# Ex 8.1, 4 - Chapter 8 Class 12 Application of Integrals (Term 2)

Last updated at Aug. 20, 2021 by Teachoo

Last updated at Aug. 20, 2021 by Teachoo

Transcript

Ex 8.1, 4 Find the area of the region bounded by the ellipse 𝑥216+ 𝑦29=1 Equation Of Given Ellipse is :- 𝑥216+ 𝑦29=1 𝑥2 42+ 𝑦2 32=1 Area of ellipse = Area of ABCD = 2 × [Area Of ABC] = 2 × −44𝑦.𝑑𝑥 Finding y We know that 𝑥216+ 𝑦29=1 𝑦29=1− 𝑥216 𝑦29= 16− 𝑥216 𝑦2= 916 16− 𝑥2 Taking square root on both sides y = ± 916 16− 𝑥2 y = ± 34 16− 𝑥2 Since , ABC is above x-axis y will be positive ∴ 𝑦= 34 16− 𝑥2 Now, Area of ellipse = 2 × −44𝑦.𝑑𝑥 = 2 × −44 34 16− 𝑥2𝑑𝑥 = 2 × 34 −44 16− 𝑥2𝑑𝑥 = 32 −44 42− 𝑥2𝑑𝑥 = 32 𝑥2 42− 𝑥2+ 422 sin−1 𝑥4−44 = 32 42 42− 42− −42 42− −42+ 162 sin−1 44− 162 sin−1 −44 = 32 2 0+2 0+8 sin−1(1)− 8 sin−1 −1 = 32 0+8 sin−1 1−8 𝒔𝒊𝒏−𝟏 −𝟏 = 32 8 sin−1 1−8(− 𝒔𝒊𝒏−𝟏 𝟏) = 32 8 sin−1 1+8 sin−1 1 = 32 ×16 sin−1 1 = 3 × 8 × 𝜋2 = 12π ∴ Area of Ellipse = 12π Square units

Chapter 8 Class 12 Application of Integrals (Term 2)

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.