Check sibling questions

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Transcript

Example 1 Find the area enclosed by the circle π‘₯2 + 𝑦2 = π‘Ž2Given π‘₯^2 + 𝑦^2= π‘Ž^2 This is a circle with Center = (0, 0) Radius = π‘Ž Since radius is a, OA = OB = π‘Ž A = (π‘Ž, 0) B = (0, π‘Ž) Now, Area of circle = 4 Γ— Area of Region OBAO = 4 Γ— ∫1_𝟎^π’‚β–’γ€–π’š 𝒅𝒙〗 Here, y β†’ Equation of Circle We know that π‘₯^2 + 𝑦^2 = π‘Ž^2 𝑦^2 = π‘Ž^2βˆ’ π‘₯^2 y = Β± √(π‘Ž^2βˆ’π‘₯^2 ) Since AOBA lies in 1st Quadrant y = √(𝒂^πŸβˆ’π’™^𝟐 ) Now, Area of circle = 4 Γ— ∫1_0^π‘Žβ–’γ€–π‘¦ 𝑑π‘₯γ€— = 4 Γ— ∫1_0^π‘Žβ–’γ€–βˆš(π‘Ž^2βˆ’π‘₯^2 ) 𝑑π‘₯γ€— = 4[𝒙/𝟐 √(𝒂^πŸβˆ’π’™^𝟐 )+𝒂^𝟐/𝟐 γ€–"sin" γ€—^(βˆ’πŸ) 𝒙/𝒂]_𝟎^𝒂 = 4[π‘Ž/2 √(π‘Ž^2βˆ’π‘Ž^2 )+π‘Ž^2/2 γ€–"sin" γ€—^(βˆ’1) π‘Ž/π‘Žβˆ’0/2 √(π‘Ž^2βˆ’0)βˆ’0^2/2 γ€–"sin" γ€—^(βˆ’1) (0)] = 4[0+π‘Ž^2/2 γ€–"sin" γ€—^(βˆ’1) (1)βˆ’0βˆ’0] = 4.π‘Ž^2/2. πœ‹/2 = 𝝅𝒂^𝟐 Using: √(π‘Ž^2βˆ’π‘₯^2 )dx = 1/2 √(π‘Ž^2βˆ’π‘₯^2 ) + π‘Ž^2/2 γ€–"sin" γ€—^(βˆ’1) π‘₯/4 + c

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.