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Area between curve and curve
Misc 13
Ex 8.2, 4 Important Deleted for CBSE Board 2023 Exams
Ex 8.2, 5 Important Deleted for CBSE Board 2023 Exams
Misc 14 Important
Misc 4 Important
Misc 11 Important
Ex 8.2,1 Important Deleted for CBSE Board 2023 Exams
Misc 15 Deleted for CBSE Board 2023 Exams
Example 7 Important Deleted for CBSE Board 2023 Exams
Misc 18 (MCQ) Deleted for CBSE Board 2023 Exams
Example 15 Important
Ex 8.2, 2 Deleted for CBSE Board 2023 Exams You are here
Example 10 Important Deleted for CBSE Board 2023 Exams
Example 6 Important Deleted for CBSE Board 2023 Exams
Misc 19 (MCQ) Important Deleted for CBSE Board 2023 Exams
Area between curve and curve
Last updated at May 29, 2018 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