# Example 10 - 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 10 Find the area of the region enclosed between the two circles: 𝑥2+𝑦2=4 and 𝑥 –22+𝑦2=4 First we find center and radius of both circles Drawing figure Area required = Area OACA’ Area required = Area ACA’D + Area OADA’ Finding point of intersection, A & A’ Solving 𝑥2 + 𝑦2 = 4 …(1) ( 𝑥−2)2 + 𝑦2 = 4 …(2) Comparing (1) & (2) 𝑥2 + 𝑦2 = (x − 2)2 + 𝑦2 𝑥2 = (x − 2)2 𝑥2 − (x − 2)2 = 0 (x − (x − 2) (x + (x − 2)) = 0 (x − x + 2) (x + x − 2) = 0 2(2x − 2) = 0 (2x − 2) = 02 2x − 2 = 0 2x = 2 x = 22 x = 1 Put x = 1 in (1) 𝑥2 + 𝑦2 = 4 12 + 𝑦2 = 4 1 + 𝑦2 = 4 𝑦2 = 4 − 1 𝑦2 = 3 y = ± 3 So, points are (1, 3) & (1, – 3) Hence A = (1, 3) & A’ = (1, – 3) Also, D = (1, 0) Area required Area required = Area ACA’D + Area OADA’ Area ACA’D Area ACA’D = 2 Area ADC = 2 12𝑦 𝑑𝑥 Here, 𝑥2+ 𝑦2=4 𝑦2=4− 𝑥2 𝑦=± 4− 𝑥2 Since Area ADC is in 1st quadrant, y = 4− 𝑥2 Area ACA’D = 2 12𝑦 𝑑𝑥 = 2 12 4− 𝑥2 𝑑𝑥 =2 12 22− 𝑥2 𝑑𝑥 =2 𝑥2 22− 𝑥2+ 222 sin−1 𝑥212 = 2 𝑥2 4− 𝑥2+2 sin−1 𝑥212 =2 22 4− 22+2 sin−1 22− 12 4− 12+2 sin−1 12 =2 1. 4−4+2 sin−11− 12 4−1+2 sin−1 12 =2 1.0+ 2𝜋21− 12 3+ 2𝜋6 =2 𝜋− 32− 𝜋3 =2 2𝜋3− 32 = 𝟒𝝅𝟑− 𝟑 Area OADA’ Area OADA’ =2 × Area OAD = 2 01𝑦 𝑑𝑥 Here, 𝑥−22+ 𝑦2=4 𝑦2=4− 𝑥−22 𝑦=± 4− 𝑥−22 Since OAD is in 1st quadrant, 𝑦= 4− 𝑥−22 Hence, Area OADA’ = 2 01𝑦 𝑑𝑥 = 2 01 4− 𝑥−22 𝑑𝑥 Putting t = 𝑥−2 Diff. w.r.t. 𝑥 𝑑𝑡𝑑𝑥=1 𝑑𝑡 =𝑑𝑥 So, = 2 01 4− 𝑥−22 𝑑𝑥 =2 − 2− 1 4− 𝑡2 𝑑𝑡 =2 𝑡2 22− 𝑡2+ 222 sin−1 𝑡2−2−1 = 2 −1 2 22− −12+2 sin−1 −1 2− −2 2 22− −12+2 sin−1 −2 2 = 2 −1 2 4−1+2 sin−1 −1 2− −1 22− 22+2 sin−1(−1) = 2 −1 2 3− 2𝜋6−0+ 2𝜋2 =2 − 3 2+ 2𝜋3 = – 3 + 4𝜋3 Therefore, Area required = Area ACA’D + Area OADA’ = 4𝜋3− 3 + – 3 + 4𝜋3 = 8𝜋3−2 3

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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.