Check sibling questions

Ex 8.1, 4 - Find area bounded by ellipse x2/16 + y2/9 = 1 - Area under curve

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

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Transcript

Ex 8.1, 4 Find the area of the region bounded by the ellipse 𝑥﷮2﷯﷮16﷯+ 𝑦﷮2﷯﷮9﷯=1 Equation Of Given Ellipse is :- 𝑥﷮2﷯﷮16﷯+ 𝑦﷮2﷯﷮9﷯=1 𝑥﷮2﷯﷮ 4﷯﷮2﷯﷯+ 𝑦﷮2﷯﷮ 3﷯﷮2﷯﷯=1 Area of ellipse = Area of ABCD = 2 × [Area Of ABC] = 2 × −4﷮4﷮𝑦.﷯𝑑𝑥 Finding y We know that 𝑥﷮2﷯﷮16﷯+ 𝑦﷮2﷯﷮9﷯=1 𝑦﷮2﷯﷮9﷯=1− 𝑥﷮2﷯﷮16﷯ 𝑦﷮2﷯﷮9﷯= 16− 𝑥﷮2﷯﷮16﷯ 𝑦﷮2﷯= 9﷮16﷯ 16− 𝑥﷮2﷯﷯ Taking square root on both sides y = ± ﷮ 9﷮16﷯ 16− 𝑥﷮2﷯﷯﷯ y = ± 3﷮4﷯ ﷮16− 𝑥﷮2﷯﷯ Since , ABC is above x-axis y will be positive ∴ 𝑦= 3﷮4﷯ ﷮16− 𝑥﷮2﷯﷯ Now, Area of ellipse = 2 × −4﷮4﷮𝑦.﷯𝑑𝑥 = 2 × −4﷮4﷮ 3﷮4﷯ ﷮16− 𝑥﷮2﷯﷯﷯𝑑𝑥 = 2 × 3﷮4﷯ −4﷮4﷮ ﷮16− 𝑥﷮2﷯﷯﷯𝑑𝑥 = 3﷮2﷯ −4﷮4﷮ ﷮ 4﷯﷮2﷯− 𝑥﷮2﷯﷯﷯𝑑𝑥 = 3﷮2﷯ 𝑥﷮2﷯ ﷮ 4﷯﷮2﷯− 𝑥﷮2﷯﷯+ 4﷯﷮2﷯﷮2﷯ sin﷮−1﷯﷮ 𝑥﷮4﷯﷯﷯﷮−4﷮4﷯ = 3﷮2﷯ 4﷮2﷯ ﷮ 4﷯﷮2﷯− 4﷯﷮2﷯﷯− −4﷯﷮2﷯ ﷮ 4﷯﷮2﷯− −4﷯﷮2﷯﷯+ 16﷮2﷯ sin﷮−1﷯﷮ 4﷮4﷯﷯− 16﷮2﷯﷯ sin﷮−1﷯ −4﷮4﷯﷯﷯ = 3﷮2﷯ 2 0﷯+2 0﷯+8 sin﷮−1﷯(1)﷮− 8 sin﷮−1﷯﷮ −1﷯﷯﷯﷯ = 3﷮2﷯ 0+8 sin﷮−1﷯﷮ 1﷯−8 𝒔𝒊𝒏﷮−𝟏﷯﷮ −𝟏﷯﷯﷯﷯ = 3﷮2﷯ 8 sin﷮−1﷯﷮ 1﷯−8(− 𝒔𝒊𝒏﷮−𝟏﷯﷮ 𝟏﷯﷯)﷯﷯ = 3﷮2﷯ 8 sin﷮−1﷯﷮ 1﷯+8 sin﷮−1﷯﷮ 1﷯﷯﷯﷯ = 3﷮2﷯ ×16 sin﷮−1﷯﷮ 1﷯﷯ = 3 × 8 × 𝜋﷮2﷯ = 12π ∴ Area of Ellipse = 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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.