# Ex 8.1, 12

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 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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CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .