# Misc 18 - Chapter 8 Class 12 Application of Integrals

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Misc 18 The area of the circle 𝑥2+𝑦2 = 16 exterior to the parabola 𝑦2=6𝑥 is (A) 43 (4𝜋− 3 ) (B) 43 (4𝜋+ 3) (C) 43 (8𝜋− 3) (D) 43 (8𝜋+ 3) Step 1: Draw the Figure 𝑥2+𝑦2 = 16 𝑥2+𝑦2= 42 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 𝑦2=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 × 04𝑦 𝑑𝑥 y → equation of circle x2 + y2 = 16 y2 = 16 − x2 y = ± 16− 𝑥2 Since we have to find area of first quadrant, we take y positive ∴ y = 16− 𝑥2 Therefore Area of circle = 4 × 04 16− 𝑥2𝑑𝑥 = 4 × 04 42− 𝑥2𝑑𝑥 = 4 × 𝑥2 42− 𝑥2+ 422 sin−1 𝑥404 = 4 × 𝑥2 16− 𝑥2+8 sin−1 𝑥404 = 4 × 42 16− 42+8 sin−1 44− 02 16− 02+8 sin−1 04 = 4 × 0+8 sin−11−0−8 sin−10 = 4 × 0+8× 𝜋2−0−0 = 4 × 4𝜋 = 16𝜋 Area OACB Since OACB is symmetric about y – axis, Area OACB = 2 × Area OAC Area OAC Area ACD Area ACD = 24𝑦 𝑑𝑥 y → equation of circle x2 + y2 = 16 y2 = 16 − x2 y = ± 16 − x2 Since ACD is 1st quadrant y = 16− 𝑥2 Therefore, Area ACD = 24 16− 𝑥2 𝑑𝑥 = 24 42− 𝑥2 = 24 42− 𝑥2 = 𝑥2 42− 𝑥2+ 424 sin−1 𝑥424 = 𝑥2 16− 𝑥2+8 sin−1 𝑥424 = 42 16− 42+8 sin−1 44 − 22 16− 22+8 sin−1 24 = 8 × 𝜋2 − 2 3 − 8 × 𝜋6 = 4𝜋 − 4𝜋3 − 2 3 = 8𝜋3 − 2 3 Area OAD Area OAD = 02𝑦 𝑑𝑥 y → equation of parabola y2 = 6x y = 6𝑥 Therefore, Area OAD = 02 6𝑥 𝑑𝑥 = 6 02 𝑥 12 𝑑𝑥 = 6 × 𝑥 32 3202 = 2 63 2 32− 0 32 = 2 63 × 2 2 = 4 2× 3× 23 = 8 33 Area OAC = Area ACD + Area OAD = 8𝜋3−2 3+ 83 3 = 8𝜋3+ 23 3 Thus, Area OACB = 2 × Area OAC = 2 × 8𝜋3+ 2 33 = 16𝜋3+ 4 33 Therefore, Required Area = Area of circle − Area OACB = 16𝜋 − 4 33+ 163𝜋 = 32𝜋3− 4 33 = 43(8𝜋 − 3) square units. So, C is the correct answer

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