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  1. Chapter 8 Class 12 Application of Integrals
  2. Serial order wise

Transcript

Example 1 Find the area enclosed by the circle ๐‘ฅ2 + ๐‘ฆ2 = ๐‘Ž2 Drawing circle ๐‘ฅ^2 + ๐‘ฆ^2= ๐‘Ž^2 Center = (0, 0) Radius = ๐‘Ž Hence OA = OB = Radius = ๐‘Ž A = (๐‘Ž, 0) B = (0, ๐‘Ž) Since Circle is symmetric about x-axis and y-axis Area of circle = 4 ร— Area of Region OBAO = 4 ร— โˆซ1_0^๐‘Žโ–’ใ€–๐‘ฆ ๐‘‘๐‘ฅใ€— We know that ๐‘ฅ^2 + ๐‘ฆ^2 = ๐‘Ž^2 ๐‘ฆ^2 = ๐‘Ž^2โˆ’ ๐‘ฅ^2 y = ยฑโˆš(๐‘Ž^2โˆ’๐‘ฅ^2 ) Since AOBA lies in 1st Quadrant, Value of y is positive y = โˆš(๐‘Ž^2โˆ’๐‘ฅ^2 ) Now, Area of circle = 4 ร— โˆซ1_0^๐‘Žโ–’ใ€–โˆš(๐‘Ž^2โˆ’๐‘ฅ^2 ) ๐‘‘๐‘ฅใ€— Using: โˆš(๐‘Ž^2โˆ’๐‘ฅ^2 )dx = 1/2 โˆš(๐‘Ž^2โˆ’๐‘ฅ^2 ) + ๐‘Ž^2/2 ใ€–"sin" ใ€—^(โˆ’1) ๐‘ฅ/๐‘Ž + c = 4[๐‘ฅ/2 โˆš(๐‘Ž^2โˆ’๐‘ฅ^2 )+๐‘Ž^2/2 ใ€–"sin" ใ€—^(โˆ’1) ๐‘ฅ/๐‘Ž]_0^๐‘Ž = 4[(๐‘Ž/2 โˆš(๐‘Ž^2โˆ’๐‘Ž^2 )+๐‘Ž^2/2 ใ€–"sin" ใ€—^(โˆ’1) ๐‘Ž/๐‘Ž)โˆ’(0/2 โˆš(๐‘Ž^2โˆ’0)+0^2/2 ใ€–"sin" ใ€—^(โˆ’1) (0))] = 4[0+๐‘Ž^2/2 ใ€–"sin" ใ€—^(โˆ’1) (1)โˆ’0โˆ’0] = 4 ร— ๐‘Ž^2/2 ใ€–"sin" ใ€—^(โˆ’1) (1) = 4 ร— ๐‘Ž^2/2ร— ๐œ‹/2 = ๐…๐’‚^๐Ÿ

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Davneet Singh
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.