# Example 1 - Chapter 8 Class 12 Application of Integrals (Term 2)

Last updated at Dec. 12, 2019 by Teachoo

Last updated at Dec. 12, 2019 by Teachoo

Transcript

Example 1 Find the area enclosed by the circle ๐ฅ2 + ๐ฆ2 = ๐2 Drawing circle ๐ฅ^2 + ๐ฆ^2= ๐^2 Center = (0, 0) Radius = ๐ Hence OA = OB = Radius = ๐ A = (๐, 0) B = (0, ๐) Since Circle is symmetric about x-axis and y-axis Area of circle = 4 ร Area of Region OBAO = 4 ร โซ1_0^๐โใ๐ฆ ๐๐ฅใ We know that ๐ฅ^2 + ๐ฆ^2 = ๐^2 ๐ฆ^2 = ๐^2โ ๐ฅ^2 y = ยฑโ(๐^2โ๐ฅ^2 ) Since AOBA lies in 1st Quadrant, Value of y is positive y = โ(๐^2โ๐ฅ^2 ) Now, Area of circle = 4 ร โซ1_0^๐โใโ(๐^2โ๐ฅ^2 ) ๐๐ฅใ Using: โ(๐^2โ๐ฅ^2 )dx = 1/2 โ(๐^2โ๐ฅ^2 ) + ๐^2/2 ใ"sin" ใ^(โ1) ๐ฅ/๐ + c = 4[๐ฅ/2 โ(๐^2โ๐ฅ^2 )+๐^2/2 ใ"sin" ใ^(โ1) ๐ฅ/๐]_0^๐ = 4[(๐/2 โ(๐^2โ๐^2 )+๐^2/2 ใ"sin" ใ^(โ1) ๐/๐)โ(0/2 โ(๐^2โ0)+0^2/2 ใ"sin" ใ^(โ1) (0))] = 4[0+๐^2/2 ใ"sin" ใ^(โ1) (1)โ0โ0] = 4 ร ๐^2/2 ใ"sin" ใ^(โ1) (1) = 4 ร ๐^2/2ร ๐/2 = ๐ ๐^๐

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.