Misc 15 - Find area {(x, y): y2 < 4x, 4x2 + 4y2 < 9} - Miscellaneous

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  1. Chapter 8 Class 12 Application of Integrals
  2. Serial order wise
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Misc 15 Find the area of the region { 𝑥, 𝑦﷯ :𝑦2≤4𝑥, 4𝑥2+ 4𝑦2 ≤ 9} Drawing figures Now, our figure will be Circle is 4𝑥2+ 4𝑦2 ≤ 9 𝑥2+ 𝑦2 ≤ 9﷮4﷯ 𝑥2+ 𝑦2 ≤ 3﷮2﷯﷯﷮2﷯ ∴ Radius of circle = 3﷮2﷯ So, Point C ( 3﷮2﷯, 0) Finding point of intersection A & B Solving 𝑦﷮2﷯=4𝑥 …(1) & 4𝑥2+ 4𝑦2= 9 …(2) Putting (1) in (2) 4𝑥2+ 4𝑦2= 9 4𝑥2+ 4 4𝑥﷯=9 4𝑥2+16𝑥−9=0 4𝑥2+18𝑥−2𝑥−9=0 2𝑥(2𝑥+9) −1(2𝑥+9)=0 2𝑥−1﷯ 2𝑥+9﷯=0 So, 𝑥= 1﷮2﷯ & 𝑥= −9﷮ 2﷯ Area OAC Area OAC = Area ACD + Area OAD Area ACD Area ACD = 1﷮2﷯﷮ 3﷮2﷯﷮𝑦 𝑑𝑥﷯ 𝑦→ Equation of circle 𝑥﷮2﷯+ 𝑦﷮2﷯= 9﷮4﷯ 𝑦﷮2﷯= 9﷮4﷯− 𝑥﷮2﷯ 𝑦= ﷮ 9﷮4﷯− 𝑥﷮2﷯﷯ Therefore Area ACD = 1﷮2﷯﷮ 3﷮2﷯﷮ ﷮ 9﷮4﷯− 𝑥﷮2﷯﷯ 𝑑𝑥﷯ = 1﷮2﷯﷮ 3﷮2﷯﷮ ﷮ 3﷮2﷯﷯﷮2﷯− 𝑥﷮2﷯﷯ 𝑑𝑥﷯ = 𝑥﷮2﷯ ﷮ 3﷮2﷯﷯﷮2﷯− 𝑥﷮2﷯﷯+ 3﷮2﷯﷯﷮2﷯﷮2﷯ sin﷮−1﷯﷮ 𝑥﷮ 3﷮2﷯﷯﷯﷯﷮ 1﷮2﷯﷮ 3﷮2﷯﷯ = 𝑥﷮2﷯ ﷮ 9﷮4﷯− 𝑥﷮2﷯﷯+ 9﷮8﷯ sin﷮−1﷯﷮ 2𝑥﷮3﷯﷯﷯﷮ 1﷮2﷯﷮ 3﷮2﷯﷯ = 3﷮2﷯﷯﷮2﷯ ﷮ 9﷮4﷯− 3﷮2﷯﷯﷮2﷯﷯+ 9﷮8﷯ sin﷮−1﷯﷮ 2 3﷮2﷯﷯﷮3﷯﷯﷯ – 1﷮2﷯﷯﷮2﷯ ﷮ 9﷮4﷯− 1﷮2﷯﷯﷮2﷯﷯+ 9﷮8﷯ sin﷮−1﷯﷮ 2 1﷮2﷯﷯﷮3﷯﷯﷯ = 3﷮4﷯ ﷮ 9﷮4﷯− 9﷮4﷯﷯+ 9﷮8﷯ sin﷮−1﷯﷮1﷯﷯ – 1﷮4﷯ ﷮ 9﷮4﷯− 1﷮4﷯﷯+ 9﷮8﷯ sin﷮−1﷯﷮ 1﷮3﷯﷯﷯ = 9﷮8﷯ sin﷮−1﷯﷮1﷯ – ﷮2﷯﷮4﷯− 9﷮8﷯ sin﷮−1﷯﷮ 1﷮3﷯﷯ = 9𝜋﷮16﷯ – ﷮2﷯﷮4﷯− 9﷮8﷯ sin﷮−1﷯﷮ 1﷮3﷯﷯ Area OAD Area OAD = 0﷮ 1﷮2﷯﷮𝑦 𝑑𝑥﷯ 𝑦→ Equation of parabola 𝑦﷮2﷯=4𝑥 𝑦= ﷮4𝑥 ﷯ Therefore Area OAD = 0﷮ 1﷮2﷯﷮ ﷮4𝑥 ﷯ 𝑑𝑥﷯ =2 0﷮ 1﷮2﷯﷮ ﷮𝑥 ﷯ 𝑑𝑥﷯ =2 0﷮ 1﷮2﷯﷮ 𝑥﷮ 1﷮2﷯﷯ 𝑑𝑥﷯ =2 𝑥﷮ 3﷮2﷯﷯﷮ 3﷮2﷯﷯﷯﷮0﷮ 1﷮2﷯﷯ = 4﷮3﷯ [ 𝑥﷮ 3﷮2﷯﷯]﷮0﷮ 1﷮2﷯﷯ = 4﷮3﷯ 1﷮2﷯﷯﷮ 3﷮2﷯﷯− 0﷮ 3﷮2﷯﷯﷯ = 4﷮3﷯ × 1﷮2 ﷮2﷯﷯ = 4﷮3﷯ × 1﷮2 ﷮2﷯﷯ × ﷮2﷯﷮ ﷮2﷯﷯ = ﷮2﷯﷮3﷯ Area OAC = Area ACD + Area OAD = 9𝜋﷮16﷯ – ﷮2﷯﷮4﷯− 9﷮8﷯ sin﷮−1﷯﷮ 1﷮3﷯﷯ + ﷮2﷯﷮3﷯ = 9𝜋﷮16﷯ − 9﷮8﷯ sin﷮−1﷯﷮ 1﷮3﷯﷯ + ﷮2﷯﷮12﷯ Required Area = 2 × Area OAC = 2 × 9𝜋﷮16﷯ − 9﷮8﷯ sin﷮−1﷯﷮ 1﷮3﷯﷯ + ﷮2﷯﷮12﷯﷯ = 9𝜋﷮8﷯ − 9﷮4﷯ sin﷮−1﷯﷮ 1﷮3﷯﷯ + ﷮2﷯﷮6﷯ = 9𝜋﷮8﷯ − 9﷮4﷯ sin﷮−1﷯﷮ 1﷮3﷯﷯ + ﷮2﷯﷮6﷯ × ﷮2﷯﷮ ﷮2﷯﷯ = 9𝜋﷮8﷯ − 9﷮4﷯ sin﷮−1﷯﷮ 1﷮3﷯﷯ + 2﷮6 ﷮2﷯﷯ = 𝟗𝝅﷮𝟖﷯ − 𝟗﷮𝟒﷯ 𝒔𝒊𝒏﷮−𝟏﷯﷮ 𝟏﷮𝟑﷯﷯ + 𝟏﷮𝟑 ﷮𝟐﷯﷯

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