# Example 1 - Chapter 8 Class 12 Application of Integrals

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Example 1 Find the area enclosed by the circle 2 + 2 = 2 2 + 2 = 2 Drawing graph Radius = Hence OA = OB = A = ( , 0) B = (0, ) Now, area of circle = 4 Area of Region OBAO = 4 0 So, we need to calculate 0 We know that 2 + 2 = 2 2 = 2 2 y = 2 2 Since AOBA lies in I Quadrant y = 2 2 Now, Area of circle = 4 0 = 4 0 2 2 = 4 2 2 2 + 2 2 sin 1 0 = 4 2 2 2 + 2 2 sin 1 0 2 2 0 0 2 2 sin 1 (0) = 4 0+ 2 2 sin 1 1 0 0 = 4. 2 2 . 2 =

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.