# Example 7 - Chapter 8 Class 12 Application of Integrals

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Example 7 Find the area lying above x-axis and included between the circle 𝑥2 +𝑦2=8𝑥 and inside of the parabola 𝑦2=4𝑥 Since equation of circle is of form 𝑥−𝑎2+ 𝑦−𝑏2= 𝑟2 , We convert our equation 𝑥2+ 𝑦2=8𝑥 𝑥2−8𝑥+ 𝑦2=0 𝑥2−2 ×4 ×𝑥+ 𝑦2=0 𝑥2−2 ×4 ×𝑥+ 42− 42+ 𝑦2=0 𝑥−42+ 𝑦2= 42 So, Circle has center 4 , 0 & Radius =4 We need to find Area OPQC First we find point P Point P is point of intersection of circle and parabola Now, equation of circle is 𝑥2+ 𝑦2=8𝑥 Putting 𝑦2=4𝑥 𝑥2+4𝑥=8𝑥 𝑥2=8𝑥−4𝑥 𝑥2=4𝑥 𝑥2−4𝑥=0 𝑥 𝑥−4=0 𝑥=0 & 𝑥=4 Since point P is in 1st quadrant, Coordinates of P = (4, 4) Note that 𝑥-coordinate same as that of center (4, 0) ∴ P lies above point C So, we need to change the figure New figure Area Required Area Required = Area OPC + Area PCQ Area OPC Area OPC = 04𝑦 𝑑𝑥 For parabola y2 = 4x y = ± 4𝑥 y = ± 2 𝑥 Since OPC is in 1st quadrant y = 2 𝑥 ∴ Area OPC = 042 𝑥𝑑𝑥 = 2 04 𝑥 12 𝑑𝑥 = 2 𝑥 12+1 12+104 = 2 𝑥 32 3204 = 2 × 23 4 32− 0 32 = 43 8−0 = 323 Area PCQ Area PCQ = 48𝑦 𝑑𝑥 For circle x2 + y2 = 8x y2 = 8x – x2 y = ± 8𝑥− 𝑥2 Since PCQ is in 1st quadrant y = 8𝑥− 𝑥2 ∴ Area PCQ = 48 8𝑥− 𝑥2𝑑𝑥 = 48 −( 𝑥2−8𝑥)𝑑𝑥 = 48 −( 𝑥2−8𝑥+16−16)𝑑𝑥 = 48 −( 𝑥2−8𝑥+16)−(−16)𝑑𝑥 = 48 16−( 𝑥2−8𝑥+16)𝑑𝑥 = 48 16− 𝑥−42𝑑𝑥 = 48 42− 𝑥−42𝑑𝑥 = (𝑥 − 4)2 42− (𝑥−4)2+ 422 𝑠𝑖𝑛−1 (𝑥 − 4)4 48 = (𝑥 − 4)2 16−( 𝑥2−8𝑥+ 42)+ 162 𝑠𝑖𝑛−1 (𝑥 − 4)4 48 = (𝑥 − 4)2 −( 𝑥2−8𝑥)+8 𝑠𝑖𝑛−1 (𝑥 − 4)4 48 = (8 − 4)2 −( 82−8(8))+8 𝑠𝑖𝑛−1 (8 − 4)4 – (4 − 4)2 −( 42−8(4))+8 𝑠𝑖𝑛−1 (4 − 4)4 = 42 0+8 𝑠𝑖𝑛−1 1 – 0+8 𝑠𝑖𝑛−1 0 = 8 𝑠𝑖𝑛−1 1 – 8 𝑠𝑖𝑛−1 0 = 8 𝜋2 – 8 × 0 = 4𝜋 Thus, Area Required = Area OPC + Area PCQ = 323 + 4𝜋 = 43 (8 + 3𝜋)

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.