# Example 10

Last updated at March 11, 2017 by Teachoo

Last updated at March 11, 2017 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 provides courses for Mathematics from Class 9 to 12. You can ask questions here.