Example 10 - Find area enclosed between two circles: x2+y2=4 - Area between curve and curve

Slide15.JPG
Slide16.JPG Slide17.JPG Slide18.JPG Slide19.JPG Slide20.JPG Slide21.JPG Slide22.JPG Slide23.JPG

  1. Class 12
  2. Important Question for exams Class 12
Ask Download

Transcript

Example 10 Find the area of the region enclosed between the two circles: 𝑥2+𝑦2=4 and 𝑥 –2﷯2+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 𝑥﷮2﷯ + 𝑦﷮2﷯ = 4 …(1) ( 𝑥−2)﷮2﷯ + 𝑦﷮2﷯ = 4 …(2) Comparing (1) & (2) 𝑥﷮2﷯ + 𝑦﷮2﷯ = (x − 2)﷮2﷯ + 𝑦﷮2﷯ 𝑥﷮2﷯ = (x − 2)﷮2﷯ 𝑥﷮2﷯ − (x − 2)﷮2﷯ = 0 (x − (x − 2) (x + (x − 2)) = 0 (x − x + 2) (x + x − 2) = 0 2(2x − 2) = 0 (2x − 2) = 0﷮2﷯ 2x − 2 = 0 2x = 2 x = 2﷮2﷯ x = 1 Put x = 1 in (1) 𝑥﷮2﷯ + 𝑦﷮2﷯ = 4 1﷮2﷯ + 𝑦﷮2﷯ = 4 1 + 𝑦﷮2﷯ = 4 𝑦﷮2﷯ = 4 − 1 𝑦﷮2﷯ = 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 1﷮2﷮𝑦 𝑑𝑥﷯ Here, 𝑥﷮2﷯+ 𝑦﷮2﷯=4 𝑦﷮2﷯=4− 𝑥﷮2﷯ 𝑦=± ﷮4− 𝑥﷮2﷯﷯ Since Area ADC is in 1st quadrant, y = ﷮4− 𝑥﷮2﷯﷯ Area ACA’D = 2 1﷮2﷮𝑦 𝑑𝑥﷯ = 2 1﷮2﷮ ﷮4− 𝑥﷮2﷯﷯ 𝑑𝑥﷯ =2 1﷮2﷮ ﷮ 2﷮2﷯− 𝑥﷮2﷯﷯ 𝑑𝑥﷯ =2 𝑥﷮2﷯ ﷮ 2﷮2﷯− 𝑥﷮2﷯﷯+ 2﷮2﷯﷮2﷯ sin﷮−1﷯﷮ 𝑥﷮2﷯﷯﷯﷮1﷮2﷯ = 2 𝑥﷮2﷯ ﷮4− 𝑥﷮2﷯﷯+2 sin﷮−1﷯﷮ 𝑥﷮2﷯﷯﷯﷮1﷮2﷯ =2 2﷮2﷯ ﷮4− 2﷮2﷯﷯+2 sin﷮−1﷯﷮ 2﷮2﷯− 1﷮2﷯ ﷮4− 1﷮2﷯﷯+2 sin﷮−1﷯﷮ 1﷮2﷯﷯﷯﷯﷯ =2 1. ﷮4−4﷯+2 sin﷮−1﷯﷮1− 1﷮2﷯ ﷮4−1﷯+2 sin﷮−1﷯﷮ 1﷮2﷯﷯﷯﷯﷯ =2 1.0+ 2𝜋﷮2﷯﷮1− 1﷮2﷯ ﷮3﷯+ 2𝜋﷮6﷯﷯﷯﷯ =2 𝜋− ﷮3﷯﷮2﷯− 𝜋﷮3﷯﷯ =2 2𝜋﷮3﷯− ﷮3﷯﷮2﷯﷯ = 𝟒𝝅﷮𝟑﷯− ﷮𝟑﷯ Area OADA’ Area OADA’ =2 × Area OAD = 2 0﷮1﷮𝑦 𝑑𝑥﷯ Here, 𝑥−2﷯﷮2﷯+ 𝑦﷮2﷯=4 𝑦﷮2﷯=4− 𝑥−2﷯﷮2﷯ 𝑦=± ﷮4− 𝑥−2﷯﷮2﷯﷯ Since OAD is in 1st quadrant, 𝑦= ﷮4− 𝑥−2﷯﷮2﷯﷯ Hence, Area OADA’ = 2 0﷮1﷮𝑦 𝑑𝑥﷯ = 2 0﷮1﷮ ﷮4− 𝑥−2﷯﷮2﷯﷯ 𝑑𝑥﷯ Putting t = 𝑥−2﷯ Diff. w.r.t. 𝑥 𝑑𝑡﷮𝑑𝑥﷯=1 𝑑𝑡 =𝑑𝑥 So, = 2 0﷮1﷮ ﷮4− 𝑥−2﷯﷮2﷯﷯ 𝑑𝑥﷯ =2 − 2﷮− 1﷮ ﷮4− 𝑡﷮2﷯﷯ 𝑑𝑡﷯ =2 𝑡﷮2﷯ ﷮ 2﷮2﷯− 𝑡﷮2﷯﷯+ 2﷮2﷯﷮2﷯ sin﷮−1﷯﷮ 𝑡﷮2﷯﷯﷯﷮−2﷮−1﷯ = 2 −1﷮ 2﷯ ﷮ 2﷮2﷯− −1﷯﷮2﷯﷯+2 sin﷮−1﷯ −1﷮ 2﷯﷯− −2﷮ 2﷯ ﷮ 2﷮2﷯− −1﷯﷮2﷯﷯+2 sin﷮−1﷯ −2﷮ 2﷯﷯ = 2 −1﷮ 2﷯ ﷮4−1﷯+2 sin﷮−1﷯ −1﷮ 2﷯﷯﷯− −1 ﷮ 2﷮2﷯− 2﷮2﷯﷯+2 sin﷮−1﷯(−1)﷯ = 2 −1﷮ 2﷯ ﷮3﷯− 2𝜋﷮6﷯−0+ 2𝜋﷮2﷯﷯ =2 − ﷮3﷯﷮ 2﷯+ 2𝜋﷮3﷯﷯ = – ﷮3﷯ + 4𝜋﷮3﷯ Therefore, Area required = Area ACA’D + Area OADA’ = 4𝜋﷮3﷯− ﷮3﷯﷯ + – ﷮3﷯ + 4𝜋﷮3﷯﷯ = 8𝜋﷮3﷯−2 ﷮3﷯

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
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.