# Ex 8.1, 12 - Class 12

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Ex 8.1, 12 Area lying in the first quadrant and bounded by the circle 𝑥2+𝑦2=4 and the lines 𝑥 = 0 and 𝑥 = 2 is (A) π (B) 𝜋2 (C) 𝜋3 (D) 𝜋4 Equation of Given Circle :- 𝑥2+ 𝑦2=4 𝑥2+ 𝑦2= 22 ∴ 𝑟𝑎𝑑𝑖𝑢𝑠 , 𝑟=2 Line 𝑥=0 is y-axis & Line x = 2 passes through point A 2 , 0 So, Required area = Area of shaded region = Area OAB = 02𝑦.𝑑𝑥 We know that, 𝑥2+ 𝑦2=4 𝑦2=4− 𝑥2 ∴ 𝑦=± 4− 𝑥2 As, OBA is in 1st Quadrant ∴ 𝑦= 4− 𝑥2 ∴ Required area = 02𝑦.𝑑𝑥 = 02 4− 𝑥2 𝑑𝑥 = 02 22− 𝑥2 𝑑𝑥 = 𝑥 2 22− 𝑥2+2 sin−1 𝑥2 02 = 22 22− 22+2 sin−1 22 − 02 22− 02−2 sin−1 02 = 0+2 sin−1 1−0 4−2 sin−1 0 = 2 sin−1 1−2 sin−1 0−0 = 2 sin−1 1− sin−1 0 = 2 𝜋2−0 = 2 . 𝜋2 = π ∴ Area Required = π square units Hence, Option (A) is correct

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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.