𝟐/𝟐 〖𝒔𝒊𝒏〗^(−𝟏)⁡〖𝒙/𝒂〗 ]_𝟎^𝒂 = 4𝑏/𝑎 [(𝑎/2 √(𝑎^2−𝑎^2 )+𝑎^2/2 sin^(−1)⁡〖𝑎/𝑎〗 )−(0/2 √(𝑎^2−0)−𝑎^2/2 sin^(−1)⁡(0) )] = 4𝑏/𝑎 [0+𝑎^2/2 sin^(−1)⁡〖(1)〗−0−0] It is of form √(𝑎^2−𝑥^2 ) 𝑑𝑥=1/2 𝑥√(𝑎^2−𝑥^2 )+𝑎^2/2 〖𝑠𝑖𝑛〗^(−1)⁡〖 𝑥/𝑎+𝑐〗 = 𝟒𝒃/𝒂 × 𝒂^𝟐/𝟐 〖𝐬𝐢𝐧〗^(−𝟏)⁡(𝟏) = 2𝑎𝑏 ×sin^(−1)⁡(1) = 2𝑎𝑏 × 𝜋/2 = 𝝅𝒂𝒃 ∴ Required Area = 𝝅𝒂𝒃 square units