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Misc 18 - Area of circle x2 + y2 = 16 exterior to parabola - Area between curve and curve

Misc 18 - Chapter 8 Class 12 Application of Integrals - Part 2
Misc 18 - Chapter 8 Class 12 Application of Integrals - Part 3 Misc 18 - Chapter 8 Class 12 Application of Integrals - Part 4 Misc 18 - Chapter 8 Class 12 Application of Integrals - Part 5 Misc 18 - Chapter 8 Class 12 Application of Integrals - Part 6 Misc 18 - Chapter 8 Class 12 Application of Integrals - Part 7 Misc 18 - Chapter 8 Class 12 Application of Integrals - Part 8 Misc 18 - Chapter 8 Class 12 Application of Integrals - Part 9 Misc 18 - Chapter 8 Class 12 Application of Integrals - Part 10 Misc 18 - Chapter 8 Class 12 Application of Integrals - Part 11


Transcript

Misc 18 The area of the circle 𝑥2+𝑦2 = 16 exterior to the parabola 𝑦2=6𝑥 is (A) 4﷮3﷯ (4𝜋− ﷮3﷯ ) (B) 4﷮3﷯ (4𝜋+ ﷮3﷯) (C) 4﷮3﷯ (8𝜋− ﷮3﷯) (D) 4﷮3﷯ (8𝜋+ ﷮3﷯) Step 1: Draw the Figure 𝑥2+𝑦2 = 16 𝑥2+𝑦2= 4﷮2﷯ It is a circle with center 0 , 0﷯ & radius 4 And y2 = 6x is a parabola Step 2: Finding point of intersection A & B Solving 𝑦﷮2﷯=6𝑥 …(1) & 𝑥2+ 𝑦2=16 …(2) Putting (1) in (2) 𝑥2+ 𝑦2=16 𝑥2+6𝑥=16 𝑥2+6𝑥−16=0 𝑥2+8𝑥−2𝑥−16=0 𝑥(𝑥+8) −2(𝑥+8)=0 𝑥−2﷯ 𝑥+8﷯=0 So, 𝑥=2 & 𝑥=−8 Area of circle Area of circle = 4 × Area of 1st quadrant = 4 × 0﷮4﷮𝑦 𝑑𝑥﷯ y → equation of circle x2 + y2 = 16 y2 = 16 − x2 y = ± ﷮16− 𝑥﷮2﷯﷯ Since we have to find area of first quadrant, we take y positive ∴ y = ﷮16− 𝑥﷮2﷯﷯ Therefore Area of circle = 4 × 0﷮4﷮ ﷮16− 𝑥﷮2﷯﷯﷯𝑑𝑥 = 4 × 0﷮4﷮ ﷮ 4﷮2﷯− 𝑥﷮2﷯﷯﷯𝑑𝑥 = 4 × 𝑥﷮2﷯ ﷮ 4﷮2﷯− 𝑥﷮2﷯﷯+ 4﷮2﷯﷮2﷯ sin﷮−1﷯﷮ 𝑥﷮4﷯﷯﷯﷮0﷮4﷯ = 4 × 𝑥﷮2﷯ ﷮16− 𝑥﷮2﷯﷯+8 sin﷮−1﷯﷮ 𝑥﷮4﷯﷯﷯﷮0﷮4﷯ = 4 × 4﷮2﷯ ﷮16− 4﷮2﷯﷯+8 sin﷮−1﷯﷮ 4﷮4﷯﷯﷯− 0﷮2﷯ ﷮16− 0﷮2﷯﷯+8 sin﷮−1﷯﷮ 0﷮4﷯﷯﷯﷯﷯ = 4 × 0+8 sin﷮−1﷯﷮1−0−8 sin﷮−1﷯﷮0﷯﷯﷯ = 4 × 0+8× 𝜋﷮2﷯−0−0﷯ = 4 × 4𝜋 = 16𝜋 Area OACB Since OACB is symmetric about y – axis, Area OACB = 2 × Area OAC Area OAC Area ACD Area ACD = 2﷮4﷮𝑦 𝑑𝑥﷯ y → equation of circle x2 + y2 = 16 y2 = 16 − x2 y = ± ﷮16 − x2 ﷯ Since ACD is 1st quadrant y = ﷮16− 𝑥﷮2﷯﷯ Therefore, Area ACD = 2﷮4﷮ ﷮16− 𝑥﷮2﷯﷯ 𝑑𝑥﷯ = 2﷮4﷮ ﷮ 4﷮2﷯− 𝑥﷮2﷯﷯ ﷯ = 2﷮4﷮ ﷮ 4﷮2﷯− 𝑥﷮2﷯﷯ ﷯ = 𝑥﷮2﷯ ﷮ 4﷮2﷯− 𝑥﷮2﷯﷯+ 4﷮2﷯﷮4﷯ sin﷮−1﷯﷮ 𝑥﷮4﷯﷯﷯﷮2﷮4﷯ = 𝑥﷮2﷯ ﷮16− 𝑥﷮2﷯﷯+8 sin﷮−1﷯﷮ 𝑥﷮4﷯﷯﷯﷮2﷮4﷯ = 4﷮2﷯ ﷮16− 4﷮2﷯﷯+8 sin﷮−1﷯﷮ 4﷮4﷯﷯﷯ − 2﷮2﷯ ﷮16− 2﷮2﷯﷯+8 sin﷮−1﷯﷮ 2﷮4﷯﷯﷯ = 8 × 𝜋﷮2﷯ − 2 ﷮3﷯ − 8 × 𝜋﷮6﷯ = 4𝜋 − 4𝜋﷮3﷯ − 2 ﷮3﷯ = 8𝜋﷮3﷯ − 2 ﷮3﷯ Area OAD Area OAD = 0﷮2﷮𝑦 𝑑𝑥﷯ y → equation of parabola y2 = 6x y = ﷮6𝑥﷯ Therefore, Area OAD = 0﷮2﷮ ﷮6𝑥﷯ 𝑑𝑥﷯ = ﷮6﷯ 0﷮2﷮ 𝑥﷯﷮ 1﷮2﷯﷯ 𝑑𝑥﷯ = ﷮6﷯ × 𝑥﷯﷮ 3﷮2﷯﷯﷮ 3﷮2﷯﷯﷯﷯﷮0﷮2﷯ = 2 ﷮6﷯﷮3﷯ 2﷮ 3﷮2﷯﷯− 0﷮ 3﷮2﷯﷯﷯ = 2 ﷮6﷯﷮3﷯ × 2 ﷮2﷯ = 4 ﷮2﷯× ﷮3﷯× ﷮2﷯﷮3﷯ = 8 ﷮3﷯﷮3﷯ Area OAC = Area ACD + Area OAD = 8𝜋﷮3﷯−2 ﷮3﷯+ 8﷮3﷯ ﷮3﷯ = 8𝜋﷮3﷯+ 2﷮3﷯ ﷮3﷯ Thus, Area OACB = 2 × Area OAC = 2 × 8𝜋﷮3﷯+ 2 ﷮3﷯﷮3﷯﷯ = 16𝜋﷮3﷯+ 4 ﷮3﷯﷮3﷯ Therefore, Required Area = Area of circle − Area OACB = 16𝜋 − 4 ﷮3﷯﷮3﷯+ 16﷮3﷯𝜋﷯ = 32𝜋﷮3﷯− 4 ﷮3﷯﷮3﷯ = 4﷮3﷯(8𝜋 − ﷮3﷯) square units. So, C is the correct answer

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Davneet Singh

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.