Get live Maths 1-on-1 Classs - Class 6 to 12

Ex 8.2

Ex 8.2,1
Important
Deleted for CBSE Board 2023 Exams

Ex 8.2, 2 Deleted for CBSE Board 2023 Exams You are here

Ex 8.2, 3 Important Deleted for CBSE Board 2023 Exams

Ex 8.2, 4 Important Deleted for CBSE Board 2023 Exams

Ex 8.2, 5 Important Deleted for CBSE Board 2023 Exams

Ex 8.2, 6 (MCQ) Deleted for CBSE Board 2023 Exams

Ex 8.2 , 7 (MCQ) Important Deleted for CBSE Board 2023 Exams

Chapter 8 Class 12 Application of Integrals

Serial order wise

Last updated at March 16, 2023 by Teachoo

Ex 8.2 , 2 Find the area bounded by curves 𝑥 – 12 + 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 𝑥2+ 𝑦2=1 𝑥−12+ 𝑦2=1 From equation (1) 𝑥2+ 𝑦2=1 𝑦2=1− 𝑥2 Put 𝑦2=1− 𝑥2 in equation (2) 𝑥−12+ 𝑦2=1 𝑥−12+1− 𝑥2=1 𝑥−12− 𝑥2=0 𝑥2−2𝑥+1− 𝑥2=0 1=2𝑥 𝑥= 12 Putting 𝑥= 12 in (1) 𝑥2 + 𝑦2 = 1 122+ 𝑦2=1 𝑦2=1− 14 𝑦2= 34 𝑦= ± 32 So, 𝑦= 32 , − 32 So, intersecting points are A = 12 , 32 & B = 12 , − 32 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 121𝑦 𝑑𝑥 𝑦 → equation of 1st circle 𝑥2+ 𝑦2=1 𝑦2=1− 𝑥2 𝑦=± 1− 𝑥2 Since Area ACD is in 1st quadrant, we take positive value So, 𝑦= 1− 𝑥2 Hence, Area ACBD = 2 121𝑦 𝑑𝑥 = 2 121 1− 𝑥2 𝑑𝑥 = 2 121 12− 𝑥2 𝑑𝑥 = 2 𝑥2 1− 𝑥2+ 122 sin−1 𝑥1 121 = 2 𝑥2 1− 𝑥2+ 12 sin−1𝑥 121 = 2 12 1−1+ 12 sin−11− 122 1− 122+ 12 sin−1 12 = 2 12 0+ 12 × 𝜋2− 14 1− 14 − 12 𝜋6 = 2 𝜋4− 𝜋12− 14 4 − 14 = 2 3𝜋 − 𝜋12− 14 34 = 2 2𝜋12 − 14 32 = 2 × 𝜋6 − 38 = 𝜋3 − 34 Area OADB Since OADB is symmetric about 𝑥−𝑎𝑥𝑖𝑠 So, Area OADB = 2 × Area AOD = 2 0 12𝑦 𝑑𝑥 𝑦 → equation of 2nd circle 𝑥−12+ 𝑦2=1 𝑦2=1− 𝑥−12 𝑦=± 1− 𝑥−12 Since Area AOD is in 1st quadrant, we take positive value So, 𝑦= 1− 𝑥−12 Hence, Area OADB = 2 0 12𝑦 𝑑𝑥 = 2 0 12 1− 𝑥−12 𝑑𝑥 Putting t = 𝑥−1 Diff w.r.t 𝑥 𝑑𝑡𝑑𝑥=1 𝑑𝑡=𝑑𝑥 So, = 2 0 12 1− 𝑥−12 𝑑𝑥 =2 −1 −1 2 1− 𝑡2 𝑑𝑡 =2 𝑡2 1− 𝑡2+ 122 sin−1 𝑡1−1 −1 2 =2 −1 22 1− −1 22+ 122 sin−1 −1 21− −12 1− −12+ 12 sin−1 −1 = 2 −1 2 1− 14+ 12 × −𝜋6+ 0− 12 −𝜋 2 = 2 −1 4 34− 𝜋12+ 𝜋4 = 2 − 34 ×2+ 𝜋 4− 𝜋12 = 2 − 3 8+ 3𝜋 − 𝜋12 = 2 − 38+ 2𝜋12 = − 34+ 𝜋3 Area Required = Area ACBD + Area OADB = 𝜋 3− 3 4− 3 4+ 𝜋3 = 𝟐𝝅𝟑− 𝟑 𝟐