Check sibling questions

Ex 8.2, 2 - Find area bounded by (x-1)2 +y2 = 1 and x2 + y2=1 - Ex 8.2

Ex 8.2, 2 - Chapter 8 Class 12 Application of Integrals - Part 2
Ex 8.2, 2 - Chapter 8 Class 12 Application of Integrals - Part 3
Ex 8.2, 2 - Chapter 8 Class 12 Application of Integrals - Part 4
Ex 8.2, 2 - Chapter 8 Class 12 Application of Integrals - Part 5
Ex 8.2, 2 - Chapter 8 Class 12 Application of Integrals - Part 6
Ex 8.2, 2 - Chapter 8 Class 12 Application of Integrals - Part 7
Ex 8.2, 2 - Chapter 8 Class 12 Application of Integrals - Part 8
Ex 8.2, 2 - Chapter 8 Class 12 Application of Integrals - Part 9
Ex 8.2, 2 - Chapter 8 Class 12 Application of Integrals - Part 10
Ex 8.2, 2 - Chapter 8 Class 12 Application of Integrals - Part 11

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Transcript

Ex 8.2 , 2 Find the area bounded by curves 𝑥 – 1﷯﷮2﷯ + 𝑦2=1 𝑎𝑛𝑑 𝑥2+𝑦2=1 First we find center and radius of both circles Drawing figure Area required = Area OACB First, we find intersection points A and B 𝑥﷮2﷯+ 𝑦﷮2﷯=1 𝑥−1﷯﷮2﷯+ 𝑦﷮2﷯=1 From equation (1) 𝑥﷮2﷯+ 𝑦﷮2﷯=1 𝑦﷮2﷯=1− 𝑥﷮2﷯ Put 𝑦﷮2﷯=1− 𝑥﷮2﷯ in equation (2) 𝑥−1﷯﷮2﷯+ 𝑦﷮2﷯=1 𝑥−1﷯﷮2﷯+1− 𝑥﷮2﷯=1 𝑥−1﷯﷮2﷯− 𝑥﷮2﷯=0 𝑥﷮2﷯−2𝑥+1− 𝑥﷮2﷯=0 1=2𝑥 𝑥= 1﷮2﷯ Putting 𝑥= 1﷮2﷯ in (1) 𝑥2 + 𝑦2 = 1 1﷮2﷯﷯﷮2﷯+ 𝑦﷮2﷯=1 𝑦﷮2﷯=1− 1﷮4﷯ 𝑦﷮2﷯= 3﷮4﷯ 𝑦= ± ﷮3﷯﷮2﷯ So, 𝑦= ﷮3﷯﷮2﷯ , − ﷮3﷯﷮2﷯ So, intersecting points are A = 1﷮2﷯ , ﷮3﷯﷮2﷯﷯ & B = 1﷮2﷯ , − ﷮3﷯﷮2﷯﷯ Now, finding area Area Required = Area ACBD + Area OADB Area ACBD Since ACBD is symmetric about 𝑥−𝑎𝑥𝑖𝑠 So, Area ACBD = 2 × Area ACD = 2 1﷮2﷯﷮1﷮𝑦 𝑑𝑥﷯ 𝑦 → equation of 1st circle 𝑥﷮2﷯+ 𝑦﷮2﷯=1 𝑦﷮2﷯=1− 𝑥﷮2﷯ 𝑦=± ﷮1− 𝑥﷮2﷯﷯ Since Area ACD is in 1st quadrant, we take positive value So, 𝑦= ﷮1− 𝑥﷮2﷯﷯ Hence, Area ACBD = 2 1﷮2﷯﷮1﷮𝑦 𝑑𝑥﷯ = 2 1﷮2﷯﷮1﷮ ﷮1− 𝑥﷮2﷯﷯﷯ 𝑑𝑥 = 2 1﷮2﷯﷮1﷮ ﷮ 1﷮2﷯− 𝑥﷮2﷯﷯﷯ 𝑑𝑥 = 2 𝑥﷮2﷯ ﷮1− 𝑥﷮2﷯﷯+ 1﷮2﷯﷮2﷯ sin﷮−1﷯﷮ 𝑥﷮1﷯﷯﷯﷮ 1﷮2﷯﷮1﷯ = 2 𝑥﷮2﷯ ﷮1− 𝑥﷮2﷯﷯+ 1﷮2﷯ sin﷮−1﷯﷮𝑥﷯﷯﷮ 1﷮2﷯﷮1﷯ = 2 1﷮2﷯ ﷮1−1﷯+ 1﷮2﷯ sin﷮−1﷯1﷯− 1﷮2﷯﷯﷮2﷯ ﷮1− 1﷮2﷯﷯﷮2﷯﷯+ 1﷮2﷯ sin﷮−1﷯﷮ 1﷮2﷯﷯﷯﷯﷯ = 2 1﷮2﷯ ﷮0﷯+ 1﷮2﷯ × 𝜋﷮2﷯− 1﷮4﷯ ﷮1− 1﷮4﷯﷯ − 1﷮2﷯ 𝜋﷮6﷯﷯ = 2 𝜋﷮4﷯− 𝜋﷮12﷯− 1﷮4﷯ ﷮ 4 − 1﷮4﷯﷯﷯ = 2 3𝜋 − 𝜋﷮12﷯− 1﷮4﷯ ﷮ 3﷮4﷯﷯﷯ = 2 2𝜋﷮12﷯ − 1﷮4﷯ ﷮3﷯﷮2﷯﷯ = 2 × 𝜋﷮6﷯ − ﷮3﷯﷮8﷯﷯ = 𝜋﷮3﷯ − ﷮3﷯﷮4﷯ Area OADB Since OADB is symmetric about 𝑥−𝑎𝑥𝑖𝑠 So, Area OADB = 2 × Area AOD = 2 0﷮ 1﷮2﷯﷮𝑦 𝑑𝑥﷯ 𝑦 → equation of 2nd circle 𝑥−1﷯﷮2﷯+ 𝑦﷮2﷯=1 𝑦﷮2﷯=1− 𝑥−1﷯﷮2﷯ 𝑦=± ﷮1− 𝑥−1﷯﷮2﷯﷯ Since Area AOD is in 1st quadrant, we take positive value So, 𝑦= ﷮1− 𝑥−1﷯﷮2﷯﷯ Hence, Area OADB = 2 0﷮ 1﷮2﷯﷮𝑦 𝑑𝑥﷯ = 2 0﷮ 1﷮2﷯﷮ ﷮1− 𝑥−1﷯﷮2﷯﷯﷯ 𝑑𝑥 Putting t = 𝑥−1 Diff w.r.t 𝑥 𝑑𝑡﷮𝑑𝑥﷯=1 𝑑𝑡=𝑑𝑥 So, = 2 0﷮ 1﷮2﷯﷮ ﷮1− 𝑥−1﷯﷮2﷯﷯ 𝑑𝑥﷯ =2 −1﷮ −1﷮ 2﷯﷮ ﷮1− 𝑡﷮2﷯﷯ 𝑑𝑡﷯ =2 𝑡﷮2﷯ ﷮1− 𝑡﷮2﷯﷯+ 1﷮2﷯﷮2﷯ sin﷮−1﷯﷮ 𝑡﷮1﷯﷯﷯﷮−1﷮ −1﷮ 2﷯﷯ =2 −1﷮ 2﷯﷯﷮2﷯ ﷮1− −1﷮ 2﷯﷯﷮2﷯﷯+ 1﷮2﷯﷮2﷯ sin﷮−1﷯﷮ −1﷮ 2﷯﷯﷮1﷯﷯− −1﷯﷮2﷯ ﷮1− −1﷯﷮2﷯﷯+ 1﷮2﷯ sin﷮−1﷯﷮ −1﷯﷯﷯﷯ = 2 −1﷮ 2﷯ ﷮1− 1﷮4﷯﷯+ 1﷮2﷯ × −𝜋﷮6﷯﷯+ ﷮0﷯− 1﷮2﷯ −𝜋﷮ 2﷯﷯﷯ = 2 −1﷮ 4﷯ ﷮ 3﷮4﷯﷯− 𝜋﷮12﷯+ 𝜋﷮4﷯﷯ = 2 − ﷮3﷯﷮4 ×2﷯+ 𝜋 ﷮4﷯− 𝜋﷮12﷯﷯ = 2 − ﷮3﷯﷮ 8﷯+ 3𝜋 − 𝜋﷮12﷯﷯ = 2 − ﷮3﷯﷮8﷯+ 2𝜋﷮12﷯﷯ = − ﷮3﷯﷮4﷯+ 𝜋﷮3﷯ Area Required = Area ACBD + Area OADB = 𝜋 ﷮3﷯− ﷮3﷯﷮ 4﷯− ﷮3﷯﷮ 4﷯+ 𝜋﷮3﷯ = 𝟐𝝅﷮𝟑﷯− ﷮𝟑﷯﷮ 𝟐﷯

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.